Chapter 7

Linear Momentum

Objectives: The student will be able to:

• Perform several investigations in order to make conclusions about the total momentum of a system.

• Analyze a problem and choose a system to determine if the forces are internal or external to that system.

• State the law of conservation of momentum and use it to solve one-dimensional explosion and collision problems using an equation.

Inquiry for Total Momentum Before and After a Collision/Explosion

• Design an experiment to demonstrate the effect of a collision/explosion on total momentum of the objects before and after using the same mass for each cart and then for a second experiment change one of the cart’s mass by adding mass to it.

• Make a prediction on the effect of some condition on the total momentum before and after a collision.

• Materials– Dynamics cart with spring bumper or plunger– Meter stick– Stop watch– Masses– Large white boards

Inquiry for Total Momentum Before and After a Collision/Explosion

• Your group will present using the whiteboards your design and findings.

• Did anything unusual happen? Were there any special insights you gained and want to make a note of? Were there any changes made to your procedure? What new questions arose?

• Were any of the results NOT what you expected?• Which of your pre-lab ideas have you decided are now incorrect? Why?• Did the data support your original hypothesis?• If not, what hypothesis does the data support?

Evaluation• Individually, you will make a claim about each investigation with supporting

evidence, and then explain how the conservation of momentum can be applied to these investigations.

• What conclusion(s) did you reach due to the results of this experiment?• What evidence supports your conclusion(s)?• Are your results reliable? How did you compensate for sources of error in the

experiment?• Can you test the predictions? If so, do results agree with your conclusion(s)?• What new problems/questions does the experiment bring up?RELATE• What are some possible applications of your conclusions to the real world

situations?• Do the results of your experiment fit any laws/theories of physics?

• This will be due .

7-2 Conservation of MomentumDuring a collision, measurements show that the total momentum does not change:

(7-3)

7-2 Conservation of Momentum

More formally, the law of conservation of momentum states:

The total momentum of an isolated system of objects remains constant.

7-2 Conservation of Momentum

Example 7-3: Railroad cars collide: momentum conserved.

A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the collision, what is their common speed immediately after the collision?

7-2 Conservation of Momentum

Momentum conservation works for a rocket as long as we consider the rocket and its fuel to be one system, and account for the mass loss of the rocket.

7-2 Conservation of Momentum

Example 7-4: Rifle recoil.

Calculate the recoil velocity of a 5.0-kg rifle that shoots a 0.020-kg bullet at a speed of 620 m/s.

Conceptual Example 7-5: Falling on or off a sled.

(a) An empty sled is sliding on frictionless ice when Susan drops vertically from a tree above onto the sled. When she lands, does the sled speed up, slow down, or keep the same speed?

(b) Later: Susan falls sideways off the sled. When she drops off, does the sled speed up, slow down, or keep the same speed?

Conservation of Momentum in 1-DWhenever two objects collide (or when they exert forces on each other without colliding, such as gravity) momentum of the system (both objects together) is conserved. This mean the total momentum of the objects is the same before and after the collision.

before: p = m1 v1 - m2 v2

after: p = - m1 va + m2 vb

m1 m2

v1v2

(Choosing right as the +

direction, m2 has - momentum.)

m1 m2

va vb

m1 v1 - m2 v2 = - m1 va + m2 vb

Directions after a collision

On the last slide the boxes were drawn going in the opposite direction after colliding. This isn’t always the case. For example, when a bat hits a ball, the ball changes direction, but the bat doesn’t. It doesn’t really matter, though, which way we draw the velocity vectors in “after” picture. If we solved the conservation of momentum equation (red box) for vb and got a negative answer, it would mean that m2 was still moving to the left after the collision. As long as we interpret our answers correctly, it matters not how the velocity vectors are drawn.

m1 v1 - m2 v2 = - m1 va + m2 vb

m1 m2

v1v2

m1 m2

va vb

Proof of Conservation of MomentumThe proof is based on Newton’s 3rd Law. Whenever two objects collide (or exert forces on each other from a distance), the forces involved are an action-reaction pair, equal in strength, opposite in direction. This means the net force on the system (the two objects together) is zero, since these forces cancel out.

M mFF

force on m due to Mforce on M due to m

For each object, F = (mass) (a) = (mass) (v / t ) = (mass v) / t = p / t. Since the force applied and the contact time is the same for each mass, they each undergo the same change in momentum, but in opposite directions. The result is that even though the momenta of the individual objects changes, p for the system is zero. The momentum that one mass gains, the other loses. Hence, the momentum of the system before equals the momentum of the system after.

Conservation of Momentum applies only in the absence of external forces!

In the first two sample problems, we dealt with a frictionless surface. We couldn’t simply conserve momentum if friction had been present because, as the proof on the last slide shows, there would be another force (friction) in addition to the contact forces. Friction wouldn’t cancel out, and it would be a net force on the system.

The only way to conserve momentum with an external force like friction is to make it internal by including the tabletop, floor, or the entire Earth as part of the system. For example, if a rubber ball hits a brick wall, p for the ball is not conserved, neither is p for the ball-wall system, since the wall is connected to the ground and subject to force by it. However, p for the ball-Earth system is conserved!

