Lecture 12 Momentum, Energy, and Collisions. Announcements Assignment 6 due Wednesday, Oct 5...

Lecture 12

Momentum, Energy, and Collisions

Announcements

Assignment 6 due Wednesday, Oct 5 (11:59pm)

EXAM: October 13 (through Chapter 9)

Look for messages regarding special TA office hours

Reminder that formula sheet for Test 2 now on Collab under Midterms

Practice problems will be attached to posted slides (this afternoon, I promise)

Practice test from last year on Collab under Resources

Reading and Review

Linear Momentum

Impulse

With no net external force:

A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v along the floor, but at right angles to one another. Find the speed and direction of the third piece.

A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v along the floor, but at right angles to one another. Find the speed and direction of the third piece.

We know that px=0, py = 0 in initial stateand no external forces act in the horizontal

An 85-kg lumberjack stands at one end of a 380-kg floating log, as shown in the figure. Both the log and the lumberjack are at rest initially. (a) If the lumberjack now trots toward the other end of the log with a speed of 2.7 m/s relative to the log, what is the lumberjack’s speed relative to the shore? Ignore friction between the log and the water. (b) If the mass of the log had been greater, would the lumberjack’s speed relative to the shore be greater than, less than, or the same as in part (a)? Explain. (c) Check (b) by assuming log has mass of 450 kg.

Rolling in the RainRolling in the Rain

a) speeds up

b) maintains constant speed

c) slows down

d) stops immediately

An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)

Because the rain falls in vertically,

it adds no momentum to the box,

thus the box’s momentum is

conserved. However, because the

mass of the box slowly increases

with the added rain, its velocity has

to decrease.

Rolling in the RainRolling in the Rain

a) speeds up

b) maintains constant speed

c) slows down

d) stops immediately

An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)

When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly (whereas it is safe to hold the gun while it is fired)?

a) it is much sharper than the gun

b) it is smaller and can penetrate your body

c) it has more kinetic energy than the gun

d) it goes a longer distance and gains speed

e) it has more momentum than the gun

Gun ControlGun Control

When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly (whereas it is safe to hold the gun while it is fired)?

a) it is much sharper than the gun

b) it is smaller and can penetrate your body

c) it has more kinetic energy than the gun

d) it goes a longer distance and gains speed

e) it has more momentum than the gun

Even though it is true that the magnitudes of the momenta of the gun and the bullet are equal, the bullet is less massive and so it has a much higher velocity. Because KE is related to v2, the bullet has considerably more KE and therefore can do more damage on impact.

Gun ControlGun Control

Two objects collide... and stick

A completely inelastic collision: no “bounce back”

No external forces... so momentum of system is conservedinitial px = mv0

final px = (2m)vf

mv0 = (2m)vf

vf = v0 / 2

mass m mass m

What about energy?

mass m mass m vf = v0 / 2

Kinetic energy is lost! KEfinal = 1/2 KEinitial

initial

final

Inelastic Collisions

This is an example of an “inelastic collision”

Collision: two objects striking one another

“Elastic” collision <=> “things bounce back”

Completely inelastic collision: objects stick together afterwards... no thing “bounces back”

Time of collision is short enough that external forces may be ignored so momentum is conserved

Inelastic collision: momentum is conserved but kinetic energy is not

Elastic vs. Inelastic

Inelastic collision: momentum is conserved but kinetic energy is not

Completely inelastic collision: colliding objects stick together, maximal loss of kinetic energy

Elastic collision: momentum and kinetic energy is conserved.

Completely Inelastic Collisions in One Dimension

Solving for the final momentum in terms of initial velocities and masses, for a 1-dimensional, completely inelastic collision between unequal masses:

Completely inelastic only(objects stick together, so have same final velocity)

KEfinal < KEinitial

Momentum Conservation:

Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block.

vf = m v0 / (m+M)

momentum conservation in inelastic collision

KE = 1/2 (mv0)2 / (m+M)

energy conservationafterwards

PE = (m+M) g h

hmax = (mv0)2 / [2 g (m+M)2]

Binomial expansion

If <<1 then

(1+ )n ~1+n

Ex.

