Momentum and Collision

Chapter 6

Momentum and Impulse

Linear Momentum

Momentum is defined as mass times velocity.

Momentum is represented by the symbol p, and is a vector quantity.

p = mvmomentum = mass velocity

Momentum

Momentum has the dimensions of mass x length/time (kg x m/s)

When do you use the word momentum? “gaining momentum” or “picking up

speed” Coasting on a bike going down a hill You accelerate and velocity increases

with time.

Bowling ball or basketball?

Picture two lanes at a bowling alley. One with a bowling ball the other with a

basketball going at the same speed. Which will exert more force on the pins? Why? More momentum

Example

A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck?

Solution

p = mv (2250 kg)(25 m/s east) 5.6 x 104 kg x m/s to the east

Your Turn A deer with a mass of 146 kg is running

head-on toward you at a speed of 17 m/s. You are going north find the momentum of the deer.

A 21 kg child on a 5.9 kg bike is riding with a velocity of 4.5 m/s to the northwest. What is the total momentum of the child

and the bike together? What is the momentum of the child? What is the momentum of the bike?

Linear Momentum

Impulse The product of the force and the time over which

the force acts on an object is called impulse.

The impulse-momentum theorem states that when a net force is applied to an object over a certain time interval, the force will cause a change in the object’s momentum.

F∆t = ∆p = mvf – mvi

force time interval = change in momentum

Linear Momentum

Stopping times and distances depend on the impulse-momentum theorem.

Force is reduced when the time interval of an impact is increased.

Example

A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the force exerted on the car during the collision.

Solution

F∆t = ∆p = mvf – mvi

F = mvf – mvi

∆t F = (1400kg)(0 m/s) – (1400 kg)(-15 m/s)

0.30 s F = 7.0 x 104 N to the east

Your Turn II A 0.50 kg football is thrown with a velocity of 15

m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.020 s. What is the force exerted on the ball by the receiver?

An 82 kg man drops from rest on a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What is the net force on the diver as he is brought to rest?

A 0.40 kg soccer ball approaches a player horizontally with a velocity of 18 m/s to the north. The player strikes the ball and cause it to move in the opposite direction with a velocity of 22 m/s. What is the impulse delivered to the ball by the player?

Impulse-Momentum Theorem

Stopping times and distances depend on the impulse-momentum theorem.

Highway safety engineers use the impulse-momentum theorem to determine stopping distances and safe following distances for cars and trucks.

Example

A 2240 kg car traveling to the west slows down uniformly from 20.0 m/s to 5.00 m/s. How long does it take the car to decelerate if the force on the car is 8410 N to the east? How far does the car travel during deceleration?

Solution

Givens: Mass = 2240 kg vi = 20.0 m/s to the west = - 20.0 m/s vf = 5.00 m/s to the west = - 5.00 m/s F = 8410 N to the east

Equation to use: F∆t = ∆p ∆t = ∆p / F ∆t = mvf – mvi / F

Plug and Chug

∆t = mvf – mvi / F∆t = (2240 kg)(-5.00 m/s)-(2240 kg)(-20.0 m/s)

8410 kg x m/s2

∆t = 4.00 s∆x = ½ (vi + vf)∆t

∆x = ½ (-20.0 m/s – 5.00 m/s)(4.00 s)∆x = - 50.0 m = 50.0 m to the west

Your Turn III

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20.0 m/s by a 6250 N braking force acting opposite the car’s motion. Use the impulse-momentum theorem to answer the following questions: What is the car’s velocity after 2.50 s? How far does the car move during 2.50 s? How long does it take the car to come to a

complete stop?

Impulse-Momentum Theorem

Force is reduced when the time interval of an impact is increased.

Examples Nets or giant air mattresses used to catch people.

Page 203 in your book shows an image of a girl being tossed in the air and caught in a blanket.

The blanket “gives way” and extends the time of collision to change the momentum over a longer period of time.

Consider the egg on the next slide and explaining what is happening

Impulse-Momentum Theorem

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Page 204 Physics 1-3 Honors 1-5

Conservation of Momentum

Momentum is Conserved

So far we only have considered the momentum of only one object at a time.

Now we will look at two or more objects interacting with each other.

