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AP Physics 1 Momentum
2015shy12shy02
wwwnjctlorg
3
Table of Contents
bull Momentum
bull ImpulseshyMomentum Equationbull The Momentum of a System of Objects
bull Types of Collisions
Click on the topic to go to that section
bull Conservation of Momentum
bull Collisions in Two Dimensions
4
Return toTable ofContents
Momentum
5
Newtonrsquos First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia)
In our experience
bull When objects of different mass travel with the same velocity the one with more mass is harder to stop
bull When objects of equal mass travel with different velocities the faster one is harder to stop
Momentum Defined
6
Define a new quantity momentum (p) that takes these observations into account
momentum = mass times velocity
Momentum Defined
p =mv
click here for a introductory video on momentum from Bill Nye
7
Sincebull mass is a scalar quantitybull velocity is a vector quantitybull the product of a scalar and a vector is a vector
andbull momentum = mass times velocity
therefore
Momentum is a vector quantity shy it has magnitude and direction
Momentum is a Vector Quantity
8
There is no specially named unit for momentum shy so there is an opportunity for it to be named after a renowned physicist
We use the product of the units of mass and velocity
mass x velocity
kgsdotms
SI Unit for Momentum
9
1 Which has more momentum
A A large truck moving at 30 ms
B A small car moving at 30 ms
C Both have the same momentum Answ
er
10
2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms
Answ
er
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
2
AP Physics 1 Momentum
2015shy12shy02
wwwnjctlorg
3
Table of Contents
bull Momentum
bull ImpulseshyMomentum Equationbull The Momentum of a System of Objects
bull Types of Collisions
Click on the topic to go to that section
bull Conservation of Momentum
bull Collisions in Two Dimensions
4
Return toTable ofContents
Momentum
5
Newtonrsquos First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia)
In our experience
bull When objects of different mass travel with the same velocity the one with more mass is harder to stop
bull When objects of equal mass travel with different velocities the faster one is harder to stop
Momentum Defined
6
Define a new quantity momentum (p) that takes these observations into account
momentum = mass times velocity
Momentum Defined
p =mv
click here for a introductory video on momentum from Bill Nye
7
Sincebull mass is a scalar quantitybull velocity is a vector quantitybull the product of a scalar and a vector is a vector
andbull momentum = mass times velocity
therefore
Momentum is a vector quantity shy it has magnitude and direction
Momentum is a Vector Quantity
8
There is no specially named unit for momentum shy so there is an opportunity for it to be named after a renowned physicist
We use the product of the units of mass and velocity
mass x velocity
kgsdotms
SI Unit for Momentum
9
1 Which has more momentum
A A large truck moving at 30 ms
B A small car moving at 30 ms
C Both have the same momentum Answ
er
10
2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms
Answ
er
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
3
Table of Contents
bull Momentum
bull ImpulseshyMomentum Equationbull The Momentum of a System of Objects
bull Types of Collisions
Click on the topic to go to that section
bull Conservation of Momentum
bull Collisions in Two Dimensions
4
Return toTable ofContents
Momentum
5
Newtonrsquos First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia)
In our experience
bull When objects of different mass travel with the same velocity the one with more mass is harder to stop
bull When objects of equal mass travel with different velocities the faster one is harder to stop
Momentum Defined
6
Define a new quantity momentum (p) that takes these observations into account
momentum = mass times velocity
Momentum Defined
p =mv
click here for a introductory video on momentum from Bill Nye
7
Sincebull mass is a scalar quantitybull velocity is a vector quantitybull the product of a scalar and a vector is a vector
andbull momentum = mass times velocity
therefore
Momentum is a vector quantity shy it has magnitude and direction
Momentum is a Vector Quantity
8
There is no specially named unit for momentum shy so there is an opportunity for it to be named after a renowned physicist
We use the product of the units of mass and velocity
mass x velocity
kgsdotms
SI Unit for Momentum
9
1 Which has more momentum
A A large truck moving at 30 ms
B A small car moving at 30 ms
C Both have the same momentum Answ
er
10
2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms
Answ
er
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
4
Return toTable ofContents
Momentum
5
Newtonrsquos First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia)
In our experience
bull When objects of different mass travel with the same velocity the one with more mass is harder to stop
bull When objects of equal mass travel with different velocities the faster one is harder to stop
Momentum Defined
6
Define a new quantity momentum (p) that takes these observations into account
momentum = mass times velocity
Momentum Defined
p =mv
click here for a introductory video on momentum from Bill Nye
7
Sincebull mass is a scalar quantitybull velocity is a vector quantitybull the product of a scalar and a vector is a vector
andbull momentum = mass times velocity
therefore
Momentum is a vector quantity shy it has magnitude and direction
Momentum is a Vector Quantity
8
There is no specially named unit for momentum shy so there is an opportunity for it to be named after a renowned physicist
We use the product of the units of mass and velocity
mass x velocity
kgsdotms
SI Unit for Momentum
9
1 Which has more momentum
A A large truck moving at 30 ms
B A small car moving at 30 ms
C Both have the same momentum Answ
er
10
2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms
Answ
er
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
5
Newtonrsquos First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia)
In our experience
bull When objects of different mass travel with the same velocity the one with more mass is harder to stop
bull When objects of equal mass travel with different velocities the faster one is harder to stop
Momentum Defined
6