3

33

1.02 1.061208

1.02 1.00 0.02 1.00 3 0.02 1.06

correct with 0.1%

Velocity of the ballistic pendulum

Pellet Mass (m): 1.84x10-3 kgPendulum Mass (M): 3.81 kgWire length (L): 4.00 m

30 3.24 10 /v s x

Crash Cars ICrash Cars I

a) I

b) II

c) I and II

d) II and III

e) all three

If all three collisions below

are totally inelastic, which

one(s) will bring the car on

the left to a complete halt?

Crash Cars ICrash Cars I

In case I, the solid wall clearly stops the car.

In cases II and III, because

ptot = 0 before the

collision, then ptot must

also be zero after the collision, which means that the car comes to a halt in all three cases.

a) I

b) II

c) I and II

d) II and III

e) all three

If all three collisions below

are totally inelastic, which

one(s) will bring the car on

the left to a complete halt?

Crash Cars IICrash Cars II

If all three collisions below are

totally inelastic, which one(s)

will cause the most damage

(in terms of lost energy)?

a) I

b) II

c) III

d) II and III

e) all three

Crash Cars IICrash Cars II

If all three collisions below are

totally inelastic, which one(s)

will cause the most damage

(in terms of lost energy)?

a) I

b) II

c) III

d) II and III

e) all three

The car on the left loses the same KE in all three cases, but in case III, the car on the right loses the most KE because KE = mv2 and the car in case III has the largest velocity.

Inelastic Collisions in 2 Dimensions

Energy is not a vector equation: there is only 1 conservation of energy equation

Momentum is a vector equation: there is 1 conservation of momentum equation per dimension

For collisions in two dimensions, conservation of momentum is applied separately along each axis:

Elastic CollisionsIn elastic collisions, both kinetic

energy and momentum are conserved.

One-dimensional elastic collision:

Elastic Collisions in 1-dimension

For given case of v2i = 0, solving for the final speeds:

Note: relative speed is conserved for head-on (1-D) elastic collision

We have two equations: conservation of momentum conservation of energy

and two unknowns (the final speeds).

Limiting cases

Limiting cases

Limiting cases

Toy Pendulum

Could two balls recoil and conserve both momentum and energy?

Incompatible!

Elastic Collisions IElastic Collisions I

v 2v

1at rest

at rest

a) situation 1

b) situation 2

c) both the same

Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on. In which case does the golf ball have the greater speed after the collision?

Remember that the magnitude of the relative velocity has to be equal before and after the collision!

Elastic Collisions IElastic Collisions I

v1

In case 1 the bowling ball will almost remain at rest, and the golf ball will bounce back with speed close to v.

v 22v

In case 2 the bowling ball will keep going with speed close to v, hence the golf ball will rebound with speed close to 2v.

a) situation 1

b) situation 2

c) both the same

Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on. In which case does the golf ball have the greater speed after the collision?

Elastic Collisions IIElastic Collisions II

v

v

m

M

Carefully place a small rubber ball (mass m) on

top of a much bigger basketball (mass M) and

drop these from the same height h so they

arrive at the ground with the speed v. What is

the velocity of the smaller ball after the

basketball hits the ground, reverses direction,

and then collides with the small rubber ball?

a) zero

b) v

c) 2v

d) 3v

e) 4v

• Remember that relative velocity has to be equal before and after collision! Before the collision, the basketball bounces up with v and the rubber ball is coming down with v, so their relative velocity is –2v. After the collision, it therefore has to be +2v!!

Elastic Collisions IIElastic Collisions II

v

v

v

v

3v

v

(a) (b) (c)

m

M

Carefully place a small rubber ball (mass m) on

top of a much bigger basketball (mass M) and

drop these from the same height h so they

arrive at the ground with the speed v. What is

the velocity of the smaller ball after the

basketball hits the ground, reverses direction,

and then collides with the small rubber ball?

a) zero

b) v

c) 2v

d) 3v

e) 4v

Elastic Collisions in 2-D

Two-dimensional collisions can only be solved if some of the final information is known, such as the final velocity of one object

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A proton collides elastically with another proton that is initially at rest.

The incoming proton has an initial speed of 3.5x105 m/s and makes

a glancing collision with the second proton. After the collision one

proton moves at an angle of 37o to the original direction of motion,

the other recoils at 53o to that same axis. Find the final speeds of the

two protons.

v0 = 3.5x105 m/s

initial

37o

53o

v2

v1

final

A proton collides elastically with another proton that is initially at rest.

