Chapter 6

70
Chapter 6 Momentum and Collisions

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Chapter 6. Momentum and Collisions. Momentum and Impulse. Objectives: 1. Compare the momentum of different moving objects. 2. Compare the momentum of the same object moving with different velocities. 3. Identify some examples of change in the momentum of an object. - PowerPoint PPT Presentation

Transcript of Chapter 6

Page 1: Chapter 6

Chapter 6

Momentum and Collisions

Page 2: Chapter 6

Momentum and Impulse

Objectives:1. Compare the momentum of different moving

objects.2. Compare the momentum of the same object

moving with different velocities.3. Identify some examples of change in the

momentum of an object.4. Describe changes in momentum in terms of force

and time.

Page 3: Chapter 6

Linear Momentum

Momentum (p) is a quantity defined as the product of the mass and velocity of an object.

p = mv

momentum = mass x velocitySI units of kg▪m/sThe faster an object moves, the more momentum it

has and the more difficult it is to stop the object.

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Concept Check

A deer with a mass of 146 kg is running head-on towards you with a speed of 17 m/s. You are going

north. Find the momentum of the deer.

Page 5: Chapter 6

Concept Check

2.5 x 103 kg▪m/s south

Page 6: Chapter 6

Concept Check

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

a. What is the total momentum of the child and the bike together?

b. What is the momentum of the child?c. What is the momentum of the bike?

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Concept Check

a. 1.2 x 102 kg▪m/s northwestb. 94 kg▪m/s northwestc. 27 kg▪m/s northwest

Page 8: Chapter 6

Concept Check

What velocity must a 1210 kg car have in order to have a momentum of 5.6 x 104 kg▪m/s east?

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Concept Check

46 m/s east.

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Changes in Momentum

A change in momentum requires force and time.It requires more force to stop an object with a

greater momentum than to stop an object with a smaller momentum.

When Newton first expressed his second law, he wrote: F = ∆p

∆tforce = change in momentum

time interval

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Impulse-Momentum TheoremWe can rearrange Newton's second law such that

F∆t = ∆p or F∆t = ∆p = mvf – mvi

force x time interval = change in momentum

Thus, a small force acting over a long time can cause the same change in momentum as a large force acting over a short time.

Assume all forces are constant unless stated otherwise.

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Changes in Momentum● The expression F∆t is the impulse of the force over the time

interval.– The product of the force and the time over which the force

acts on an object.

Page 13: Chapter 6

Concept Check

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?

Page 14: Chapter 6

Concept Check

3.8 x 102 N to the left

Page 15: Chapter 6

Concept Check

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?

Page 16: Chapter 6

Concept Check

1.1 x 103 N upward

Page 17: Chapter 6

Concept Check

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 causes it to move in the opposite direction with a velocity of 22 m/s.

What impulse was delivered to the ball by the player?

Page 18: Chapter 6

Concept Check

16 kg▪m/s to the south

Page 19: Chapter 6

Concept Check

A 0.50 kg object is at rest. A 3.00 N force to the right acts on the object during a time interval of

1.50 s.a. What is the velocity of the object at the end of

this time interval?b. At the end of this interval, a constant force of 4.00 N to the left is applied for 3.00 s. What is the

velocity at the end of the 3.00 s?

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Concept Check

a. 9.0 m/s to the right.b. 15 m/s to the left.

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Stopping Times and DistancesStopping distances and safe following distances are

determined using the impulse-momentum theorem.Stopping time is determined by solving the impulse-

momentum theorem for time.∆t = ∆ p = mvf-mvi

F FStopping distance can then be determined by the

kinematic equation∆x = ½ (vi + vf)∆t

Page 22: Chapter 6

Concept Check

If the maximum coefficient of kinetic friction between a 2300 kg car and a road is 0.50, what is

the minimum stopping distance for a car entering a skid at 29 m/s?

Page 23: Chapter 6

Concept Check

86 m

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Concept Check

How long would it take a 2240 kg car to come to a full stop from 20.0 m/s to the west if the force on

the car 8410 N to the east? How far would the car go before stopping? Assume a constant

acceleration.

Page 25: Chapter 6

Concept Check

5.33 s; 53.3 m to the west

Page 26: Chapter 6

Concept Check

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:a. What is the car's velocity after 2.50 s?

b. How far does the car move during 2.50 s?c. How long does it take the car to come to a

complete stop?