Define a new quantity momentum (p) that takes these observations into account
momentum = mass times velocity
Momentum Defined
p =mv
click here for a introductory video on momentum from Bill Nye
7
Sincebull mass is a scalar quantitybull velocity is a vector quantitybull the product of a scalar and a vector is a vector
andbull momentum = mass times velocity
therefore
Momentum is a vector quantity shy it has magnitude and direction
Momentum is a Vector Quantity
8
There is no specially named unit for momentum shy so there is an opportunity for it to be named after a renowned physicist
We use the product of the units of mass and velocity
mass x velocity
kgsdotms
SI Unit for Momentum
9
1 Which has more momentum
A A large truck moving at 30 ms
B A small car moving at 30 ms
C Both have the same momentum Answ
er
10
2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms
Answ
er
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
6
Define a new quantity momentum (p) that takes these observations into account
momentum = mass times velocity
Momentum Defined
p =mv
click here for a introductory video on momentum from Bill Nye
7
Sincebull mass is a scalar quantitybull velocity is a vector quantitybull the product of a scalar and a vector is a vector
andbull momentum = mass times velocity
therefore
Momentum is a vector quantity shy it has magnitude and direction
Momentum is a Vector Quantity
8
There is no specially named unit for momentum shy so there is an opportunity for it to be named after a renowned physicist
We use the product of the units of mass and velocity
mass x velocity
kgsdotms
SI Unit for Momentum
9
1 Which has more momentum
A A large truck moving at 30 ms
B A small car moving at 30 ms
C Both have the same momentum Answ
er
10
2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms
Answ
er
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
7
Sincebull mass is a scalar quantitybull velocity is a vector quantitybull the product of a scalar and a vector is a vector
andbull momentum = mass times velocity
therefore
Momentum is a vector quantity shy it has magnitude and direction
Momentum is a Vector Quantity
8
There is no specially named unit for momentum shy so there is an opportunity for it to be named after a renowned physicist
We use the product of the units of mass and velocity
mass x velocity
kgsdotms
SI Unit for Momentum
9
1 Which has more momentum
A A large truck moving at 30 ms
B A small car moving at 30 ms
C Both have the same momentum Answ
er
10
2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms
Answ
er
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
8
There is no specially named unit for momentum shy so there is an opportunity for it to be named after a renowned physicist
We use the product of the units of mass and velocity
mass x velocity
kgsdotms
SI Unit for Momentum
9
1 Which has more momentum
A A large truck moving at 30 ms
B A small car moving at 30 ms
C Both have the same momentum Answ
er
10
2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms
Answ
er
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
9
1 Which has more momentum
A A large truck moving at 30 ms
B A small car moving at 30 ms
C Both have the same momentum Answ
er
10
2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms
Answ
er
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
10
2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 ms
Answ
er
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
11
3 What is the momentum of a 20 kg object with a velocity of 50 ms to the left
Answ
er
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
12
4 What is the velocity of a 5 kg object whose momentum is minus15 kgshyms
Answ
er
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
13
5 What is the mass of an object that has a momentum of 35 kgshyms with a velocity of 7 ms
Answ
er
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
14
ImpulseshyMomentum Equation
Return toTable ofContents
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
15
Suppose that there is an event that changes an objects momentum
bull from p0 shy the initial momentum (just before the event)
bull by Δp shy the change in momentum
bull to pf shy the final momentum (just after the event)
The equation for momentum change is
Change in Momentum
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
16
Momentum change equation
Newtons First Law tells us that the velocity (and so the momentum) of an object wont change unless the object is affected by an external force
Momentum Change = Impulse
Look at the above equation Can you relate Newtons First Law to the Δp term
Δp would represent the external force
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
17
Lets now use Newtons Second Law and see if we can relate Δp to the external force in an explicit manner
Weve found that the momentum change of an object is equal to an external Force applied to the object over a period of time
Momentum Change = Impulse
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
18
The force acting over a period of time on an object is now defined as Impulse and we have the ImpulseshyMomentum equation
Impulse Momentum Equation
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
19
There no special unit for impulse We use the product of the units of force and time
force x time
Nsdots
Recall that N=kgsdotms2 so
Nsdots=kgsdotms 2 x s
= kgsdotms
SI Unit for Impulse
This is also the unit for momentum which is a good thing since Impulse is the change in momentum
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
20
6 There is a battery powered wheeled cart moving towards you at a constant velocity You want to apply a force to the cart to move it in the opposite direction Which combination of the following variables will result in the greatest change of momentum for the cart Select two answers
A Increase the applied force
B Increase the time that the force is applied
C Maintain the same applied force
D Decrease the time that the force is applied
Answ
er
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
21
7 From which law or principle is the ImpulseshyMomentum equation derived from
A Conservation of Energy
B Newtons First Law
C Newtons Second Law
D Conservation of Momentum Answ
er
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
22
8 Can the impulse applied to an object be negative Why or why not Give an example to explain your answer
