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5.2 Representations of Changes in Momentum

7 min readmay 18, 2023

Kanya Shah

Kanya Shah

Daniella Garcia-Loos

Daniella Garcia-Loos

Kanya Shah

Kanya Shah

Daniella Garcia-Loos

Daniella Garcia-Loos

Two Object Problems

When you’re given a two-object problem and you’re asked to calculate the , take into account the initial momentum and the final momentum individually first of each object. pinitial = pfinal. When you solve for the final momentum, decide whether you should use an individual value or the sum of all masses in the system.

If you’re solving for the final of the system, then you should absolutely use the sum of both masses regardless of how many objects there are. If an object is initially at rest, its initial momentum will be zero since its in the beginning was zero (p = m*v). 

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2Fph.PNG?alt=media&token=267dc7f9-c55a-4947-a2e3-ee2381051e28

is proportional to the force and time so if you decrease T or F, you decrease the .

Image Credit: ponderisd.net

Example Problem #1:

A tennis ball of 0.05 kg is hit with a racket, causing it to travel at a of 40 m/s. The racket has a of 0.5 kg and is traveling at a of 10 m/s. What is the total momentum of the ball and the racket before and after the collision?

Solution:

Before the collision, the momentum of the ball is given by the formula:

The of the ball is 0.05 kg, and its is 40 m/s.

Therefore, the momentum of the ball before the collision is: momentum = 0.05 kg * 40 m/s = 2 kg*m/s

Before the collision, the momentum of the racket is given by the formula:

The of the racket is 0.5 kg, and its is 10 m/s.

Therefore, the momentum of the racket before the collision is: momentum = 0.5 kg * 10 m/s = 5 kg*m/s

Before the collision, the total momentum of the ball and the racket is: 2 kgm/s + 5 kgm/s = 7 kg*m/s

After the collision, the total momentum of the ball and the racket remains constant, since there are no acting on the system.

Therefore, the total momentum of the ball and the racket after the collision is also: 7 kg*m/s

Example Problem #2:

Two cars are traveling on a highway. Car A has a of 2000 kg and is traveling at a of 60 m/s. Car B has a of 1000 kg and is traveling at a of 40 m/s. The cars collide head-on and stick together. What is the total momentum of the cars before and after the collision?

Solution:

Before the collision, the momentum of car A is given by the formula:

The of car A is 2000 kg, and its is 60 m/s.

Therefore, the momentum of car A before the collision is: momentum = 2000 kg * 60 m/s = 120000 kg*m/s

Before the collision, the momentum of car B is given by the formula:

The of car B is 1000 kg, and its is 40 m/s.

Therefore, the momentum of car B before the collision is: momentum = 1000 kg * 40 m/s = 40000 kg*m/s

Before the collision, the total momentum of the cars is: 120000 kgm/s + 40000 kgm/s = 160000 kg*m

After the collision, the combined of the cars is 2000 kg + 1000 kg = 3000 kg.

The combined of the cars is not given, so we will assume it to be v.

The total momentum of the cars after the collision is therefore: momentum = 3000 kg * v m/s

Since the total momentum of the cars before and after the collision must be equal, we can set these two expressions equal to each other:

120000 kgm/s + 40000 kgm/s = 3000 kg * v m/s

160000 kg*m/s = 3000 kg * v m/s

v = 160000 kg*m/s / 3000 kg = 53.333 m/s

Therefore, the combined of the cars after the collision is 53.333 m/s.

Example Problem #3:

A ball of 0.1 kg is thrown with a of 10 m/s. A catcher catches the ball with a of 5 m/s. What is the of the ball?

Solution:

The of the ball is given by the formula: = * change in

The of the ball is 0.1 kg, and its change in is: 10 m/s - 5 m/s = 5 m/s

Therefore, the of the ball is: = 0.1 kg * 5 m/s = 0.5 kg*m/s

Example Problem #4:

A rocket of 100 kg is launched from the surface of the Earth with a of 100 m/s. The rocket carries a payload of 50 kg. What is the total momentum of the rocket and the payload before and after the launch?