Picture this. . . You are playing pool. You strike the cue ball it

hits the 8 ball. The 8 ball had no momentum before they collided.

During the collision the cue ball loses momentum and the 8 ball gains momentum.

The momentum the cue ball loses is the same amount that the 8 ball gained.

Momentum is Conserved

The Law of Conservation of Momentum:

The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between the objects.

m1v1,i + m2v2,i = m1v1,f + m2v2,f

total initial momentum = total final momentum

Momentum is Conserved

The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between the objects.

Go back to the pool table example. The cue ball and the 8 ball do not have a constant momentum, but the total momentum is constant.

Momentum is Conserved

Consider objects pushing away from each other.

You jump up. You push off the Earth. Take you mv.

Let’s say 60 kg m/s upward. That means that the Earth must have a corresponding momentum of 60 kg m/s downward. However, the Earth has an enormous mass which means its velocity is very small.

Momentum is Conserved

Picture this . . . Two people on skates facing one

another. They push away from one another. Initially, they are both at rest with a momentum of 0. When they push away, they move in opposite directions with equal but opposite momentum, so that the total momentum is 0.

Example A 76 kg boater, initially at rest in a A 76 kg boater, initially at rest in a

stationary 45 kg boat, steps out of the stationary 45 kg boat, steps out of the boat and onto the dock. If the boater boat and onto the dock. If the boater moves out of the boat with a velocity of moves out of the boat with a velocity of 2.5 m/s to the right, what is the final 2.5 m/s to the right, what is the final velocity of the boat?velocity of the boat?

Solution

Given: m1 = 76 kg m2 = 45 kg

v1,i = 0 v2,i = 0

v1,f = 2.5 m/s to the right

Unknown: v2,f = ?

Solution

Choose an equation or situation: Because the total momentum of an isolated system remains constant, the total initial momentum of the boater and the boat will be equal to the total final momentum of the boater and the boat.

m1v1,i + m2v2,i = m1v1,f + m2v2,f

Solution

Because the boater and the boat are initially at rest, the total initial momentum of the system is equal to zero. Therefore, the final momentum of the system must also be equal to zero.

m1v1,f + m2v2,f = 0

Rearrange the equation to solve for the final velocity of the boat.

2 2, 1 1,

12, 1,

2

–

–

f f

f f

m m

m

m

v v

v v

Solution

Substitute the values into the equation and solve:

2,

2,

76 kg– 2.5 m/s to the right

45 kg

–4.2 m/s to the right

f

f

v

v

Solution

The negative sign for v2,f indicates that the boat is moving to the left, in the direction opposite the motion of the boater. Therefore,

v2,f = 4.2 m/s to the left

Your Turn IV A 63.0 kg astronaut is on a space walk when

the tether line to the shuttle breaks. The astronaut is able to throw a spare 10.0 kg oxygen tank in a direction away from the shuttle with a speed of 12.0 m/s, propelling the astronaut back to the shuttle. Assuming that the astronaut starts from rest with respect to the shuttle, find the astronaut’s final speed with respect to the shuttle after the tank is thrown.

An 85.0 kg fisherman jumps from a dock into a 135.0 kg rowboat at rest on the west side of the dock. If the velocity of the fisherman is 4.30 m/s to the west as he leaves the dock, what is the final velocity of the fisherman and the boat?

Your Turn IV Each croquet ball in a set has a mass of 0.50

kg. The green ball, traveling at 12.0 m/s, strikes the red ball, which is at rest. Assuming that the croquet ball slide on a frictionless surface and all collisions are head-on, find the final speed of the red ball in each of the following situations: The green ball stops moving after it strikes the red

ball. The green ball continues moving after the collision

at 2.4 m/s in the same direction.

Momentum is Conserved

Newton’s third law leads to conservation of momentum

During the collision, the force exerted on each bumper car causes a change in momentum for each car.

The total momentum is the same before and after the collision.

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Page 211 Physics 1-3 Honors 1-4

Elastic and Inelastic Collisions

Collisions

Perfectly inelastic collision A collision in which two objects stick together after

colliding and move together as one mass is called a perfectly inelastic collision.

Example: The collision between two football players during a tackle.