Answ
er
Yes since Impulse is a vector quantity it has magnitude and direction Direction can be described as negative An example would be if a baseball is thrown in the positive direction and you are catching it with a mitt The mitt changes the baseballs momentum from positive to zero shy hence applying a negative impulse to it
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
23
Effect of Collision Time on Force
∆t (seconds)
F (newtons) Since force is inversely
proportional to Δt changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object
Impulse = F(∆t) = F(∆t)
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
24
Every Day Applications
Impulse = F(∆t) = F(∆t) The inverse relationship of Force and time interval leads to many interesting applications of the ImpulseshyMomentum equation to everyday experiences such as
bull car structural safety designbull car air bagsbull landing after parachutingbull martial artsbull hitting a baseballbull catching a basebal
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
25
Every Day Applications
I=FΔt=Δp Lets analyze two specific cases from the previous list
bull car air bagsbull hitting a baseballWhenever you have an equation like the one above you have to decide which values will be fixed and which will be varied to determine the impact on the third valueFor the car air bags well fix Δp vary Δt and see its impact on F For the bat hitting a ball well fix F vary Δt and see the impact on Δp
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
26
Car Air Bags
I=FΔt=Δp
In the Dynamics unit of this course it was shown how during an accident seat belts protect passengers from the effect of Newtons First Law by stopping the passenger with the car and preventing them from striking the dashboard and window
They also provide another benefit explained by the ImpulseshyMomentum equation But this benefit is greatly enhanced by the presence of air bags Can you see what this benefit is
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
27
Car Air Bags
I=FΔt=Δp The seat belt also increases the time interval that it takes the passenger to slow down shy there is some play in the seat belt that allows you to move forward a bit before it stops you
The Air bag will increase that time interval much more than the seat belt by rapidly expanding letting the passenger strike it then deflating
Earlier it was stated that for the Air bag example Δp would be fixed and Δt would be varied So weve just increased Δt Why is Δp fixed
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
28
Car Air Bags
I=FΔt=Δp Δp is fixed because as long as the passenger remains in the car the car (and the passengers) started with a certain velocity and finished with a final velocity of zero independent of seat belts or air bags
Rearranging the equation we have
F represents the Force delivered to the passenger due to the accident
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
29
Car Air Bags
I=FΔt=Δp Since Δp is fixed by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag) the force F delivered to the passenger is smaller
Less force on the passenger means less physical harm Also another benefit needs a quick discussion of Pressure
Pressure is Force per unit area By increasing the area of the body that feels the force (the air bag is big) less pressure is delivered to parts of the body shy reducing the chance of puncturing the body Also good
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
30
9 If a car is in a front end collision which of the below factors will help reduce the injury to the driver and passengers Select two answers
A An absolutely rigid car body that doesnt deform
B Deployment of an air bag for each adult in the car
C Deployment of an air bag only for the driver
D The proper wearing of a seatbelt or child seat for each person in the car
Answ
er
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
31
10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest
A Force delivered to the car
B Interval of time for the collision
C Momentum change
D Acceleration of the car Answ
er
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
32
Hitting a Baseball
I=FΔt=Δp Now were going to take a case where there is a given force and the time interval will be varied to find out what happens to the change of momentum
This is different from the Air Bag example just worked (Δp was constant Δt was varied and its impact on F was found)
Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat The ball is approaching the batter with a positive momentum
What is the goal of the batter
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
33
Hitting a Baseball
I=FΔt=Δp The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing The greater the Δp the faster the ball will fly off his bat which will result in it going further hopefully to the seats
Hitting a baseball is way more complex than the analysis that will follow If youre interested in more information please check out the book The Physics of Baseball written by Robert Adair a Yale University physicist
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
34
Hitting a Baseball
I=FΔt=Δp In this case the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball This means he should follow through with his swing
The batter needs to apply a large impulse to reverse the balls large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers)
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
35
Every Day Applications
I=FΔt=Δp Now discuss the other examples Make sure you decide which object in the collision is more affected by the force or the change in momentum and which variables are capable of being varied
Consider the Air Bag example shy the car experiences the same impulse as the passenger during an accident but a car is less valuable than a human being shy so it is more important for the passenger that less force is delivered to his body shy and more force is absorbed by the car
bull car structural safety designbull landing after parachutingbull martial artsbull catching a basebal
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
36
11 An external force of 25 N acts on a system for 100 s What is the magnitude of the impulse delivered to the system
Answ
er
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
37
