Physics 1D03 - Lecture 25 TEST 1 Tuesday May 19 9:30 am CNH-104 Kinematics, Dynamics & Momentum

Physics 1D03 - Lecture 25

TEST 1

Tuesday May 19

9:30 am

CNH-104

Kinematics, Dynamics

& Momentum


Momentum

• Newton’s original “quantity of motion” • a conserved quantity• a vector

-Newton’s Second Law in another form

-Momentum and Momentum Conservation


(We say “linear” momentum to distinguish it from angular momentum, a different physical quantity.)

Definition: The linear momentum p of a particle is its mass times its velocity:

Momentum is a vector, since velocity is a vector. Units: kg m/s (no special name).

p mv


Example 1 & 2

• A car of mass 1500kg is moving with a velocity of 72km/h. What is its momentum ?

• Two cars, one of mass 1000kg is moving at 36km/h and the other of mass 1200kg is moving at 72km/h in the opposite direction. What is the momentum of the system ?


Concept Quiz

Which object will have the smallest momentum ?

a) A 1.6x10-19kg particle moving at 1x103m/s

b) A 1000kg car moving at 2m/s

c) A 100kg person moving at 10m/s

d) A 5000kg truck at rest


dt)vd(m

dtvd

mamF

dtpd

FΣ

Newton’s Second Law

If mass is constant, then the rate of change of (mv) is equal to m times the rate of change of v. We can rewrite Newton’s Second Law:

or

This is how Newton wrote the Second Law. It remains true in cases where the mass is not constant.

Net external force = rate of change of momentum


Example 3

Rain is falling vertically into an open railroad car which moves along a horizontal track at a constant speed. The engine must exert an extra force on the car as the water collects in it (the water is initially stationary, and must be brought up to the speed of the train).

v

F

Calculate this extra force if:

v = 20 m/s

The water collects in the car at the rate of 6 kg per minute


Solution

Plan: The momentum of the car increases as it gains mass (water). Use Newton’s second law to find F.

v

F

v is constant, and dm/dt is 6 kg/min or 0.1 kg/s (change to SI units!),

so F = (0.1 kg/s) (20 m/s) = 2.0 N

F and p are vectors; we get the horizontal force from the rate of increase of the horizontal component of momentum.

dtd

mdtdm

dtmd

dtd v

vvp

F )(


ptotal = p1 + p2 + ... = m1v1 + m2v2 + ...

The total momentum of a system of particles is the vector sum of the momenta of the individual particles:

Since we are adding vectors, we can break this up into components so that:

px,Tot = p1x + p2x + ….

Etc.


Example 4

• A particle of mass 2kg is moving with a velocity of 5m/s and an angle of 45o to the horizontal. Determine the components of its momentum.


Newton’s 3rd Law and Momentum Conservation

Two particles interact:

Newton’s 3rd Law: F21 = -F12

The momentum changes are equal and opposite; the total momentum:

p = p1 + p2=0

doesn’t change.

The fine print: Only internal forces act. External forces would transfer momentum into or out of the system.

eg: particles moving through a cloud of gas

p1= F12 t p2= F21 t

F12

F21 = -F12

m1

m2


Conservation of momentum simply says that the initial and final momenta are equal:

pi=pf

Since momentum is a vector, we can also express it in terms of the components. These are independently

conserved:

pix=pfx piy=pfy piz=pfz


Concept Quiz

A subatomic particle may decay into two (or more) different particles. If the total momentum before is zero before the decay before, what is the total after?

a) 0 kg m/s

b) depends on the masses

c) depends on the final velocities

d) depends on b) and c) together

Before

After


10 min rest


Collisions

• Conservation of Momentum • Elastic and inelastic collisions


Collisions

A collision is a brief interaction between two (or more) objects. We use the word “collision” when the interaction time Δt is short relative to the rest of the motion.

During a collision, the objects exert equal and opposite forces on each other. We assume these “internal” forces are much larger than any external forces on the system.

We can ignore external forces if we compare velocities just before and just after the collision, and if the interaction force is much larger than any external force.

m1 m2

v1,i v2,i

v1,f v2,f

F1

F2 = -F1


Elastic and Inelastic Collisions

Momentum is conserved in collisions. Kinetic energy is sometimes conserved; it depends on the nature of the interaction force.

A collision is called elastic if the total kinetic energy is the same before and after the collision. If the interaction force is conservative, a collision between particles will be elastic (eg: billiard balls).

If kinetic energy is lost (converted to other forms of energy), the collision is called inelastic (eg: tennis ball and a wall).

