# Collisions and Coefficient of Restitution

> Collisions and coefficient of restitution for JEE Physics: momentum conservation, elastic and inelastic cases, energy loss and bouncing heights.

- Canonical URL: https://prepelephant.com/topics/jee/physics/collisions-and-coefficient-of-restitution
- Exam / course: JEE · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Collisions and Coefficient of Restitution", PrepElephant, https://prepelephant.com/topics/jee/physics/collisions-and-coefficient-of-restitution

## Direct answer

In every collision between isolated bodies, linear momentum is conserved; kinetic energy survives only if the collision is elastic. The coefficient of restitution, e = (relative velocity of separation)/(relative velocity of approach) = |v2 − v1|/|u1 − u2|, grades the bounce: e = 1 elastic, e = 0 perfectly inelastic (the bodies coalesce), 0 < e < 1 inelastic. Two conservation laws then solve any one-dimensional collision: momentum plus the restitution equation. For a perfectly inelastic collision the kinetic energy lost is (1/2) mu^2 (with mu the reduced mass m1 m2/(m1 + m2) and u the approach speed); for a ball dropped from height h, the height after the nth bounce is e^(2n) h.

## What you must remember

- **Always conserve momentum:** with no external force during the brief impact, m1 u1 + m2 u2 = m1 v1 + m2 v2 holds for every e; kinetic energy conservation holds only for e = 1.
- **Restitution values:** e = 1 for an ideal elastic collision; steel balls on steel approach 0.9-plus, a cricket ball on pitch sits near 0.6, and putty on a wall gives 0.
- **Equal-mass elastic 1D rule:** the bodies exchange velocities exactly; if the second was at rest, the first stops dead — billiards in one line.
- **Energy loss, perfectly inelastic:** loss = (1/2) (m1 m2/(m1 + m2)) u(rel)^2; if the target is very heavy (wall, earth), the light body loses almost all its kinetic energy.
- **Ballistic pendulum logic:** during embedding only momentum is conserved; the subsequent swing conserves energy — never mix the two stages into one equation.
- **Two-dimensional elastic collision:** equal masses with one initially at rest scatter at 90 degrees to each other — a JEE staple proved by vector addition of momenta.
- **Bounce heights:** h(after nth bounce) = e^(2n) h; total distance travelled before stopping is h(1 + e^2)/(1 − e^2).
- **Pattern note:** Main asks direct before-after numericals; Advanced composes collisions with springs, variable e and centre-of-mass reasoning.

## One formula answers them all

A 2 kg block at 6 m/s strikes a 4 kg block at rest; e = 0.5. Momentum: 12 = 2v1 + 4v2. Restitution: 0.5 = (v1 − v2)/(0 − 6) with sign care — separation speed is v2 − v1 when the heavy block moves forward, giving v2 − v1 = 3. Solving, v1 = 0 m/s and v2 = 3 m/s. The energy check: initial KE = 36 J, final = 18 J, so exactly half the kinetic energy left as heat and sound — the restitution equation quietly guarantees this is consistent.

The ballistic pendulum shows why laws must be staged. A bullet of mass m at speed u embeds in a block M hanging at rest: during embedding, momentum gives the combined body V = m u/(m + M); the rise to height h then obeys (1/2)(m + M)V^2 = (m + M)g h. Writing (1/2)m u^2 = (m + M)g h in one step is the perennial wrong answer, and it overestimates h by the large factor (m + M)/m.

## Where students slip

Sign discipline in the restitution ratio causes most losses: e uses relative speeds, so both numerator and denominator must be separation and approach speeds as positive quantities — the safest habit is |v2 − v1|/|u1 − u2|. The second slip is conserving kinetic energy during an inelastic event; the phrase "perfectly inelastic" is a signal to conserve momentum alone and then compute the loss from the reduced-mass formula rather than assume zero final kinetic energy. Third, in explosion problems (the reverse of collision), momentum before equals momentum after — usually zero — while kinetic energy increases, drawn from chemical or nuclear store; candidates who conserve KE across an explosion find impossible answers.

## Frequently asked questions

### What does the coefficient of restitution physically measure?

The ratio of relative separation speed to relative approach speed, e = |v2 − v1|/|u1 − u2|, measuring how much relative motion a collision restores: 1 fully, 0 not at all.

### Why is kinetic energy not conserved in an inelastic collision?

Part of the macroscopic kinetic energy converts into heat, sound and permanent deformation; total energy is conserved, but the kinetic part alone is not.

### How much kinetic energy is lost when two bodies stick together?

(1/2) (m1 m2/(m1 + m2)) u(rel)^2, the reduced mass times half the square of the approach speed, which is maximal loss for the given approach speed.

### To what height does a ball rise after several bounces?

After the nth bounce it reaches e^(2n) times the original height h, since each bounce multiplies speed by e and height by e^2.

### Why does a bullet-block pendulum need two conservation stages?

Momentum is conserved during embedding, then mechanical energy during the rise; applying either across both stages ignores the inelastic conversion in between.