12 In the previous problem an external force of 25 N acted on a system for 100 s The impulse delivered was 250 Nshys What was the change in momentum of the system
Answ
er
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
38
13 The momentum change of an object is equal to the ______
A force acting on itB impulse acting on itC velocity change of the objectD objects mass times the force acting on it
Answ
er
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
39
14 Air bags are used in cars because they
B Answ
er
A increase the force with which a passenger hits the dashboard
B increase the duration (time) of the passengers impactC decrease the momentum of a collisionD decrease the impulse in a collision
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
40
15 One car crashes into a concrete barrier Another car crashes into a collapsible barrier at the same speed What is the difference between the two crashes Select two answers
A change in momentum
B force on the car
C impact time
D final momentum
Answ
er
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
41
16 In order to increase the final momentum of a golf ball the golfer should Select two answers
A maintain the speed of the golf club after the impact (follow through)
B Hit the ball with a greater force
C Decrease the time of contact between the club and the ball
D Decrease the initial momentum of the golf club Answ
er
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
42
17 An external force acts on an object for 00020 s During that time the objects momentum increases by 400 kgshyms What was the magnitude of the force
Answ
er
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
43
18 A 50000 N force acts for 0030 s on a 25 kg object that was initially at rest What is its final velocity
Answ
er
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
44
19 A 1200 kg car slows from 400 ms to 50 ms in 20 s What force was applied by the brakes to slow the car
Answ
er
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
45
Graphical Interpretation of ImpulseGraphs are a very valuable method to solve physics problems
So far weve dealt with a constant force exerted over a given time interval But forces are not always constant shy most of the time they are changing over time
In the baseball example the force of the bat starts out very small as it makes initial contact with the ball It then rapidly rises to its maximum value then decreases as the ball leaves the bat This would be a very difficult problem to handle without a graph
Start with representing a constant force over a time interval Δt and plot Force vs time on a graph
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
Return toTable ofContents
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
46
Graphical Interpretation of Impulse
Note the area under the ForceshyTime graph is
height times base = F ∆t = I (impulse) = ∆p (change in momentum)
F [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
47
Graphical Interpretation of ImpulseF [N]
t [s]
F
∆t
Area =
Impulse (Change in Momentum)
By taking the area under a Forceshytime graph we have found both Impulse and Δp If the force is not constant the area can be found by breaking it into solvable shapes (rectangles triangles) and summing them up
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
48
20 Using the Fshyt graph shown what is the change in momentum during the time interval from 0 to 6 s
Answ
er
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
49
21 A 5 kg object with an initial velocity of 3 ms experiences the force shown in the graph What is its velocity at 6 s
Answ
er
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
50
22 Using the Fshyt graph shown what is the change in momentum during the time interval from 6 to 10 s
Answ
er
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
51
The Momentum of a System of Objects
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52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
52
If a system contains more than one object its total momentum is the vector sum of the momenta of those objects
psystem = sum p
psystem = p1 + p2 + p3 +
psystem = m1v1 + m2v2 + m3v3 +
The Momentum of a System of Objects
Its critically important to note that
momenta add as vectors
not as scalars
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
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59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
53
psystem = m1v1 + m2v2 + m3v3 +
In order to determine the total momentum of a system Firstbull Determine a direction to be considered positive bull Assign positive values to momenta in that direction bull Assign negative values to momenta in the opposite direction
The Momentum of a System of Objects
Then Add the momenta to get a total momentum of the system
+shy
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
54
Determine the momentum of a system of two objects m1 has a mass of 6 kg and a velocity of 13 ms towards the east and m2 has a mass of 14 kg and a velocity of 7 ms towards the west
psystem = p1 + p2
psystem = m1v1 + m2v2
Example
m1 = 6 kg m2 = 14 kgv1 = 13 ms v2 = minus7 ms
6 kg 14 kg13 ms 7 ms East (+)
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
psystem = m1v1 + m2v2= (6 kg)(13 ms) + (14 kg)(minus7 ms)= (78 kgsdotms) + (minus98 kgsdotms)= minus20 kgsdotms
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
55
23 Determine the momentum of a system of two objects m1 has a mass of 60 kg and a velocity of 20 ms north and m2 has a mass of 3 kg and a velocity 20 ms south
Answ
er
psystem = m1v1 + m2v2= (6 kg)(20 ms) + (3 kg)(minus20 ms)= (120 kgsdotms) + (minus60 kgsdotms)= 60 kgsdotms (magnitude)
direction is North
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
56
24 Determine the momentum of a system of two objects the first has a mass of 80 kg and a velocity of 80 ms to the east while the second has a mass of 50 kg and a velocity of 15 ms to the west
Answ
er
psystem = m1v1 + m2v2= (8 kg)(8 ms) + (5 kg)(minus15 ms)= (64 kgsdotms) + (minus75 kgsdotms)= minus11 kgsdotms (West)
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
57
25 Determine the momentum of a system of three objects The first has a mass of 70 kg and a velocity of 23 ms north the second has a mass of 90 kg and a velocity of 7 ms north and the third has a mass of 50 kg and a velocity of 42 ms south
Answ
er
psystem = m1v1 + m2v2 + m3v3
