# Momentum and collisions

Canonical: https://duckyhelper.com/learn/physics/momentum-and-collisions/
Updated: 2026-10-01

Momentum is mass times velocity, \(p = mv\), measured in kg·m/s, and it has a direction. When objects collide with no outside net force, the total momentum before equals the total momentum after. In an elastic collision kinetic energy is also conserved. In an inelastic collision some kinetic energy turns into heat, sound and bent metal, and if the objects stick together it is perfectly inelastic.

## Key ideas

$$
p = m v \qquad J = F \Delta t = \Delta p
$$

- \(p\) is momentum, in kilogram meters per second (kg·m/s). Its direction is the direction of the velocity.
- \(m\) is mass (kg) and \(v\) is velocity (m/s), with a sign for direction.
- \(J\) is impulse, in newton seconds (N·s), which is the same unit as kg·m/s.
- \(F\) is the average force (N) acting for a time \(\Delta t\) (s). \(\Delta p\) is the change in momentum.

**Conservation of momentum** for two objects (1 and 2), before and after (primes):

$$
m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'
$$

**Types of collisions**

| Type | Momentum | Kinetic energy | Example |
| --- | --- | --- | --- |
| Elastic | Conserved | Conserved | Billiard balls (almost), atoms bouncing |
| Inelastic | Conserved | Some is lost | A tennis ball hitting a racket |
| Perfectly inelastic | Conserved | The most is lost; objects stick together | Two train cars coupling |

Impulse explains safety gear. For the same change in momentum, a longer stopping time means a smaller force. Airbags, helmets and bending your knees on landing all stretch out \(\Delta t\).

## Worked examples

**Example 1: impulse on a baseball**

Problem: A 0.145 kg baseball comes toward a bat at 38.0 m/s and leaves in the opposite direction at 46.0 m/s. The contact lasts 1.20 ms. Find the impulse and the average force.

1. Take the direction the ball leaves as positive, so \(v_1 = -38.0\ \text{m/s}\) and \(v_2 = +46.0\ \text{m/s}\).
2. Change in momentum.

   $$
   \Delta p = m(v_2 - v_1) = (0.145\ \text{kg})(46.0 - (-38.0))\ \text{m/s} = 12.18\ \text{kg·m/s}
   $$
3. Average force. Convert 1.20 ms to 0.00120 s.

   $$
   F = \frac{\Delta p}{\Delta t} = \frac{12.18\ \text{kg·m/s}}{0.00120\ \text{s}} = 10{,}150\ \text{N}
   $$

Answer: Impulse 12.2 N·s, average force \(1.02 \times 10^4\) N

**Example 2: cars that stick together**

Problem: A 1500 kg car moving at 20.0 m/s hits a stopped 1000 kg car, and they lock together. How fast do they move just after, and how much kinetic energy is lost?

1. Momentum before: only the first car moves.

   $$
   p = (1500\ \text{kg})(20.0\ \text{m/s}) + 0 = 30{,}000\ \text{kg·m/s}
   $$
2. After, both move together with mass 2500 kg.

   $$
   v' = \frac{30{,}000\ \text{kg·m/s}}{2500\ \text{kg}} = 12.0\ \text{m/s}
   $$
3. Kinetic energy before and after.

   $$
   KE_1 = \tfrac{1}{2}(1500)(20.0)^2 = 300{,}000\ \text{J} \qquad KE_2 = \tfrac{1}{2}(2500)(12.0)^2 = 180{,}000\ \text{J}
   $$
4. Lost to crumpling, heat and sound.

   $$
   300{,}000 - 180{,}000 = 120{,}000\ \text{J}
   $$

Answer: 12.0 m/s, with \(1.20 \times 10^5\) J of kinetic energy lost

**Example 3: pushing apart (recoil)**

Problem: Two skaters stand still on ice. A 60.0 kg skater and a 40.0 kg skater push off each other. The 40.0 kg skater moves away at 3.00 m/s. How fast does the other skater move?

1. Total momentum before is zero, so it is still zero after.

   $$
   0 = (40.0)(3.00) + (60.0)v'
   $$
2. Solve.

   $$
   v' = -\frac{120\ \text{kg·m/s}}{60.0\ \text{kg}} = -2.00\ \text{m/s}
   $$
3. The minus sign means the opposite direction.

Answer: 2.00 m/s in the opposite direction

**Example 4: an elastic collision of equal masses**

Problem: A ball moving at 3.0 m/s hits an identical ball at rest head-on, in a perfectly elastic collision. What are their velocities afterward?