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Concept Check

a. 14 m/s northb. 42 m north

c. 8.0 s

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Concept Check

Assume the car in problem 2 has a mass of 3250 kg.a. How much force would be required to cause the

same acceleration as in item 1? Use the impulse-momentum theorem.

b. How far would the car move before stopping?

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Concept Check

a. 1.22 x 104 N to the eastb. 53.3 m to the west

Page 30: Chapter 6

More Time = Less Force● The impulse-momentum theorem is also used to design

safety equipment that reduces the force acting on the human body during collisions. (Nets, air mattresses, air bags, etc.)

● When an egg is dropped on the floor, it breaks.● When an egg is dropped on a pillow, it doesn't break.

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Conservation of Momentum

Objectives:1. Describe the interaction between two objects in

terms of the change in momentum of each object.2. Compare the total momentum of two objects

before and after they interact.3. State the law of conservation of momentum.4. Predict the final velocities of objects after

collisions, given the initial velocities.

Page 32: Chapter 6

Conservation of Momentum

Momentum is a conserved quantity.When 2 objects collide, the initial momentum is

equal in magnitude to the final momentum.m1v1,i + m2v2,i = m1v1,f + m2v2,f

For an isolated system, the law of the conservation of momentum can be stated asThe total momentum of all objects interacting

with one another remains constant regardless of the nature of the forces between the objects.

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Collisions● Momentum is conserved in collisions.● Consider two billiard balls: one is moving at a constant

velocity and the other is at rest.– When the moving ball collides with the nonmoving ball,

the moving ball slows down and the ball that was initially at rest begins to move.

– The individual momentums of the balls have changed, but the total momentum of the system remains the same (IN THE ABSENCE OF FRICTION!!!).

● In most problems, disregard friction

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Moving Apart● Conservation of momentum also occurs when objects move

away from each other.● When you are sitting in your favorite class (Physics of

course), your momentum is zero.● When the bell rings and you jump up and run like heck to

the door, you have momentum, so how is momentum conserved?

● You have to include the Earth in your analysis of the situation.

● If you jump up at an initial momentum of 25 kg▪m/s upward

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Moving Apart● The Earth is moving away from you at a momentum of 25

kg▪m/s downward, so the total momentum remains the same (zero).

Page 36: Chapter 6

Concept Check

A 63.0 kg astronaut is on a spacewalk 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.

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Concept Check

1.90 m/s

Page 38: Chapter 6

Concept Check

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?

Page 39: Chapter 6

Concept Check

1.66 m/s west

Page 40: Chapter 6

Concept Check

Each croquet ball in a set has a mass of 0.50 kg. The green ball, traveling at 12.0 m/s, strikes the

blue ball, which is at rest. Assuming that the balls slide on a frictionless surface and all collisions are

head-on, find the final speed of the blue ball in each of the following situations:

a. The green ball stops moving after it strikes the blue ball.

b. The green ball continues moving after the collision at 2.4 m/s in the same direction.

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Concept Check

a. 12.0 m/sb. 9.6 m/s

Page 42: Chapter 6

Concept Check

A boy on a 2.0 kg skateboard initially at rest tosses a 8.0 kg jug of water in the forward direction. If

the jug has a speed of 3.0 m/s relative to the ground and the boy and the skateboard move in the opposite direction at 0.60 m/s, find the boy's mass.

Page 43: Chapter 6

Concept Check

38 kg

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Forces In Real Collisions Are Not Constant

We treat the forces involved in a collision as though they were constant, but in real-life situations the forces will vary throughout the course of a collision.

At any instant during a collision, the forces of the two colliding objects will be equal and opposite.

When solving impulse problems, use the average force over the time interval as the value for force.

Page 45: Chapter 6

Elastic and Inelastic Collisions

Objectives:1. Identify different types of collisions.2. Determine the changes in kinetic energy during

perfectly inelastic collisions.3. Compare conservation of momentum and

conservation of kinetic energy in perfectly inelastic and elastic collisions.

4. Find the final velocity of an object in perfectly inelastic and elastic collisions.

Page 46: Chapter 6

Perfectly Inelastic Collisions

When two objects collide and move together as one after the collision (stick together), the collision is a perfectly inelastic collision.