= (7kg)(23ms) + (9kg)(7ms) +(5 kg)(minus42ms)= (161 kgsdotms) + (63 kgsdotms) + (minus210 kgsdotms)
= 14 kgsdotms (North)
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
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67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
58
Conservation of Momentum
Return toTable ofContents
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
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110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
59
Some of the most powerful concepts in science are called conservation laws This was covered in detail in the Energy unit shy please refer back to that for more detail Here is a summary
Conservation laws bull apply to closed systems shy where the objects only interact with each other and nothing elsebull enable us to solve problems without worrying about the details of an event
Conservation Laws
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
60
In the last unit we learned that energy is conserved
Like energy momentum is a conserved property of nature This means
bull Momentum is not created or destroyed
bull The total momentum in a closed system is always the same
bull The only way the momentum of a system can change is if momentum is added or taken away by an outside force
Momentum is Conserved
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
61
We will use the Conservation of Momentum to explain and predict the motion of a system of objects
As with energy it will only be necessary to compare the system at two times just before and just after an event
Conservation of Momentum
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
62
The momentum of an object changes when it experiences an impulse (I=FΔt)
This impulse arises when a nonshyzero external force acts on the object
This is exactly the same for a system of objects
Conservation of Momentum
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
63
Conservation of Momentum
If there is no net external force on the system then F = 0 and since I = FΔt I is also zero
Since psystem 0 = psystem f the momentum of the system is conserved that is it does not change
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
64
26 Why dont the internal forces on a system change the momentum of the system
Answ
er
The internal forces result from theinteraction of the objects within the system By Newtons Third Law these are all actionshyreaction pairsand viewed from outside the systemall cancel out Thus there is no netforce acting on the system and the momentum is conserved
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
65
27 An external positive force acts on a system of objects Which of the following are true Select two answers
A The velocity of the system remains the same
B The velocity of the system increases
C The momentum of the system decreases
D The momentum of the system increases
Answ
er
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
66
Types of Collisions
Return toTable ofContents
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
67
Types of Collisions
Objects in an isolated system can interact with each other in two basic ways
bull They can collide
bull If they are stuck together they can explode (push apart)
In an isolated system both momentum and total energy are conserved But the energy can change from one form to another
Conservation of momentum and change in kinetic energy can help predict what will happen in these events
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
68
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form
bull inelastic collisions two objects collide converting some kinetic energy into other forms of energy such as potential energy heat or sound bull elastic collisions two objects collide and bounce off each other while conserving kinetic energy shy energy is not transformed into any other typebull explosions an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
69
Inelastic Collisions
There are two types of Inelastic Collisions
bull perfect inelastic collisions two objects collide stick together and move as one mass after the collision transferring kinetic energy into other forms of energy
bull general inelastic collisions two objects collide and bounce off each other transferring kinetic energy into other forms of energy
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
70
Elastic Collisions
There is really no such thing as a perfect elastic collision During all collisions some kinetic energy is always transformed into other forms of energy
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions
In chemistry the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law
Other examples include a steel ball bearing dropping on a steel plate a rubber superball bouncing on the ground and billiard balls bouncing off each other
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
71
Explosions
A firecracker is an example of an explosion The chemical potential energy inside the firecracker is transformed into kinetic energy light and sound
A cart with a compressed spring is a good example When the spring is against a wall and it is released the cart starts moving shy converting elastic potential energy into kinetic energy and sound
Think for a moment shy can you see a resemblance between this phenomenon and either an elastic or inelastic collision
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
72
Explosions
In both an inelastic collision and an explosion kinetic energy is transformed into other forms of energy shy such as potential energy But they are time reversed
An inelastic collision transforms kinetic energy into other forms of energy such as potential energy
An explosion changes potential energy into kinetic energy
Thus the equations to predict their motion will be inverted
The next slide summarizes the four types of collisions andexplosions
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
73
Collisions and Explosions
Event Description Momentum Conserved
Kinetic Energy Conserved
General Inelastic Collision
Objects bounce
off each otherYes
No Kinetic energy is converted to other forms of energy
Perfect Inelastic Collision
Objects stick together Yes
No Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce
off each otherYes Yes
ExplosionObject or
objects break apart
Yes
No Release of potential energy increases kinetic
energy
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