Solution:

Before the launch, the momentum of the rocket is given by the formula:

The of the rocket is 100 kg, and its is 0 m/s (since it is stationary before the launch).

Therefore, the momentum of the rocket before the launch is: momentum = 100 kg * 0 m/s = 0 kg*m/s

Before the launch, the momentum of the payload is also zero, since its is zero.

Before the launch, the total momentum of the rocket and the payload is: 0 kgm/s + 0 kgm/s = 0 kg*m/s

After the launch, the momentum of the rocket is given by the formula:

The of the rocket is 100 kg, and its is 100 m/s.

After the launch, the momentum of the rocket is: momentum = 100 kg * 100 m/s = 10000 kg*m/s

After the launch, the momentum of the payload is given by the formula:

The of the payload is 50 kg, and its is 100 m/s (since it is carried along by the rocket).

Therefore, the momentum of the payload after the launch is: momentum = 50 kg * 100 m/s = 5000 kg*m/s

After the launch, the total momentum of the rocket and the payload is: 10000 kgm/s + 5000 kgm/s = 15000 kg*m/s

Example Problem #5:

A baseball of 0.15 kg is hit with a bat, causing it to travel at a of 50 m/s. The bat has a of 1 kg and is traveling at a of -20 m/s (i.e. in the opposite direction). What is the total momentum of the ball and the bat before and after the collision?

Solution:

Before the collision, the momentum of the ball is given by the formula:

The of the ball is 0.15 kg, and its is 50 m/s.

Therefore, the momentum of the ball before the collision is: momentum = 0.15 kg * 50 m/s = 7.5 kg*m/s

Before the collision, the momentum of the bat is given by the formula:

The of the bat is 1 kg, and its is -20 m/s.

Therefore, the momentum of the bat before the collision is: momentum = 1 kg * -20 m/s = -20 kg*m/s

Before the collision, the total momentum of the ball and the bat is: 7.5 kgm/s + -20 kgm/s = -12.5 kg*m/s

After the collision, the total momentum of the ball and the bat remains constant, since there are no acting on the system.

Therefore, the total momentum of the ball and the bat after the collision is also: -12.5 kg*m/s

Interpreting Different Scenarios

You need to be able to interpret different scenarios, especially with problems that have multiple acting on the system. When you see problems like that, make sure you find the of the forces after calculating individual forces. If you’re looking for the momentum, you can either find individual momentums of the objects or find the . Understand how to apply the formula and make sure you can explain your thought process because some FRQ’s ask for conceptual explanations. 

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2Ff.PNG?alt=media&token=5a2cf595-2910-4a90-adff-9b68d850d75b

Image Credit: studyblue.com

Key Terms to Review (8)

Change in momentum

: The change in momentum refers to the difference between the initial momentum and the final momentum of an object. It is a measure of how much an object's motion has changed.

External Forces

: External forces are forces acting on an object that originate from outside the system being analyzed. These forces can cause changes in the motion or shape of the object.

FRQ (Free Response Questions)

: FRQ, or Free Response Questions, are a type of exam question that requires students to provide a written response rather than selecting from multiple choice options. These questions assess students' understanding and ability to apply concepts in a more open-ended format.

Mass

: Mass refers to the amount of matter an object contains. It is a measure of the inertia or resistance to changes in motion.

Momentum = Mass * Velocity

: Momentum is defined as the product of an object's mass and its velocity. It represents how difficult it is to stop or change the motion of an object.

Total Momentum of the System

: The total momentum of a system refers to the combined momentum of all objects within that system. It takes into account both mass and velocity.

Vector Sum

: The vector sum is the result of adding two or more vectors together using vector addition. It represents both magnitude and direction.

Velocity

: Velocity refers to the rate at which an object changes its position in a specific direction. It includes both speed and direction.