The total momentum of the the two objects after the collision is equal to the total momentum of the two objects before the collision.

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

Page 47: Chapter 6
Page 48: Chapter 6

Concept Check

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?

Page 49: Chapter 6

Concept Check

3.8 m/s to the south

Page 50: Chapter 6

Concept Check

A grocery shopper tosses a 9.0 kg bag of rice into a stationary 18.0 kg grocery 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 bag?

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Concept Check

1.8 m/s

Page 52: Chapter 6

Concept Check

A 1.50 x 104 kg railroad car moving at 7.00 m/s to the north collides with and sticks to another

railroad car of the same mass that is moving in the same direction at 1.50 m/s. What is the velocity of

the joined cars after the collision?

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Concept Check

4.25 m/s to the north

Page 54: Chapter 6

Concept Check

A dry cleaner throws a 22 kg bag of laundry onto a stationary 9.0 kg cart. The cart and laundry bag begins moving at 3.0 m/s to the right. Find the velocity of the laundry bag before the collision.

Page 55: Chapter 6

Concept Check

4.2 m/s to the right

Page 56: Chapter 6

Concept Check

A 47.4 kg student runs down the sidewalk and jumps with a horizontal speed of 4.20 m/s onto a

stationary skateboard. The student and skateboard move down the sidewalk with a speed of 3.95 m/s.

Find the following:a. the mass of the skateboard

b. how fast the student would have to jump to have a final speed of 5.00 m/s.

Page 57: Chapter 6

Concept Check

3.00 kg5.32 m/s

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Kinetic Energy

Kinetic energy is not conserved in inelastic collisions.

Some of the kinetic energy is converted to sound energy and internal energy as objects are deformed during the collision.

The decrease in kinetic energy can be calculated by using the formula for kinetic energy.

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Concept Check

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.a. What is the final velocity of the combined mass?b. What is the decrease in kinetic energy during the

collision?

Page 60: Chapter 6

Concept Check

a. 0.43 m/s to the westb. 17 J

Page 61: Chapter 6

Concept Check

During practice, a student kicks a 0.40 kg soccer ball with a velocity of 8.5 m/s to the south into a

0.15 kg bucket lying on its side. The bucket travels with the ball after the collision.

a. What is the final velocity of the combined mass?b. What is the decrease in kinetic energy during the

collision?

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Concept Check

a. 6.2 m/s to the southb. 3 J

Page 63: Chapter 6

Concept Check

A 56 kg ice skater travels 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

with joined hands.a. What is the final velocity of the two skaters?

b. What is the decrease in kinetic energy during the collision?

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Concept Check

a. 4.6 m/s to the southb. 3.9 x 103 J

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Elastic Collisions● In an elastic collision two objects collide and return to their

original shapes with no loss of kinetic energy (they don't stick together.)– Both momentum and kinetic energy are

conserved.● In the real world, most collisions are not perfectly inelastic or

elastic but somewhere in between.● We consider all collisions in which the objects do not stick

together to be elastic.

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Elastic Collisions

The total momentum is always constant throughout the collision.

The total kinetic energy is also conserved.m1v1,i +m2 v2,i = m1 v1,f + m2 v2.f

 ½ m1 v1,i

2 + ½ m2 v2,i2 = ½ m1 v1,f 2+ ½ m2 v2,f 2

Page 67: Chapter 6

Concept Check

A 0.015 kg marble sliding to the right at 22.5 cm/s on a frictionless surface makes an elastic head-on

collision with a 0.015 kg marble moving to the left at 18.0 cm/s. After the collision, the first marble

moves to the left at 18.0 cm/s.a. Find the velocity of the second marble after the

collision.b. Verify your answer by calculating the total

kinetic energy before and after the collision.

Page 68: Chapter 6

Concept Check

a. 22.5 cm/s to the rightb. KEi = 6.2 x10-4 J = KEf

Page 69: Chapter 6

Concept Check

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 any effects of the water.a. Find the velocity of the canoe after the collision.

b. Verify your answer by calculating the total kinetic energy before and after the collision.

Page 70: Chapter 6

Concept Check

a. 14.1 m/s to the rightb. 3.04 x 103 J