74
28 In _______ collisions momentum is conserved
A Elastic
B Inelastic
C All
Answ
er
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
75
29 In ______ collisions kinetic energy is conserved
A Elastic
B Inelastic
C All
Answ
er
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
76
30 In an inelastic collision kinetic energy is transformed into what after the collision Select two answers
A Nothing kinetic energy is conserved
B More kinetic energy
C Thermal energy
D Light energy Answ
er
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
77
Conservation of MomentumDuring a collision or an explosion measurements show that the total momentum of a closed system does not change The diagram below shows the objects approaching colliding and then separating
A BmAvA mBvB
A B
A BmAvA mBvB
+xthe prime means after
If the measurements dont show that the momentum is conserved then this would not be a valid law Fortunately they do and it is
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
78
Perfect Inelastic CollisionsIn a perfect inelastic collision two objects collide and stick together moving afterwards as one object
A B
After (moving together)
pA+pB=(mA+ mB)v
A B
pA=mAvA
Before (moving towards each other)
pB=mBvB
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
79
31 A 13500 kg railroad freight car travels on a level track at a speed of 45 ms It collides and couples with a 25000 kg second car initially at rest and with brakes released No external force acts on the system What is the speed of the two cars after colliding
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (13500kg)(45ms) (13500+25000)kg= 16 ms
in the same direction as the first cars initial velocity
m1 = 13500kgm2 = 25000kgv1 = 45msv2 = 0 msv1 = v2 = v
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
80
32 A cannon ball with a mass of 1000 kg flies in horizontal direction with a speed of 250 ms and strikes a ship initially at rest The mass of the ship is 15000 kg Find the speed of the ship after the ball becomes embedded in it
Answ
er
m1v1+m2v2 = m1v1+m2v2m1v1+0 = (m1+m2) vv = m1v1(m1+m2)
= (100kg)(250ms) (100+15000)kg= 17 ms
in the same direction as cannon balls initial velocity
m1 = 100kgm2 = 15000kgv1 = 800msv2 = 0 msv1 = v2 = v
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
81
33 A 40 kg girl skates at 55 ms on ice toward her 70 kg friend who is standing still with open arms As they collide and hold each other what is their speed after the collision
Answ
er
m1 = 40kgm2 = 70kgv1 = 55msv2 = 0 msv1 = v2 = v
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
82
Explosions
In an explosion one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objectsTo make the problems solvable at this math level we will assumebull the object (or a coupled pair of objects) breaks into two pieces bull the explosion is along the same line as the initial velocity
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
83
Explosions
A B
Before (moving together)
pA+pB=(mA+ mB)v
A BpA=mAvA
After (moving apart)
pB=mBvB
Heres a schematic of an explosion Can you see how its the time reversed picture of a perfectly inelastic collision
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
84
34 A 5 kg cannon ball is loaded into a 300 kg cannon When the cannon is fired it recoils at 5 ms What is the cannon balls velocity after the explosion
Answ
er
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
85
35 Two railcars one with a mass of 4000 kg and the other with a mass of 6000 kg are at rest and stuck together To separate them a small explosive is set off between them The 4000 kg car is measured travelling at 6 ms How fast is the 6000 kg car going
Answ
er
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
86
Elastic CollisionsIn an elastic collision two objects collide and bounce off each other as shown below and both momentum and kinetic energy are conserved
This will give us two simultaneous equations to solve to predict their motion after the collision
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
87
Elastic Collisions
A B
pA=mAvA
Before (moving towards) After (moving apart)
pB=mBvB
A B
pA=mAvA pB=mBvB
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
88
Elastic Collisions
Since both of these equations describe the motion of the same system they must both be true shy hence they can be solved as simultaneous equations That will be done on the next slide and an interesting result will appear
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
89
Elastic Collision Simultaneous Equations
m1v1 + m2v2 = m1v1 +m2v2m1v1 shy m1v1 = m2v2 shy m2v2 m1(v1 shy v1) = m2(v2 shy v2)
frac12m1v12 + frac12m2v22 = frac12m1v12 +frac12m2v22
m1v12 + m2v22 = m1v12 +m2v22
m1v12 shy m1v12 = m2v22 shy m2v22 m1(v12 shy v12) = m2(v22 shy v22)m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)
m1(v1 + v1)(v1 shy v1) = m2(v2 + v2)(v2 shy v2)m1(v1 shy v1) = m2(v2 shy v2)
v1 + v1 = v2 + v2
Conservation of Momentum Conservation of Kinetic Energy
v1 shy v2 = shy(v1 shy v2)
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
90
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously the following result appeared
Do you recognize the terms on the left and right of the above equation And what does it mean
v1 shy v2 = shy(v1 shy v2)
The terms are the relative velocities of the two objects before and after the collision It means that for all elastic collisions shy regardless of mass shy the relative velocity of the objects is the same before and after the collision
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
91
36 Two objects have an elastic collision Before they collide they are approaching with a velocity of 4 ms relative to each other With what velocity do they move apart from one another after the collision
Answ
er
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
92
37 Two objects have an elastic collision Object m1 has an initial velocity of +40 ms and m2 has a velocity of shy30 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
93
38 Two objects have an elastic collision Object m1 has an initial velocity of +60 ms and m2 has a velocity of 20 ms After the collision m1 has a velocity of 10 ms What is the velocity of m2
Answ
er
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
94
Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations and solving them in a slightly different manner (the algebra is a little more intense) the following equations result