5.2 Representations of Changes in Momentum

7 min readmay 18, 2023

Kanya Shah

Kanya Shah

Daniella Garcia-Loos

Daniella Garcia-Loos

Kanya Shah

Kanya Shah

Daniella Garcia-Loos

Daniella Garcia-Loos

Two Object Problems

When you’re given a two-object problem and you’re asked to calculate the , take into account the initial momentum and the final momentum individually first of each object. pinitial = pfinal. When you solve for the final momentum, decide whether you should use an individual value or the sum of all masses in the system.

If you’re solving for the final of the system, then you should absolutely use the sum of both masses regardless of how many objects there are. If an object is initially at rest, its initial momentum will be zero since its in the beginning was zero (p = m*v). 

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2Fph.PNG?alt=media&token=267dc7f9-c55a-4947-a2e3-ee2381051e28

is proportional to the force and time so if you decrease T or F, you decrease the .

Image Credit: ponderisd.net

Example Problem #1:

A tennis ball of 0.05 kg is hit with a racket, causing it to travel at a of 40 m/s. The racket has a of 0.5 kg and is traveling at a of 10 m/s. What is the total momentum of the ball and the racket before and after the collision?

Solution:

Before the collision, the momentum of the ball is given by the formula:

The of the ball is 0.05 kg, and its is 40 m/s.

Therefore, the momentum of the ball before the collision is: momentum = 0.05 kg * 40 m/s = 2 kg*m/s

Before the collision, the momentum of the racket is given by the formula:

The of the racket is 0.5 kg, and its is 10 m/s.

Therefore, the momentum of the racket before the collision is: momentum = 0.5 kg * 10 m/s = 5 kg*m/s

Before the collision, the total momentum of the ball and the racket is: 2 kgm/s + 5 kgm/s = 7 kg*m/s

After the collision, the total momentum of the ball and the racket remains constant, since there are no acting on the system.

Therefore, the total momentum of the ball and the racket after the collision is also: 7 kg*m/s

Example Problem #2:

Two cars are traveling on a highway. Car A has a of 2000 kg and is traveling at a of 60 m/s. Car B has a of 1000 kg and is traveling at a of 40 m/s. The cars collide head-on and stick together. What is the total momentum of the cars before and after the collision?

Solution:

Before the collision, the momentum of car A is given by the formula:

The of car A is 2000 kg, and its is 60 m/s.

Therefore, the momentum of car A before the collision is: momentum = 2000 kg * 60 m/s = 120000 kg*m/s

Before the collision, the momentum of car B is given by the formula:

The of car B is 1000 kg, and its is 40 m/s.

Therefore, the momentum of car B before the collision is: momentum = 1000 kg * 40 m/s = 40000 kg*m/s

Before the collision, the total momentum of the cars is: 120000 kgm/s + 40000 kgm/s = 160000 kg*m

After the collision, the combined of the cars is 2000 kg + 1000 kg = 3000 kg.

The combined of the cars is not given, so we will assume it to be v.

The total momentum of the cars after the collision is therefore: momentum = 3000 kg * v m/s

Since the total momentum of the cars before and after the collision must be equal, we can set these two expressions equal to each other:

120000 kgm/s + 40000 kgm/s = 3000 kg * v m/s

160000 kg*m/s = 3000 kg * v m/s

v = 160000 kg*m/s / 3000 kg = 53.333 m/s

Therefore, the combined of the cars after the collision is 53.333 m/s.

Example Problem #3:

A ball of 0.1 kg is thrown with a of 10 m/s. A catcher catches the ball with a of 5 m/s. What is the of the ball?

Solution:

The of the ball is given by the formula: = * change in

The of the ball is 0.1 kg, and its change in is: 10 m/s - 5 m/s = 5 m/s

Therefore, the of the ball is: = 0.1 kg * 5 m/s = 0.5 kg*m/s

Example Problem #4:

A rocket of 100 kg is launched from the surface of the Earth with a of 100 m/s. The rocket carries a payload of 50 kg. What is the total momentum of the rocket and the payload before and after the launch?