Example

EarthM

apple

An apple is originally at rest and then dropped. After falling a short time, it’s moving pretty fast, say at a speed V. Obviously, momentum is not conserved for the apple, since it didn’t have any at first. How can this be?

m

F

F

answer: Gravity is an external force on the apple, so momentum for it alone is not conserved. To make gravity “internal,” we must define a system that includes the other object responsible for the gravitational force--Earth. The net force on the apple-Earth system is zero, and momentum is conserved for it. During the fall the Earth attains a very small speed v. So, by conservation of momentum:

V

v

m V = M v

Internal and External Forces

Internal forces: the forces that the particles of the system exert on each other (for any system)

Forces exerted on any part of the system by some object outside it called external forces

For the system of two astronauts, the internal forces are FBonA, exerted by particle B on particle A, and FAonB, exerted by particle A on particle B

There are NO external forces: we have isolated system

Total momentum of two particles is the vector sum of the momenta of individual particles

What is an isolated system?• A system is a collection of two or more objects. An isolated system is

a system that F is free from the influence of a net external force that alters the momentum of the system. There are two criteria for the presence of a net external force; it must be...

• a force that originates from a source other than the two objects of the system

• a force that is not balanced by other forces.

• A system in which the only forces that

contribute to the momentum change of an

individual object are the forces acting

between the objects themselves can be

considered an isolated system.

Read the following description of a collision and evaluate whether or not the collision occurs in an

isolated system. If it is not an isolated system, then identify the net external force.

• Two cars collide on a gravel roadway on which frictional forces are large.

• No. What is the net external force?

• The friction between the cars and the road is an external force. This force contributes to a change in total momentum of the system.

Read the following description of a collision and evaluate whether or not the collision occurs in an

isolated system. If it is not an isolated system, then identify the net external force.

• Hans Full is doing the annual vacuuming. Hans is pushing the Hoover vacuum cleaner across the living room carpet.

• No. What is the net external force?

• The friction between the cleaner and the floor and the applied force exerted by Hans are both external forces. These forces contribute to a change in total momentum of the system.

Read the following description of a collision and evaluate whether or not the collision occurs in an

isolated system. If it is not an isolated system, then identify the net external force.

• Two air track gliders collide on a friction-free air track.

• Yes. It is an isolated system.

• There are no external forces; the system is isolated. The total system momentum would be expected to be conserved.

Sample Problem 1

7 kg

v = 0700 m/s

A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates the butter, but while passing through it, the bullet pushes the butter to the left, and the butter pushes the bullet just as hard to the right, slowing the bullet down. If the butter skids off at 4 cm/s after the bullet passes through it, what is the final speed of the bullet?(The mass of the rifle matters not.)

35 g

7 kgv = ?

35 g

4 cm/s

continued on next slide

Sample Problem 1 (cont.)

7 kg

v = 0

700 m/s

35 g

7 kgv = ?

35 g

4 cm/s

p before = 7 (0) + (0.035) (700)

= 24.5 kg · m /s

Let’s choose left to be the + direction & use conservation of momentum, converting all units to meters and kilograms.

p after = 7 (0.04) + 0.035 v

= 0.28 + 0.035 v

p before = p after 24.5 = 0.28 + 0.035 v v = 692 m/s

v came out positive. This means we chose the correct direction of the bullet in the “after” picture.

Sample Problem 2

7 kg

v = 0

700 m/s

35 g

Same as the last problem except this time it’s a block of wood rather than butter, and the bullet does not pass all the way through it. How fast do they move together after impact? Assume frictionless surface.

v

Sample Problem 2

7 kg

v = 0

700 m/s

35 g

Same as the last problem except this time it’s a block of wood rather than butter, and the bullet does not pass all the way through it. How fast do they move together after impact?

v7. 035 kg

(0.035) (700) = 7.035 v v = 3.48 m/s

Note: Once again we’re assuming a frictionless surface, otherwise there would be a frictional force on the wood in addition to that of the bullet, and the “system” would have to include the table as well.

Sample Problem 3

before

after

3 kg 15 kg10 m/s 6 m/s

3 kg 15 kg4.5 m/s v

A crate of raspberry donut filling collides with a tub of lime Kool Aid on a frictionless surface. Which way on how fast does the Kool Aid rebound?

Sample Problem 3

before

after

3 kg 15 kg10 m/s 6 m/s

3 kg 15 kg4.5 m/s v

A crate of raspberry donut filling collides with a tub of lime Kool Aid on a frictionless surface. Which way on how fast does the Kool Aid rebound? answer: Let’s draw v to the right in the after picture.

3 (10) - 6 (15) = -3 (4.5) + 15 v v = -3.1 m/sSince v came out negative, we guessed wrong in drawing v to the right, but that’s OK as long as we interpret our answer correctly. After the collision the lime Kool Aid is moving 3.1 m/s to the left.

Homework

• Evaluation argument from inquiry labs.• Problems 4, 6, 8, 10

Conclusions

• Based on the investigations, what do you think is meant by conservation of momentum?

• Kahoot 7-2