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
95
Properties of Elastic Collisions
Using these equations we will predict the motion of three specific cases and relate them to observation
bull Two same mass particles collidebull A very heavy particle collides with a light particle at restbull A light particle collides head on with a heavy particle at rest
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
96
Collision of same mass particles
v1 = v2 v2 = v1
m mm
v1
m
v2
v1 = v2 and v2 = v1
For a collision of two particles of the same mass m = m1 = m2 When this substitution is made into the two equations on the left (try it) they simplify to
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
97
Collision of same mass particles
The particles exchange velocities A good example of this is playing billiards The cue ball and the striped and solid balls are all the same mass When the cue ball hits a solid ball the cue ball stops and the solid ball takes off with the cue balls velocity
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
98
Collision of same mass particles
The particles exchange velocities Another good example is found in nuclear reactor engineering When Uranium 235 fissions it releases 3 high speed neutrons These neutrons are moving too fast to cause another Uranium 235 nucleus to fission so they have to be slowed down What type of atom would be of most use here
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
99
Collision of same mass particles
The particles exchange velocities An atom with a similar mass to neutrons A Hydrogen atom moving slower than the neutrons would be great shy the neutron would exchange velocities slowing down while speeding up the hydrogen atom Operational reactors use water for this purpose (heavier than hydrogen but has many other benefits shy nuclear engineering is not as simple as elastic collisions)
v1 = v2 and v2 = v1
v1 = v2 v2 = v1
m mm
v1
m
v2
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
100
Large mass striking lighter mass at rest
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
v1 = v1 and v2 = 2v1
For this case m1 gtgt m2 (gtgt is the symbol for much greater) which means that m1 plusmn m2 asymp m1 The equations simplify to
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
101
Large mass striking lighter mass at rest
The large mass particle continues on its way with a slightly smaller velocity The small particle moves at twice the initial speed of the large mass particle Picture a bowled bowling ball striking a beach ball at rest
v1 = v1 and v2 = 2v1
m1 m1
m2 m2
v1 v2 = 0 v1 asymp v1 v2 = 2v1
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
102
Small mass striking heavier mass at rest
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
v1 = shyv1 and v2 asymp 0
For this case m1 ltlt m2 (ltlt is the symbol for much less than) which means that m1 plusmn m2 asymp m2 The equations simplify to
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
103
Small mass striking heavier mass at rest
v1 = shyv1 and v2 asymp 0
The large mass particle remains mostly at rest The small particle reverses direction with the same magnitude as its initial velocity Picture a moving golf ball hitting a stationary bowling ball
m2
m1
m2
v2 = 0
m1 v1v1
v2 = 0
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
104
39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest The velocity of the bowling ball is virtually unaffected by the collision What will be the speed of the ping pong ball
Answ
er
v1shyv2 = shy(v1shyv2)
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
105
40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of shy2v The bat barely changes velocity during the collision How fast is the baseball going after its hit
Answ
er
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
106
41 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m1 after the collision
Answ
er
When identical mass objects experience an elastic collision they swap their initial velocitiesv1 = v2 = shy30 msv2 = v1 = 60 msSo the velocity of m1 is shy30 ms
v1 = +6msv2 = shy3msv1 = v2 =
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
107
42 Two objects with identical masses have an elastic collision the initial velocity of m1 is +60 ms and m2 is shy30 ms What is the velocity of m2 after the collision
Answ
er
v1 = +6msv2 = shy3msv1 = v2 =
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
108
43 A golf ball is hit against a solid cement wall and experiences an elastic collsion The golf ball strikes the wall with a velocity of +35 ms What velocity does it rebound with
Answ
er
When a light object strikes a very massive object the light object rebounds with the opposite velocity shy in this case the golf ball will leave the wall with a velocity of shy35 ms
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
109
Collisions in Two Dimensions
Return toTable ofContents
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
110
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame
This means that the momentum conservation equation p = p can be solved independently for each component
This of course also applies to three dimensions but well stick with two for this chapter
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
111
Example Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall rebounding with the same velocity where its angle of incidence equals its angle of reflection
Is momentum conserved in this problem
m
m
θ
θ
p
p
pxpy
pxpy
The solid lines represent the momentum of the ball (blue shy prior to collision red shy after the collision) The dashed lines are the x and y components of the momentum vectors
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
112
Example Collision with a Wall
Momentum is not conserved
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis
However since we have an elastic collision the ball bounces off the wall with the same speed that it struck the wall Hence the magnitude of the initial momentum and the final momentum is equal
m
m
θ
θ
p
p
pxpy
pxpy
Now its time to resolve momentum into components along the x and y axis
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
113
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Apply the Impulse Momentum Theorem in two dimensions
