In every collision momentum is conserved. Kinetic energy is a different story, and comparing it before and after is how physicists sort collisions into kinds.
| Kind | Momentum | Kinetic energy | Example |
|---|---|---|---|
| Elastic | conserved | conserved | steel balls, air-track gliders with magnets |
| Inelastic | conserved | some lost | a tennis ball on a racket, most car bumpers |
| Perfectly inelastic | conserved | the most lost (they stick) | clay, Velcro carts, coupling train cars |
Key idea
To classify a collision, compute the total kinetic energy before and after. Equal: elastic. Less after: inelastic. Stuck together: perfectly inelastic. In an elastic collision the objects also separate as fast as they approached: $v_2' - v_1' = v_1 - v_2$.
Why it works
Where does that separation rule come from? Write momentum as $m_1(v_1 - v_1') = m_2(v_2' - v_2)$ and kinetic energy as $m_1(v_1^2 - v_1'^2) = m_2(v_2'^2 - v_2^2)$. Factor the squares: $$m_1(v_1 - v_1')(v_1 + v_1') = m_2(v_2' - v_2)(v_2' + v_2).$$ Divide by the momentum equation and you are left with $v_1 + v_1' = v_2 + v_2'$, which rearranges to "separation speed equals approach speed". For a target at rest this gives $$v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1, \qquad v_2' = \frac{2m_1}{m_1 + m_2}v_1.$$
Worked examples
Worked example: Classify it
A $2.0\ \text{kg}$ cart at $3.0\ \text{m/s}$ hits a $1.0\ \text{kg}$ cart at rest. Afterward they move at $1.0\ \text{m/s}$ and $4.0\ \text{m/s}$, both forward. What kind of collision is it?
- Momentum check: before $6.0$, after $$(2.0)(1.0) + (1.0)(4.0) = 6.0\ \text{kg·m/s}.$$ Good.
- $K$ before: $\tfrac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}$. $K$ after: $$\tfrac{1}{2}(2.0)(1.0)^2 + \tfrac{1}{2}(1.0)(4.0)^2 = 1.0 + 8.0 = 9.0\ \text{J}.$$
- Kinetic energy is unchanged, so the collision is elastic. (Separation check: $4.0 - 1.0 = 3.0$ = approach speed.)
Answer: Elastic
Worked example: A light ball hits a heavy one
A $1.0\ \text{kg}$ ball at $6.0\ \text{m/s}$ hits a $2.0\ \text{kg}$ ball at rest, elastically. Find both velocities after.
- $v_1' = \dfrac{1.0 - 2.0}{3.0}(6.0) = -2.0\ \text{m/s}$: the light ball bounces back.
- $v_2' = \dfrac{2(1.0)}{3.0}(6.0) = +4.0\ \text{m/s}$.
- Check $K$: $18\ \text{J}$ before; $\tfrac{1}{2}(1)(2)^2 + \tfrac{1}{2}(2)(4)^2 = 2 + 16 = 18\ \text{J}$ after.
Answer: $-2.0\ \text{m/s}$ and $+4.0\ \text{m/s}$
Real collisions sit between the extremes. Two steel balls come close to elastic; a tennis ball on a hard floor keeps only part of its kinetic energy; colliding cars are strongly inelastic. Try every kind with the bounciness slider in the Momentum and Collisions lab of the Science Lab.
Common mistake
Thinking kinetic energy, like momentum, is always conserved. Momentum is protected by the third law; kinetic energy has no such protection and is usually partly turned into thermal energy and sound. Only use "$K$ before = $K$ after" when a problem says the collision is elastic (or you have checked it). Quick test: if the objects stick, it cannot be elastic.