Solution:

Before the launch, the momentum of the rocket is given by the formula:

The of the rocket is 100 kg, and its is 0 m/s (since it is stationary before the launch).

Therefore, the momentum of the rocket before the launch is: momentum = 100 kg * 0 m/s = 0 kg*m/s

Before the launch, the momentum of the payload is also zero, since its is zero.

Before the launch, the total momentum of the rocket and the payload is: 0 kgm/s + 0 kgm/s = 0 kg*m/s

After the launch, the momentum of the rocket is given by the formula:

The of the rocket is 100 kg, and its is 100 m/s.

After the launch, the momentum of the rocket is: momentum = 100 kg * 100 m/s = 10000 kg*m/s

After the launch, the momentum of the payload is given by the formula:

The of the payload is 50 kg, and its is 100 m/s (since it is carried along by the rocket).

Therefore, the momentum of the payload after the launch is: momentum = 50 kg * 100 m/s = 5000 kg*m/s

After the launch, the total momentum of the rocket and the payload is: 10000 kgm/s + 5000 kgm/s = 15000 kg*m/s

Example Problem #5:

A baseball of 0.15 kg is hit with a bat, causing it to travel at a of 50 m/s. The bat has a of 1 kg and is traveling at a of -20 m/s (i.e. in the opposite direction). What is the total momentum of the ball and the bat before and after the collision?

Solution:

Before the collision, the momentum of the ball is given by the formula:

The of the ball is 0.15 kg, and its is 50 m/s.

Therefore, the momentum of the ball before the collision is: momentum = 0.15 kg * 50 m/s = 7.5 kg*m/s

Before the collision, the momentum of the bat is given by the formula:

The of the bat is 1 kg, and its is -20 m/s.

Therefore, the momentum of the bat before the collision is: momentum = 1 kg * -20 m/s = -20 kg*m/s

Before the collision, the total momentum of the ball and the bat is: 7.5 kgm/s + -20 kgm/s = -12.5 kg*m/s

After the collision, the total momentum of the ball and the bat remains constant, since there are no acting on the system.

Therefore, the total momentum of the ball and the bat after the collision is also: -12.5 kg*m/s

Interpreting Different Scenarios

You need to be able to interpret different scenarios, especially with problems that have multiple acting on the system. When you see problems like that, make sure you find the of the forces after calculating individual forces. If you’re looking for the momentum, you can either find individual momentums of the objects or find the . Understand how to apply the formula and make sure you can explain your thought process because some FRQ’s ask for conceptual explanations. 

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2Ff.PNG?alt=media&token=5a2cf595-2910-4a90-adff-9b68d850d75b

Image Credit: studyblue.com

Key Terms to Review (8)

Change in momentum

: The change in momentum refers to the difference between the initial momentum and the final momentum of an object. It is a measure of how much an object's motion has changed.

External Forces

: External forces are forces acting on an object that originate from outside the system being analyzed. These forces can cause changes in the motion or shape of the object.

FRQ (Free Response Questions)

: FRQ, or Free Response Questions, are a type of exam question that requires students to provide a written response rather than selecting from multiple choice options. These questions assess students' understanding and ability to apply concepts in a more open-ended format.

Mass

: Mass refers to the amount of matter an object contains. It is a measure of the inertia or resistance to changes in motion.

Momentum = Mass * Velocity

: Momentum is defined as the product of an object's mass and its velocity. It represents how difficult it is to stop or change the motion of an object.

Total Momentum of the System

: The total momentum of a system refers to the combined momentum of all objects within that system. It takes into account both mass and velocity.

Vector Sum

: The vector sum is the result of adding two or more vectors together using vector addition. It represents both magnitude and direction.

Velocity

: Velocity refers to the rate at which an object changes its position in a specific direction. It includes both speed and direction.


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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.


© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.