A completely inelastic collision is one in which the two colliding objects stick together after the collision (eg: alien slime and a spaceship). Kinetic energy is lost in this collision.


If there are no external forces, then the total momentum is conserved:

p1,i + p2,i = p1,f + p2,f

This is a vector equation. It applies to each component of p separately.

m1 m2

v1,i v2,i

v1,f v2,f


Elastic Collisions

In one dimension (all motion along the x-axis):

1) Momentum is conserved:

ffii 22112211 vmvmvmvm

2222

12112

12222

12112

1ffii

vmvmvmvm

In one dimension, the velocities are represented by positive or negative numbers to indicate direction.

2) Kinetic Energy is conserved:

We can solve for two variables if the other four are known.


One useful result: for elastic collisions, the magnitude of the relative velocity is the same before and after the collision:

|v1,i – v2,i | = |v1,f – v2,f |(This is true for elastic collisions in 2 and 3 dimensions as well).

An important case is a particle directed at a stationary target (v2,i = 0):

• Equal masses: If m1 = m2, then v1,f will be zero (1-D).• If m1 < m2, then the incident particle recoils in the opposite direction.• If m1 > m2, then both particles will move “forward” after the collision.

before after


Elastic collisions, stationary target (v2,i = 0):

Two limiting cases:1) If m1 << m2 , the incident particle

rebounds with nearly its original speed.

v1

-v1

2) If m1 >> m2 , the target particle moves away with (nearly) twice the original speed of the incident particle.

v1 v1 2v1


Concept Quiz

A tennis ball is placed on top of a basketball and both are dropped.The basketball hits the ground at speed v0. What is the maximum speed at which the tennis ball can bounce upward from the basketball? (For “maximum” speed, assume the basketball is much more massive than the tennis ball, and both are elastic).

a) v0

b) 2v0

c) 3v0 ?

v0

v0


Example

An angry 60.0 kg physicist standing on a frozen lake throws a 0.5 kg stone to the east with a speed of 24.0 m/s.

Find the recoil velocity of the angry physicist.


Example – inelastic collision:

A neutron, with mass m = 1 amu (atomic mass unit), travelling at speed v0, strikes a stationary deuterium nucleus (mass 2 amu), and sticks to it, forming a nucleus of tritium. What is the final speed of the tritium nucleus?


2 kg 4 kg

6 m/s 5 m/s

An elastic collision:

Two carts moving toward each other collide and bounce back. If cart 1 bounces back with v=2m/s, what is the final speed of cart 2 ?


Example

A bullet of mass m is shot into a block of mass M that is at rest on the edge of a table. If the bullet embeds in the block, determine the velocity of the bullet if they land a distance of x from the base of the table, which has a height of h.


10 min rest


Impulse

Fp dtd

Newton #2: , or dp = F dt

(Extra) In general (force not constant), we integrate:

For a constant force, p = F t . The vector quantity F t iscalled the Impulse:

J = FΔt = Δp

dtJ F

(change in p) = (total impulse from external forces)

(Newton’s Second Law again)

where the integral gives the area under a curve…


F

ftt it

Area = )( tF

F

ftt it

Impulse is the area under the curve. The average forceis the constant force which would give the same impulse.

Compare with work: W = F x ; so the work-energy theorem (derived from Newton #2) is K = F x.

The impulse – momentum theorem, J=FΔt=Δp, tells us thatwe do not need to know the details of the force/interaction,we only need to know the area under the curve=integralof the force.


Quiz

A 100g rubber and a 100g clay ball are thrown at a wall with equal speed. The rubber ball bounces back while the clay ball sticks to the wall.Which ball exerts a larger impulse on the wall ?

a) the clay bass b/c it sticks

b) the rubber ball b/c it bounces

c) they are equal because the have equal momenta

d) neither exerts an impulse b/c the wall doesn’t move


Example 1

A golf ball is launched with a velocity of 44 m/s. The ball has a mass of 50g. Determine the average force on the ball during the collision with the club, if the collision lasted 0.01 s.


Example 2

Use the impulse-momentum theorem to find how long a falling object takes to increase its speed from 5.5m/s to 10.4m/s.


Example 3

• A 150 g baseball is thrown with a speed of 20m/s. It is hit straight back toward the pitcher at a speed of 40m/s. The interaction force is shown by the graph:

What is the maximum force Fmax that the bat exerts on the ball ?

Fmax

6mst

F

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Physics 1D03 - Lecture 25 TEST 1 Tuesday May 19 9:30 am CNH-104 Kinematics, Dynamics & Momentum