Conservation of momentum for a perfectly inelastic collision:

m1v1,i + m2v2,i = (m1 + m2)vf

total initial momentum = total final momentum

Example

A 1850 kg luxury sedan stopped at a traffic light is struck from behind by a compact car with a mass of 975 kg. The two cars become entangled as a result of the collision. If the compact car was moving with a velocity of 22.0 m/s to the north before the collision, what is the velocity of the entangled mass after the collision?

Given: m1 = 1850 kg m2 = 975 kgv1,i = 0 m/s v2,i = 22.0 m/s north

Unknown: vf

Solution

Choose your equation:

m1v1,i + m2v2,i = (m1 + m2)vf

vf = m1v1,i + m2v2,i

(m1 + m2)

vf = (1850 kg)(0 m/s) + (975 kg)(22.0 m/s)

(1850 kg + 975 kg)

vf = 7.59 m/s north

Your Turn V A 1500 kg car traveling at 15.0 m/s to the south collides

with a 4500 kg truck that is initially at rest at a stoplight. The car and truck stick together and move together after the collision. What is the final velocity of the two-vehicle mass?

You are shopping at Publix and toss a 9.0 kg bag of rice into a stationary 18.0 kg shopping cart. The bag hits the cart with a horizontal speed of 5.5 m/s toward the front of the cart. What is the final speed of the cart and the bag?

A 47.4 kg student runs down the sidewalk and jumps with a horizontal velocity of of 4.20 m/s onto a stationary skateboard. The student and the skateboard move down the sidewalk with a speed of 3.95 m/s. Find the following: The mass of the skateboard How fast the student would have to jump to have a

final speed of 5.00 m/s

Kinetic Energy in Inelastic Collisions In an inelastic collision the total kinetic energy

does not remain constant when the objects collide and stick together.

Some energy is converted into sound energy and internal energy as the objects deform during the collision.

Elastic in physics refers to a material that when work is done to deform the material during a collision the same amount of work is done to return the material to its original shape.

Inelastic material does not return to its original shape and therefore some energy is converted to sound or heat.

Example

Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 0.500 kg and an initial velocity of 4.00 m/s to the right. The second ball has a mass of 0.250 kg and an initial velocity of 3.00 m/s to the left. What is the decrease in kinetic energy during the collision?

Solution

Given: m1= 0.500 kg m2 = 0.250 kg

v1,i = 4.00 m/s to the right

v1,i = +4.00 m/s

v2,i = 3.00 m/s to the left

v2,i = –3.00 m/s

Unknown: ∆KE = ?

Solution

The change in kinetic energy is simply the initial kinetic energy subtracted from the final kinetic energy.

∆KE = KEi – KEf

Determine both the initial and final kinetic energy.

2 21, 2, 1 1, 2 2,

21, 2, 1 2

1 1Initial:

2 21

Final: 2

i i i i i

f f f f

KE KE KE m v m v

KE KE KE m m v

Solution

Use the equation for a perfectly inelastic collision to calculate the final velocity.

v

f

m1v

1,i m

2v

2,i

m1 m

2

(0.500 kg)(4.00 m/s) (0.250 kg)(–3.00 m/s)

0.500 kg 0.250 kg

1.67 m/s to the right

fv

fv

Solution

Next calculate the initial and final kinetic energy.

2 2

2

1 10.500 kg 4.00 m/s 0.250 kg –3.00 m/s 5.12 J

2 21

0.500 kg 0.250 kg 1.67 m/s 1.05 J2

i

f

KE

KE

Solution

Finally, calculate the change in kinetic energy.

– 1.05 J – 5.12 J

–4.07 J

f iKE KE KE

KE

Your Turn VI A 0.25 kg arrow with a velocity of 12 m/s to the west

strikes and pierces the center of a 6.8 kg target. What is the final velocity of the combined mass? What is the decrease in kinetic energy during the collision?

A clay ball with a mass of 0.35 kg hits another 0.35 kg ball at rest, and the two stick together. The first ball has an initial speed of 4.2 m/s. What is the final speed of both balls? Calculate the decrease in kinetic energy that occurs during the collision.

A 56 kg ice skater traveling at 4.0 m/s to the north meets and joins hands with a 65 kg skater traveling at 12.0 m/s in the opposite direction. Without rotating the two skaters continue skating together. What is the final velocity of the two skaters and what is the decrease in kinetic energy durning the collision?