The incoming proton has an initial speed of 3.5x105 m/s and makes

a glancing collision with the second proton. After the collision one

proton moves at an angle of 37o to the original direction of motion,

the other recoils at 53o to that same axis. Find the final speeds of the

two protons.

v0 = 3.5x105 m/s

initial

37o

53o

v2

v1

final

Momentum conservation:

if we’d been given only 1 angle, would have needed conservation of energy also!

Center of MassTreat extended mass as a bunch of small masses:

In a uniform gravitational field you can treat gravitational force as if it acts at the “Center of Mass”

g i iF m g m g M g

Center of MassThe center of mass of a system is the point where the

system can be balanced in a uniform gravitational field.

For two objects:

The center of mass is closer to the more massive object.

Center of Mass

In general:

Symmetry often very useful in determining the

Center of Mass

1 1 1 1 1 1

1 1 1

i iCM

i

m r m rm r m r m rR

m m m m M

Center of Mass

The center of mass need not be within the object

Motion about the Center of MassThe center of mass of a complex or composite object follows a trajectory as if it were a single particle - with mass equal to the complex object, and experiencing a force equal to the sum of all external forces on that complex object

Motion of the center of mass

Action/Reaction pairs inside the system cancel out

The total mass multiplied by the acceleration of the center of mass is equal to the net external force

The center of mass accelerates just as though it were a point particle of mass M acted on by

Momentum of a composite object

Recoil SpeedRecoil Speed

a) 0 m/s

b) 0.5 m/s to the right

c) 1 m/s to the right

d) 20 m/s to the right

e) 50 m/s to the right

A cannon sits on a stationary

railroad flatcar with a total mass

of 1000 kg. When a 10-kg

cannonball is fired to the left at

a speed of 50 m/s, what is the

speed of the center of mass (of

the flatcar + cannonball)?

Recoil SpeedRecoil Speed

Because the initial momentum of the

system was zero, the final total

momentum must also be zero,

regardless of the release of internal

energy, internal forces, etc.

If no external forces act, the motion of

the center of mass does not change

a) 0 m/s

b) 0.5 m/s to the right

c) 1 m/s to the right

d) 20 m/s to the right

e) 50 m/s to the right

A cannon sits on a stationary

railroad flatcar with a total mass

of 1000 kg. When a 10-kg

cannonball is fired to the left at

a speed of 50 m/s, what is the

speed of the center of mass (of

the flatcar + cannonball)?

Recoil Speed IIRecoil Speed II

a) 0 m/s

b) 0.5 m/s to the right

c) 1 m/s to the right

d) 20 m/s to the right

e) 50 m/s to the right

A cannon sits on a stationary

railroad flatcar with a total

mass of 1000 kg. When a 10-

kg cannonball is fired to the

left at a speed of 50 m/s, what

is the recoil speed of the

flatcar?

Recoil Speed IIRecoil Speed II

Because the initial momentum of the

system was zero, the final total

momentum must also be zero. Thus, the

final momenta of the cannonball and the

flatcar must be equal and opposite.

pcannonball = (10 kg)(50 m/s) = 500 kg-

m/s

pflatcar = 500 kg-m/s = (1000 kg)(0.5

m/s)

a) 0 m/s

b) 0.5 m/s to the right

c) 1 m/s to the right

d) 20 m/s to the right

e) 50 m/s to the right

A cannon sits on a stationary

railroad flatcar with a total

mass of 1000 kg. When a 10-

kg cannonball is fired to the

left at a speed of 50 m/s, what

is the recoil speed of the

flatcar?

Center of Mass

(1)

XCM

(2)

a) higher

b) lower

c) at the same place

d) there is no definable CM in this case

The disk shown below in (1) clearly has its center of mass at the center.

Suppose the disk is cut in half and the pieces arranged as shown in (2).Where is the center of mass of (2) as compared to (1) ?

(1)

XCM

(2)

CM

Center of Mass

The CM of each half is closer to the top of the semicircle than the bottom. The CM of the whole system is located at the midpoint of the two semicircle CMs, which is higher than the yellow line.

a) higher

b) lower

c) at the same place

d) there is no definable CM in this case

The disk shown below in (1) clearly has its center of mass at the center.

Suppose the disk is cut in half and the pieces arranged as shown in (2).

Where is the center of mass of (2) as compared to (1) ?