What does this tell us
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
114
Example Collision with a Wall
m
m
θ
θ
p
p
pxpy
pxpy
Momentum is conserved in the y directionbut it is not conserved in the x direction
The initial and final momentum in the y directionare the same The initial and final momentumin the x direction are equal in magnitude butopposite in direction shy they reverse
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
115
44 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball
A shymv
B shy2mv
C shymv cosθ
D shy2mv cosθ m
m
θ
θ
p
p
pxpy
pxpy An
swer
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
116
45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially What was the change in momentum of the ball in the y direction
A shymv
B 0
C mv
D 2mv m
m
θ
θ
p
p
pxpy
pxpy An
swer
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
117
General Two Dimensional Collisions
m1 m2
Well now consider the more general case of two objects moving in random directions in the xshyy plane and colliding Since there is no absolute reference frame well line up the xshyaxis with the velocity of one of the objects
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
118
General Two Dimensional Collisions
This is not a head on collision shy note how m1 heads off with a y component of velocity after it strikes m2 Also did you see how we rotated the coordinate system so the x axis is horizontal
To reduce the amount of variables that must be specified we will also assume that mass 2 is at rest
The problem now is to find the momentum of m2 after the collision
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
119
General Two Dimensional Collisions
m1 m2
Before
p2 = 0
After
m1
m2
p2 =
This will be done by looking at the vectors first shy momentum must be conserved in both the x and the y directions
Since the momentum in the y direction is zero before the collision it must be zero after the collision
And the value that m1 has for momentum in the x direction must be shared between both objects after the collision shy and not equally shy it will depend on the masses and the separation angle
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
120
General Two Dimensional Collisions
m1
m2
Here is the momentum vector breakdown of mass 1 after the collision
m2 needs to have a component in the y direction to sum to zero with m1s final y momentum And it needs a component in the x direction to add to m1s final x momentum to equal the initial x momentum of m1
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
121
46 After the collision shown below which of the following is the most likely momentum vector for the blue ball
A
B
C
D
E
before after
m1
m1
m2
m2
Answ
er
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
122
General Two Dimensional CollisionsNow that weve seen the vector analysis lets run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision
There is a bowling ball with momentum 200 kgshyms that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above What is the final velocity of the pin
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
123
General Two Dimensional Collisions
Given
Find
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
124
General Two Dimensional Collisions
Use Conservation of Momentum in the x and y directions
x direction yshydirection
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
125
before after600deg
120 kgshyms
θ
200 kgshyms
m2 m2
m1 m1
General Two Dimensional Collisions
Now that the x and y components of the momentum of mass 2 have been found the total final momentum is calculated
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
126
47 A 50 kg bowling ball strikes a stationary bowling pin After the collision the ball and the pin move in directions as shown and the magnitude of the pins momentum is 180 kgshyms What was the velocity of the ball before the collision
before after
30deg18 kgshyms
531degpinball
Answ
er
yshydirectionp1y + p2y = p1y + p2y0 + 0 = p1y +18sin(shy30deg)0 = p1y shy 9p1y = 9 kgshyms
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
127
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθ
One common kind of inelastic collision is where two cars collide and stick at an intersectionIn this situation the two objects are traveling along paths that are perpendicular just prior to the collision
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
128
Perfect Inelastic Collisions in Two Dimensions
m1
m2Before
p1
p2
Afterm1m2
pθpx
py
pshyconservation in x in y
final momentum
final velocity
final direction
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
129
48 Object A with mass 200 kg travels to the east at 100 ms and object B with mass 500 kg travels south at 200 ms They collide and stick together What is the velocity (magnitude and direction) of the objects after the collision
Answ
er
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
130
Explosions in Two Dimensions
p3
p2
p1
The Black object explodes into 3 pieces (blue red and green)
We want to determine the momentum of the third piece
During an explosion the total momentum is unchanged since no EXTERNAL force acts on the systembull By Newtons Third Law the forces that occur between the particles within the object will add up to zero so they dont affect the momentumbull if the initial momentum is zero the final momentum is zero bull the third piece must have equal and opposite momentum to the sum of the other two
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
131
Explosions in Two Dimensions
θ
p1
p2
p3
The Black object explodes into 3 pieces (blue red and green) We want to determine the momentum of the third piece
before px = py = 0
after p1x + p2x + p3x = 0
p1y + p2y + p3y = 0
In this case the blue and red pieces are moving perpendicularly to each other so
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
132
49 A stationary cannon ball explodes in three pieces The momenta of two of the pieces is shown below What is the direction of the momentum of the third piece
A
B
C
D
Answ
er
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
133
50 A stationary 100 kg bomb explodes into three pieces A 200 kg piece moves west at 2000 ms Another piece with a mass of 300 kg moves north with a velocity of 1000 ms What is the velocity (speed and direction) of the third piece
Answ
er
Attachments
Eqn1pict
Attachments
Eqn1pict