 # Linear Momentum Speed Review  ###### Jason Kuma

Writer | Coach | USC - Physics B.S & Business B.A. | Fremont, CA

###### Article Content
This article will simplify linear momentum and address many misconceptions. By the end of the article you will be able to solve any problem using conservation of momentum or the impulse-momentum law. Note that this doesn’t cover angular (or rotational) momentum. To learn about that click this link.

### What is momentum?

Anything that has mass (which is pretty much everything) AND has velocity has momentum. Hence the equation p = mv.

Conservation of Momentum

Means that the initial momentum value should equal the final. If they are not equal then momentum is NOT conserved and there has been an impulse caused by an external force (more on this later).

Conservation of momentum: Pi = Pf

To solve problems add up the momentums’ of objects before a collision. Then set it equal to the sum of momentum of the objects after a collision.

### Three Types of Collisions:

1. Elastic – Objects collide and bounce of each other (energy is generally conserved in this type of collision)
2. Inelastic – Objects collide and stick to each other.
3. Explosion – Object starts at rest then explodes outwards in multiple pieces.

Momentum is conserved in ALL collision. Energy, however, might not be conserved. Why? When objects collide a small about of heat is dissipated and lost to the surround environment.

### Example of Conservation

Let’s try the following problem:

A 0.450 kg ice puck, moving east with a speed of 3.00 m/s, has a head-on collision with a 0.900 kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each object after the collision?

Step one: Identify if momentum is conserved. This problem deals with a elastic collision, so momentum is definitely conserved. Therefore, we can use conservation of momentum

Step two: Apply conservation of momentum – Pi = Pf –> (.45)(3) = (.9)(v1) + (.45)(v2)

Step three: notice how you have too many unknowns to be able to solve this problem. You need a second equation. This will come from the conservation of energy (we can use this because the collision is elastic). Ei = Ei –> (.45)(3)2 = (.9)(v2)2 + (.45)(v1)2

Step four: rearrange and solve the equations above. Use any method that works best for you. I would recommend substitution in this case.

Step five: final solution, v1 = -1 m/s, v2 = 2 m/s

### Impulse

Impulse is a change in momentum. This means that Pi = Pf.

Impulse = ∆p = m∆v = Ft

The equations above show: the change in momentum, which is really a change in velocity, comes from an external force being applied. This makes sense as a force, implies an acceleration, which is a change in velocity.

What is an external force? It is a force that is not within the system. For example, imagine a car traveling. The system is just the car. Suppose it hits a wall and comes to a stop. The wall applied an external force, that caused a change in velocity, thus a change in momentum.

### Example Impulse Problem

A 0.0600kg tennis ball is traveling at 30.0m/s. After being hit by the opponents racket, the ball’s velocity is 20.0 m/s in the opposite direction. Calculate the:
a) change in the ball’s momentum and
b) average force exerted by the racket if the ball and racket were in contact for 0.0400 s

Solution:

1. Check if momentum is conserved. It is NOT conserved in this problem, since the racket is applying an external force. Hence, we will use the Impulse theorm to solve this.
2. Re-arrange and apply Impulse equations –> ∆p = m∆v –> (.06)(30+20) = 3 Ns [make sure to apply signs correctly, as this is one of the most common mistakes]
3. Solve the part (b) by apply the other equation for impulse. ∆p = Ft –> 3 = F(.04) –> F = 75N

### Other Important Concepts for the AP Exam

1. Velocity of center of mass. As long as momentum is conserved, so will the velocity of the center of mass.
2. Collisions at angles. Apply the same laws and equations as collisions in one dimension, but this time split velocity into x and y components and apply conservation in both directions.

That’s a wrap! If you are still having difficulty applying these concepts be sure to ask you teachers or friends for guidance. If you want quick and easy answer to any difficult questions, consider hiring a tutor here! Nerd-Notes tutor’s guarantees boost is grades and understanding.

If you get all of these right, you have reached mastery. Score a 80% you will do well on an exam. Anything below a 70% — you may want to review concepts or problem solving strategies in detail. These should be complete within 3-5 minutes.

##### Question 1

A pool cue ball, mass 0.7 kg, is traveling at 2 m/s when it collides head on with another ball, mass 0.5 kg, traveling in the opposite direction with a speed of 1.2 m/s. After the collision, the cue ball travels in the opposite direction at 0.3 m/s. What is the velocity of the other ball?

##### Questions 2

A bullet of mass 0.0500 kg traveling at 50.0 m/s is fired horizontally into a wooden block suspended from a long rope. The mass of the wooden block is 0.300 kg and it is initially at rest. The collision is completely inelastic and after impact the bullet+ wooden block move together until the center of mass of the system rises a vertical distance h above its initial position. Calculate the a) velocity of the bullet +wooden block Just after impact and b) vertical distance h reached by the bullet and wooden block.

##### Question 3

A child in a boat throws a 6.40 kg package out horizontally with a speed of 10.0 m/s, Fig. 7-31. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 26.0 kg and that of the boat is 45.0 kg. Ignore water resistance.

##### Question 4

A baseball, mass 0.6 kg, is traveling to the right at 45 m/s when it is hit by a bat and travels the opposite direction at 60 m/s. The bat hits the ball with a force of 1,260 N (a) What is the ball’s change in momentum? (b) How long is the ball in contact with the bat?

##### Question 5

A 4kg mass is traveling at 10m/s to the right when it collides inelastically with a stationary 7kg mass. The 7kg mass then travels at 2m/s at an angle of 22° below the horizontal. What are the velocity and the angle of the 4kg mass?

##### Question 6

A 2,000kg car collides with a stationary 1,000 kg car. Afterwards, they slide 6 m before coming to a stop. The coefficient of friction between the tires and the road is 0.7. Find the initial velocity of the 2,000kg car before the collision?

##### Question 7

A 3800 kg open railroad car coasts along with a constant speed of 8.60 m/s along a level track. Snow begins to fall vertically and fills the car at rate of 3.50 kg/min. Ignoring friction with the tracks, what is the speed of the car after 90 min?

##### Question 8

A 23 g bullet traveling at 230 m/s penetrates a stationary 2.0 kg block of wood and emerges cleanly at 170m/s. How fast does it move after the bullet emerges? (Frictional forces are negligible)

##### Question 9

Consider these two situations: (1) A car moving at 75 mph collides with another car of equal mass moving at 75 mph in the opposite direction. (2) A car moving at 75 mph hits a stationary steel wall.
The collision time is the same for the two cases and the collisions are both inelastic. In which of these cases would result in the greatest impact force?

(a) Case 1
(b) Case 2
(c) Equal force in both case 1 and case 2
(e) None of the above

##### Question 10

A bowling ball moving with speed v collides head-on with a stationary tennis ball. The collision is elastic, and there is no friction. The bowling ball barely slows down. What is the speed of the tennis ball after the collision?

(a) nearly v
(b) nearly 2v
(c) nearly 3vc
(d) Infinite
(e) zero

1. 2 m/s
2. 7.1 m/s, 2.6 m
3. -.9 m/s
4. -63 Ns, .05 s
5. 6.9 m/s, 11°
6. 13.61 m/s
7. 7.9 m/s
8. .7 m/s
9. C
10. B ###### Jason Kuma

Writer | Coach | USC - Physics B.S & Business B.A. | Fremont, CA

## Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy

Unit 5 – Momentum

Unit 6 – Torque

Unit 7 – Oscillations

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