PhysicsLesson 16 of 18About 40 minutes

Conservation of momentum

When a moving billiard ball hits a resting one, the first slows down and the second speeds up. Momentum moves from one ball to the other, but the total does not change. This is one of the most reliable laws in all of physics.

By the end of this lesson you can

track how much momentum each of two interacting objects gains or loses;
state the law of conservation of momentum and say when it applies;
solve problems about collisions and objects pushing apart;
explain how Newton's third law leads to conservation of momentum.

Momentum moves between objects

Think first

Ball A rolls into ball B, which is at rest. After the hit, A has slowed down and B is moving. Do you think the momentum A lost is more than, less than or equal to the momentum B gained?

Compare your answer

Equal. Whatever momentum A loses, B gains. The simulation below lets you check this with the numbers.

Billiard balls

Simulation

Try this

Play the collision and watch the three momentum bars. What happens to the black "total" bar during the collision?
Look at the force graph. How do the force on A and the force on B compare at every moment?

The law of conservation of momentum

Conservation of momentum

The total momentum of a group of objects stays constant as long as no net external force acts on the group:

$$m_1\vec v_{1,i} + m_2\vec v_{2,i} = m_1\vec v_{1,f} + m_2\vec v_{2,f}$$

The group of objects you are tracking is called a system. Forces between objects inside the system, such as the push between two colliding balls, cannot change the total. Only forces from outside the system can. A system with no net external force is called isolated.

In practice, conservation works well for collisions and explosions even when outside forces such as friction act, because the forces between the objects during the brief collision are much larger than the outside forces.

Common mistake

Forgetting that momentum is a vector. Choose a positive direction at the start and give every velocity a sign. An object moving in the negative direction has negative momentum.

Objects pushing apart

Two skaters standing still on ice have zero total momentum. When they push off each other, they must still have zero total momentum. So their momenta must be equal in size and opposite in direction: $m_1 v_1 = -m_2 v_2$. The lighter skater moves away faster.

Skaters push off

Simulation

Try this

Change the masses and push off several times. Check that $m_1 v_1$ and $m_2 v_2$ always have the same size. Which skater moves faster?

Worked example 1

A 60.0 kg person stands in a 40.0 kg rowboat at rest near a dock. The person steps out toward the dock with a horizontal velocity of 3.00 m/s. What is the velocity of the boat? Ignore the water's drag.

System and direction. The system is the person plus the boat. Take the direction toward the dock as positive. At first, both are at rest, so the total momentum is zero.
Conservation. $0 = m_p v_p + m_b v_b$
Solve for the boat. $$v_b = -\frac{m_p v_p}{m_b} = -\frac{(60.0)(3.00)}{40.0} = -4.50\ \text{m/s}$$
4.50 m/s away from the dock. (The minus sign means opposite to the person.)

Stepping out of a boat

Simulation

Try this

Press Step out and compare the final speeds with Worked example 1. Why does the total momentum bar stay at zero?

Rockets

Think first

In empty space there is no air or ground to push against. How can a rocket speed up there?

Compare your answer

The rocket pushes its own exhaust gas backward, and the gas pushes the rocket forward. The gas gains backward momentum and the rocket gains an equal forward momentum. No outside object is needed.

A rocket in empty space

Simulation

Try this

Fire the engine. Watch the rocket's momentum bar and the gas's momentum bar. What is the total at every moment?

Worked example 2

An 85.0 kg astronaut floats at rest outside a space station. To return, she throws a 1.50 kg wrench away from the station at 6.00 m/s. How fast does she move toward the station?

Total momentum before: zero.
After: $0 = (1.50)(6.00) + (85.0)v$, with the wrench's direction as positive.
Solve. $v = -\dfrac{9.00}{85.0} = -0.106\ \text{m/s}$
0.106 m/s toward the station: slow, but it will get her there.

Worked example 3

On a frictionless track, a 2.00 kg cart rolling right at 3.00 m/s runs into a 1.00 kg cart that is standing still. Just after the hit, the 2.00 kg cart is still moving right, now at 1.00 m/s. How fast is the 1.00 kg cart moving, and in which direction?

Momentum before. $(2.00)(3.00) + (1.00)(0) = 6.00\ \text{kg·m/s}$
Momentum after. $(2.00)(1.00) + (1.00)v_2 = 2.00 + v_2$
Set them equal. $6.00 = 2.00 + v_2$, so $v_2 = 4.00\ \text{m/s}$.
4.00 m/s to the right

Why momentum is conserved

Conservation of momentum follows from Newton's third law. When two objects collide, object 1 pushes on object 2 and object 2 pushes back on object 1 with an equal and opposite force, $\vec F_1 = -\vec F_2$, at every instant. The two forces also act for exactly the same time $\Delta t$. So the impulses are equal and opposite:

$$\vec F_1\Delta t = -\vec F_2\Delta t \quad\Rightarrow\quad \Delta \vec p_1 = -\Delta \vec p_2$$

Whatever momentum one object gains, the other loses, so the total cannot change.

Bumper cars

Simulation

Try this

Collide the cars and compare the blue and pink force curves. They mirror each other at every instant.
Make one car much heavier. Do the forces still match? Do the changes in velocity?
Turn on Show average force. The real force rises and falls, but a constant average force over the same time gives the same impulse.

Check yourself

Question 1

A 50.0 kg skater standing at rest throws a 2.50 kg medicine ball forward at 4.00 m/s. How fast does the skater move backward?

Show solution

$0 = (2.50)(4.00) + (50.0)v$, so $v = -0.200\ \text{m/s}$: 0.200 m/s backward.

Question 2

A 1200 kg cannon on wheels fires a 15.0 kg shell horizontally at 300 m/s. What is the cannon's recoil speed?

Show solution

$v = \dfrac{(15.0)(300)}{1200} = 3.75\ \text{m/s}$, opposite the shell.

Question 3

A 1.50 kg cart traveling right at 2.00 m/s bumps into a 3.00 kg cart that is not moving, and rebounds to the left at 0.500 m/s. At what speed does the 3.00 kg cart move off?

Show solution

Right is positive. Before: $(1.50)(2.00) = 3.00\ \text{kg·m/s}$. After: $(1.50)(-0.500) + 3.00v = -0.750 + 3.00v$. So $3.00v = 3.75$ and $v = 1.25\ \text{m/s}$ to the right.

Question 4

A small car and a large truck collide head-on. During the collision, how do their changes in momentum compare?

Equal in size and opposite in direction
The car's change is larger.
The truck's change is larger.
It depends on which vehicle was faster.

Show solution

The forces on them are equal and opposite and act for the same time, so their impulses, and their changes in momentum, are equal and opposite. The car's change in velocity is larger, because its mass is smaller.

Question 5

A person on a frictionless frozen pond is stuck, unable to walk. Which plan will get them to the edge?

Waving their arms in circles
Jumping straight up
Leaning forward as far as possible
Throwing a shoe away from the edge they want to reach

Show solution

Throwing the shoe gives it momentum in one direction, so the person gains equal momentum in the other. The other actions do not give the person any lasting horizontal momentum.

Summary

In an interaction, the momentum one object gains equals the momentum the other loses.
The total momentum of an isolated system is conserved: $m_1\vec v_{1,i} + m_2\vec v_{2,i} = m_1\vec v_{1,f} + m_2\vec v_{2,f}$.
Objects pushing apart from rest get equal and opposite momenta; that is how rockets work.
Conservation of momentum follows from Newton's third law: equal and opposite forces for the same time give equal and opposite impulses.

Finished?

Mark the lesson as complete to keep track of your progress. Your progress is saved in this browser only.