Elastic Collisions

A collision in which the total momentum and the total kinetic energy are conserved is called an elastic collision.

Elastic means that after a collision the objects remain separated.

Two objects collide and return to their original shapes with no loss of total kinetic energy. After the collision the two objects move separately.

Both the total momentum and total kinetic energy are conserved.

Real Collisions

Most collisions are not perfectly inelastic (they don’t stick together and move as one)

Most collisions are not elastic. Even nearly elastic collisions result in

some decrease of kinetic energy. A football deforms when kicked A sound is produced (sound signifies a

decrease in kinetic energy)

For all intensive purposes

We will consider all collisions in which objects do not stick together to be elastic collisions.

Therefore, total momentum and total kinetic energy will stay the same before and after the collision.

Kinetic Energy is Conserved in Elastic Collisions

B

PB

A

PA

∆PA= F ∆t

A B

∆PB= -F ∆t

B

PB

A

PA

Elastic Collisions

The total momentum is always constant throughout the collision. In addition, if the collision is perfectly elastic, the value of the total kinetic energy after the collision is equal to the value before the collision.

m1v

1,i m

2v

2,im

1v

1,f m

2v

2,f

1

2m

1v

1,i2

1

2m

2v

2,i2

1

2m

1v

1,f2

1

2m

2v

2,f2

Example

A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head-on collision with a 0.030 kg shooter marble moving to the left at 0.180 m/s. After the collision, the smaller marble moves to the left at 0.315 m/s. Assume that neither marble rotates before or after the collision and that both marbles are moving on a frictionless surface. What is the velocity of the 0.030 kg marble after the collision?

Solution

Given: m1 = 0.015 kg m2 = 0.030 kg

v1,i = 0.225 m/s to the right, v1,i = +0.225 m/s

v2,i = 0.180 m/s to the left, v2,i = –0.180 m/s

v1,f = 0.315 m/s to the left, v1,i = –0.315 m/s

Unknown: v2,f = ?

Solution

2 , 1 , 2 , 1 ,

1 2 , 1 ,,

2

–

–

m m m m

m m m

m

2 f 1i 2 i 1 f

1,i 2 i 1 f2 f

v v v v

v v vv

2,

–3 –3 –3

2,

–3

2, –2

–2,

0.015 kg 0.225 m/s 0.030 kg –0.180 m/s – 0.015 kg –0.315 m/s

0.030 kg

3.4 10 kg m/s –5.4 10 kg m/s – –4.7 10 kg m/s

0.030 kg

2.7 10 kg m/s

3.0 10 kg

9.0 10 m/s to the right

f

f

f

v

v

v

2 fv

Solution

2 2 2 21 1, 2 2, 1 1, 2 2,

2 2

–4 2 2 –4

2 2

–4 2 2 –4

1 1 1 1

2 2 2 21 1

0.015 kg 0.225 m/s 0.030 kg –0.180 m/s2 2

8.7 10 kg m /s 8.7 10 J

1 10.015 kg 0.315 m/s 0.030 kg 0.090 m/s

2 2

8.7 10 kg m /s 8.7 10 J

i i f f

i

f

m v m v m v m v

KE

KE

Your Turn VII Two billiard balls, each with a mass of 0.35 kg, strike each

other head-on. One ball is initially moving left at 4.1 m/s and ends up moving right at 3.5 m/s. The second ball is initially moving to the right at 3.5 m/s. Assume that neither ball rotates before or after the collision and that both balls are moving on a frictionless surface. Predict the final velocity of the second ball.

A 0.015 kg marble sliding to the right at 12.5 m/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg marble moving to the left at 18.0 m/s. After the collision, the first marble moves to the left at 18.0 m/s. Find the velocity of the second marble after the collision. Verify your answer by calculating the total kinetic energy before and after the collision.

Your Turn VII (continued)

A 16.0 kg canoe moving to the left at 12.5 m/s makes an elastic head-on collision with a 14.0 kg raft moving to the right at 16.0 m/s. After the collision, the raft moves to the left at 14.4 m/s. Disregard and effects of the water. Find the velocity of the canoe after the collision. Calculate the total kinetic energy before and after the collision.

Types of Collisions

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Pg 220 Physics 1-4 Honors 1-5