Collisions and Coefficient of Restitution
On this page
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.