1. For an elastic head-on collision with the second object at rest, the standard results are:

   $$
   v_1' = \frac{m_1 - m_2}{m_1 + m_2} v_1 \qquad v_2' = \frac{2m_1}{m_1 + m_2} v_1
   $$
2. Equal masses make \(m_1 - m_2 = 0\) and \(\tfrac{2m_1}{m_1 + m_2} = 1\).

   $$
   v_1' = 0 \qquad v_2' = v_1 = 3.0\ \text{m/s}
   $$
3. Check: momentum \(m(3.0)\) before and after, and kinetic energy \(\tfrac{1}{2}m(3.0)^2\) before and after.

Answer: The first ball stops and the second moves off at 3.0 m/s

## Common mistakes and how to fix them

- **Ignoring direction.** Momentum is a vector. Fix: pick a positive direction and give velocities the other way a minus sign.
- **Assuming kinetic energy is conserved.** Only in elastic collisions. Fix: momentum is the safe tool for every collision. Use KE only when told it is elastic.
- **Using a final speed for the combined mass with only one mass.** Fix: when objects stick together, divide by the total mass.
- **Forgetting the bounce in impulse problems.** A ball that bounces back changes momentum more than one that stops. Fix: \(\Delta p = m(v_2 - v_1)\) with signs.

**Practice problems**

1. What is the momentum of a 70. kg runner moving at 8.0 m/s?
   A. 560 kg·m/s
   B. 78 kg·m/s
   C. 8.8 kg·m/s
   D. 2240 kg·m/s

   Answer: 560 kg·m/s. \(p = mv = 70. \times 8.0 = 560\) kg·m/s.

2. A 0.50 kg ball of clay moving at 6.0 m/s hits and sticks to a 1.5 kg cart at rest. How fast does the cart move?

   Answer: 1.5 m/s. Momentum before: \(0.50 \times 6.0 = 3.0\) kg·m/s. After, the total mass is 2.0 kg: \(v = 3.0 \div 2.0 = 1.5\) m/s.

3. Which quantity is conserved in every collision with no outside net force?
   A. Kinetic energy
   B. Momentum
   C. Velocity
   D. Speed of each object

   Answer: Momentum. Momentum is conserved in all collisions of an isolated system. Kinetic energy is conserved only in elastic ones.

4. A force of 180 N acts on a cart for 0.050 s. What impulse does it give?

   Answer: 9.0 N·s. \(J = F\Delta t = 180 \times 0.050 = 9.0\) N·s.

5. Why does an airbag reduce the force on a driver in a crash?
   A. It makes the change in momentum smaller
   B. It makes the stopping time longer
   C. It makes the driver's mass smaller
   D. It cancels the driver's momentum

   Answer: It makes the stopping time longer. The driver's change in momentum is the same either way. Since \(F = \Delta p / \Delta t\), a longer stopping time means a smaller force.

## Frequently asked questions

### What is the difference between momentum and kinetic energy?

Both depend on mass and speed, but momentum (\(mv\)) has a direction and kinetic energy (\(\tfrac{1}{2}mv^2\)) does not. Two equal carts moving toward each other at the same speed have zero total momentum but plenty of kinetic energy.

### When is momentum not conserved?

When an outside net force acts on the system during the event, like friction from the ground over a long time. In most collisions the forces between the objects are much larger and act for a very short time, so outside forces barely matter and momentum is conserved.

### How do I solve a collision in two dimensions?

Momentum is conserved in each direction separately. Split every velocity into x and y components, write one conservation equation for x and one for y, and solve them together. This shows up in AP Physics 1 and later courses.

## Sources

- [OpenStax College Physics 2e, 8.2 Impulse](https://openstax.org/books/college-physics-2e/pages/8-2-impulse), accessed 2026-10-01
- [OpenStax College Physics 2e, 8.3 Conservation of Momentum](https://openstax.org/books/college-physics-2e/pages/8-3-conservation-of-momentum), accessed 2026-10-01
- [OpenStax College Physics 2e, 8.4 Elastic Collisions in One Dimension](https://openstax.org/books/college-physics-2e/pages/8-4-elastic-collisions-in-one-dimension), accessed 2026-10-01
- [OpenStax College Physics 2e, 8.5 Inelastic Collisions in One Dimension](https://openstax.org/books/college-physics-2e/pages/8-5-inelastic-collisions-in-one-dimension), accessed 2026-10-01

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## Try asking Ducky

- "Check my signs on the baseball problem. Did I give the bounce-back velocity a minus sign?"
- "Is problem 3 elastic or inelastic? Help me decide before I pick an equation."
- "Quiz me on impulse with a few real-life examples like helmets and catching an egg."

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