Momentum & collisions: practice problems with step-by-step solutions

Free Momentum & collisions practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: perfectly inelastic, elastic, impulse and force.

Free practice problemEasy
4.5 kg750 m/sV = ?

A 4.5 kg rifle, free to move, fires a 12 g bullet at 750 m/s. Both start at rest. Find the recoil speed V of the rifle.

M = 4.5 kgm = 12 gv = 750 m/s
m/s
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What you practise

  1. Easyperfectly inelastic
  2. Mediumelastic
  3. Hardimpulse and force

How to solve Momentum & collisions problems

  1. Choose a positive direction and give each velocity its sign.
  2. Write the total momentum before: Σm·v.
  3. For a perfectly inelastic collision the bodies move together after; for an elastic one kinetic energy is also conserved.
  4. Set momentum before equal to momentum after and solve.

Common mistakes

  • Forgetting the minus sign of a velocity in the opposite direction.
  • Averaging velocities instead of weighting them by mass.
  • Assuming kinetic energy is conserved in an inelastic collision.

Key formulas

Momentum
p = m·v
Conservation
Σm·v (before) = Σm·v (after)
Perfectly inelastic
v′ = (m1v1 + m2v2)/(m1 + m2)
Impulse
J = F·Δt = m·Δv

Learn Momentum & collisions from scratch

Before this: Newton’s laws

1What momentum is

A lorry at 10 km/h is harder to stop than a bicycle at the same speed. A bullet is tiny but very hard to stop because it is so fast. Momentum combines both: mass and velocity.

p = m · v

Momentum p is in kg·m/s. It has a direction, the same as the velocity. Choose a positive direction: a body moving the other way has a negative velocity and a negative momentum.

Momentum matters because it is conserved in collisions and explosions. The total before equals the total after, whatever happens during the crash.

Self-check What is the momentum of a 2 kg ball moving at 3 m/s?
  1. 1.5 kg·m/s
  2. 18 kg·m/s
  3. 6 kg·m/s
  4. 9 kg·m/s

Answer: 6 kg·m/s. p = m·v = 2 · 3 = 6 kg·m/s. 9 is ½·m·v², the kinetic energy in joules, which is a different quantity.

2Conservation of momentum

When two bodies collide, they push on each other with equal and opposite forces (Newton’s third law) for the same short time. What one gains in momentum, the other loses.

So if no outside force acts during the collision, the total momentum stays the same:

m₁·v₁ + m₂·v₂ = m₁·v′₁ + m₂·v′₂

A prime (′) marks a velocity after the collision. Friction from the road is small during the short crash, so you can ignore it.

Collisions come in two kinds:

  • Perfectly inelastic: the bodies stick together and move with one common velocity v′. Kinetic energy is lost to heat, sound and dents.
  • Elastic: the bodies bounce apart and the kinetic energy ½·m·v² is also conserved.

For a perfectly inelastic collision the common velocity follows directly:

v′ = (m₁·v₁ + m₂·v₂) / (m₁ + m₂)

An explosion or a gun firing is the same idea backwards: the total momentum is zero before, so it is zero after.

Self-check A 2 kg trolley at 6 m/s hits a 1 kg trolley at rest. They stick together. What is their common velocity?
  1. 12 m/s
  2. 2 m/s
  3. 3 m/s
  4. 4 m/s

Answer: 4 m/s. v′ = (2 · 6 + 1 · 0)/(2 + 1) = 12/3 = 4 m/s. Averaging the velocities (3 m/s) ignores that the heavier trolley carries more momentum.

3Impulse

A force that acts for a time Δt changes the momentum of a body. The product F·Δt is called the impulse J, in N·s (the same as kg·m/s).

J = F · Δt = m · Δv = m · (v′ − v)

The same change in momentum over a longer time needs a smaller force. That is why airbags and crumple zones save lives.

For a ball that bounces back, the velocity changes sign. With the incoming direction positive, a ball that hits a wall at 10 m/s and returns at 5 m/s has Δv = −5 − 10 = −15 m/s.

4The method step by step

  1. Choose a positive direction and give each velocity its sign.
  2. Write the total momentum before: m₁·v₁ + m₂·v₂.
  3. Decide what happens after: one common velocity (they stick), two velocities (they bounce) or several pieces (an explosion).
  4. Set momentum before equal to momentum after and solve for the unknown.
  5. If energy is asked, compare ½·m·v² before and after.
  6. Check the sign of the answer: a negative velocity moves in the negative direction.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A 1000 kg car moving at 20 m/s collides head-on with a 1500 kg car moving at 10 m/s in the opposite direction. The cars lock together. Find their common velocity v′ just after the collision, positive in the first car’s direction.

1

Signs

+m1​1000 kg20 m/sm2​1500 kg10 m/s

Take the direction of the first car as positive. Then v₁ = +20 m/s and v₂ = −10 m/s.

2

Momentum before

1000 · 20 + 1500 · (−10) = 20000 − 15000 = 5000 kg·m/s
3

Momentum after

The cars move as one body of 2500 kg with velocity v′.

p′ = 2500 · v′
4

Solve

2500 · v′ = 5000 ⇒ v′ = 2 m/s
+m1​ + m2​2500 kgv′ = 2 m/s
5

Check the answer

v′ is positive, so the wreck moves in the first car’s direction. That fits: the first car had more momentum (20 000 against 15 000 kg·m/s). The kinetic energy drops from 275 000 J to 5000 J, as expected when the cars stick.

Answer

v′ = +2 m/s, in the direction the first car was moving.

