What implications does inertia have on an astronaut throwing a baseball inside a spacecraft moving at constant velocity?

The Astronaut Will Experience An Equal And Opposite Reaction According To Newton's Third Law As There Is No External Forces Acting On System

Fill in the blank: An object that changes direction (left to right or up to down)...

keeps its acceleration constant

What is the momentum of an object directly proportional to?

Mass and velocity of the object.

Mass and acceleration of the object.

Force exerted on the object and time.

Velocity of the object and distance traveled.

Which unit is used to measure momentum in the International System of Units?

Kilogram meter per second (kg·m/s)

If a fully loaded truck traveling at a constant velocity hits a patch of ice and starts sliding without rotating or tipping over, what happens to its linear momentum?

It remains constant since no external force changes its velocity during sliding on ice.

It increases because sliding on ice reduces frictional resistance to motion.

It decreases due to increased air resistance as it starts sliding on ice.

It becomes zero as there is no longer any engine propulsion during sliding.

When does an object have maximum kinetic energy during an inelastic collision between two objects?

Just before the collision when the objects are moving at their speeds.

Just after the collision when the objects stick together and move as a unit.

At the moment of impact when they exert equal force on each other.

After they separate, collision slows both objects down.

What happens to the gravitational force between a planet and an object on its surface when the mass of the object triples?

The gravitational force triples as well.

The gravitational force increases by a factor of nine.

The gravitational force decreases to one third its original value.

There is no change in the gravitational force.

When a lightweight ball at high speed strikes a heavy stationary sphere head-on and bounces back, what describes the impulse experienced by each object?

Both objects experience impulses of equal magnitude but opposite direction.

The lighter ball experiences less impulse since it has less mass.

The heavier sphere experiences more impulse because it starts from rest.

No impulse occurs for either object as they are different masses.

What must be true if two billiard balls collide on a frictionless table resulting in one ball stopping completely while the other moves away with increased speed?

Both balls had unequal masses with more massive ball stopping post-collision.

Both balls exchanged their velocities due to having identical masses.

Both balls continued with reduced speeds post-collision as some energy was converted into heat or sound.

One ball stopped because it was hit off-center causing it to spin instead continuing forward motion.

Two identical objects are attracting each other due to gravity; if one object's mass doubles while keeping distance constant, what effect does this have on the magnitude of attraction?

It doubles as well since only one mass changed.

There is no change because both objects must change mass for any effect on attraction magnitude.

It quadruples because doubling anything related to gravity quadruples its effects.

The attraction becomes half as strong because increasing mass has an inverse effect on attraction strength.

or

Log in

Find what you need to study

Light

5.2 Representations of Changes in Momentum

7 min read•may 18, 2023

Kanya Shah

Daniella Garcia-Loos

Kanya Shah

Daniella Garcia-Loos

Attend a live cram event

Review all units live with expert teachers & students

Two Object Problems

When you’re given a two-object problem and you’re asked to calculate the change in momentum, 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 mass value or the sum of all masses in the system.

If you’re solving for the final velocity 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 velocity in the beginning was zero (p = m*v).

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

Image Credit: ponderisd.net

Example Problem #1:

A tennis ball of mass 0.05 kg is hit with a racket, causing it to travel at a velocity of 40 m/s. The racket has a mass of 0.5 kg and is traveling at a velocity 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: momentum = mass * velocity

The mass of the ball is 0.05 kg, and its velocity 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: momentum = mass * velocity

The mass of the racket is 0.5 kg, and its velocity 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 external forces 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 mass of 2000 kg and is traveling at a velocity of 60 m/s. Car B has a mass of 1000 kg and is traveling at a velocity 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: momentum = mass * velocity

The mass of car A is 2000 kg, and its velocity 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: momentum = mass * velocity

The mass of car B is 1000 kg, and its velocity 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 mass of the cars is 2000 kg + 1000 kg = 3000 kg.

The combined velocity 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 velocity of the cars after the collision is 53.333 m/s.

Example Problem #3:

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

Solution:

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

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

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

Example Problem #4:

A rocket of mass 100 kg is launched from the surface of the Earth with a velocity 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: momentum = mass * velocity

The mass of the rocket is 100 kg, and its velocity 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 velocity 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: momentum = mass * velocity

The mass of the rocket is 100 kg, and its velocity 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: momentum = mass * velocity

The mass of the payload is 50 kg, and its velocity 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 mass 0.15 kg is hit with a bat, causing it to travel at a velocity of 50 m/s. The bat has a mass of 1 kg and is traveling at a velocity 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: momentum = mass * velocity

The mass of the ball is 0.15 kg, and its velocity 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: momentum = mass * velocity

The mass of the bat is 1 kg, and its velocity 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 external forces 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 external forces acting on the system. When you see problems like that, make sure you find the vector sum 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 total momentum of the system. Understand how to apply the formula and make sure you can explain your thought process because some FRQ’s ask for conceptual explanations.

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.