6Common mistakes

  • Forgetting the minus sign of a velocity in the opposite direction.
  • Averaging the velocities instead of weighting them by mass.
  • Assuming kinetic energy is conserved when the bodies stick together. Only momentum is.
  • For a bounce, using the difference of the speeds instead of the change in velocity: the ball reverses, so the speeds add.
  • Mixing grams and kilograms for a bullet: 12 g is 0.012 kg.
Self-check A 0.2 kg ball hits a wall at 10 m/s and bounces straight back at 5 m/s. The contact lasts 0.01 s. How large is the average force of the wall?
  1. 100 N
  2. 200 N
  3. 3 N
  4. 300 N

Answer: 300 N. Δv = −5 − 10 = −15 m/s, so F = m·|Δv|/Δt = 0.2 · 15/0.01 = 300 N. Using 10 − 5 = 5 m/s (100 N) forgets that the ball reverses. 3 N is the impulse in N·s, not divided by the time.

7How exam questions look

  • Easy: perfectly inelastic collisions: two cars that lock together, a trolley hitting another, or a rifle that recoils.
  • Medium: elastic collisions of gliders on a track, the kinetic energy lost when bodies stick, explosions, and a bullet fired into a hanging block.
  • Hard: impulse and force: a ball bouncing off a wall, cars colliding at a crossing (momentum in two directions), or a wreck that skids to a stop.

For an elastic collision with the second body at rest: v′₁ = (m₁ − m₂)/(m₁ + m₂) · v₁ and v′₂ = 2m₁/(m₁ + m₂) · v₁.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
0.5 kg7.5 m/s5 kg

A 0.5 kg trolley moving at 7.5 m/s runs into a 5 kg trolley at rest. After the collision the first trolley continues in the same direction at 0.5 m/s. Find the velocity v′2 of the second trolley.

m1 = 0.5 kgv1 = 7.5 m/sm2 = 5 kgv′1 = 0.5 m/s

Find v′2 (m/s)

Show hints
  1. Total momentum before = total momentum after.
  2. m₁·v₁ = m₁·v′₁ + m₂·v′₂.
  3. momentum transferred = 3.5 kg·m/s.
Show the answer and the step-by-step solution

Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.

Momentum balance

0.5·7.5 = 0.5·0.5 + 5·v′2
v′2 = 3.5/5 = 0.7 m/s

Answer

v′2 = 0.7 m/s
Example 2Medium
1400 kg10 m/s1500 kg

A 1400 kg car moving at 10 m/s runs into a stationary 1500 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?

m1 = 1400 kgv1 = 10 m/sm2 = 1500 kgv2 = 0

Find ΔK (J)

Show hints
  1. Find the common velocity from momentum conservation first.
  2. Then compare ½·m·v² before and after.
  3. v′ = 4.828 m/s.
Show the answer and the step-by-step solution

Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.

Common velocity

v′ = m1·v1/(m1 + m2) = 1400·10/2900 = 4.828 m/s

Kinetic energy before and after

Kbefore = ½·1400·10² = 70000 J
Kafter = ½·2900·4.828² = 33793 J

Energy lost

ΔK = 70000 − 33793 = 36207 J

Answer

ΔK = 36207 J

The lost energy goes into deformation, heat and sound.

Example 3Hard
5 kgv = ?20 g25°L = 0.7 m

A 20 g bullet is fired horizontally into a 5 kg block hanging from light strings 0.7 m long, and stays in it. The strings swing out to a largest angle of 25° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 20 gM = 5 kgL = 0.7 mθ = 25°g = 9.81 m/s²

Find v (m/s)

Show hints
  1. Height gained: h = L·(1 − cos θ).
  2. Swing: V = √(2gh). Impact: m·v = (m + M)·V.
  3. h = 0.06558 m, V = 1.134 m/s.
Show the answer and the step-by-step solution

Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.

Rise of the block

h = L·(1 − cos θ) = 0.7·(1 − cos 25°) = 0.06558 m

Speed after impact

V = √(2·9.81·0.06558) = 1.134 m/s

Collision

v = (m + M)·V/m = 5.02·1.134/0.02 = 284.7 m/s

Answer

v = 284.7 m/s

Questions and answers

What is Momentum & collisions?

A lorry at 10 km/h is harder to stop than a bicycle at the same speed. A bullet is tiny but very hard to stop because it is so fast. Momentum combines both: mass and velocity.

Momentum p is in kg·m/s. It has a direction, the same as the velocity. Choose a positive direction: a body moving the other way has a negative velocity and a negative momentum.

How do I solve Momentum & collisions problems step by step?
  1. Choose a positive direction and give each velocity its sign.
  2. Write the total momentum before: Σm·v.
  3. For a perfectly inelastic collision the bodies move together after; for an elastic one kinetic energy is also conserved.
  4. Set momentum before equal to momentum after and solve.
What are common mistakes in Momentum & collisions?
  • Forgetting the minus sign of a velocity in the opposite direction.
  • Averaging velocities instead of weighting them by mass.
  • Assuming kinetic energy is conserved in an inelastic collision.
What do the Momentum & collisions exercises cover?
  • Easy: perfectly inelastic
  • Medium: elastic
  • Hard: impulse and force

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Newton’s laws

Calculators

Collision and momentum calculator

Courses with this topic

DynamicsPhysics for EngineersPhysics 1: MechanicsAP Physics 1 practice problems

More topics in Physics

KinematicsNewton’s lawsCircular motionSprings & pendulumsCoulomb’s law & electric fieldsMagnetic fields & inductionWaves, sound & optics

Printable worksheet

Momentum & collisions worksheet with answers
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