Object A (mass m) moves at 10 m/s toward stationary object B (also mass m). After the collision, A moves at 2 m/s and B moves at 8 m/s in the same direction. What is the coefficient of restitution?
Ae = (2 + 8)/10 = 1.0 — this is a perfectly elastic collision
Be = (8 − 2)/(10 − 0) = 0.6
Ce = 10/(8 + 2) = 1.0 — same result, confirming elastic collision
De cannot be calculated without knowing the actual masses
The coefficient of restitution is e = (relative speed of separation)/(relative speed of approach). Before: A approaches B at 10 − 0 = 10 m/s relative speed. After: B moves away from A at 8 − 2 = 6 m/s relative speed. So e = 6/10 = 0.6. Option A sums the final speeds instead of taking their difference — a common error that conflates momentum with relative velocity. Momentum conservation checks out (10m = 2m + 8m ✓), confirming these are valid post-collision velocities. Note e = 0.6 < 1, consistent with a real (not perfectly elastic) collision.
Question 2 Multiple Choice
A student claims that for a collision with known e = 0.7, momentum conservation alone is sufficient to find both final velocities. Why is this incorrect, and what additional equation is needed?
AMomentum is not conserved in inelastic collisions; you need energy conservation instead
BMomentum conservation gives one equation for two unknowns; the restitution equation e = (relative separation speed)/(relative approach speed) provides the necessary second equation
CFor e < 1, the standard collision equations don't apply — you need a separate energy-loss formula
DThe student is correct — momentum conservation is sufficient for any collision problem
For a collision between two objects, there are two unknown final velocities. Momentum conservation gives exactly one equation linking them. Without a second independent equation, the system is underdetermined — infinitely many pairs of final velocities conserve momentum. The restitution equation e = v_rel,after / v_rel,before is that second equation. Together they form a solvable 2×2 linear system. This is precisely why the coefficient of restitution is useful: energy conservation fails for inelastic collisions (the energy lost is unknown), but e is a given material property that provides the needed constraint.
Question 3 True / False
A ball dropped from height h rebounds to height h'. The coefficient of restitution for this ball-floor collision equals h'/h.
TTrue
FFalse
Answer: False
e = √(h'/h), not h'/h. Height is proportional to kinetic energy (h ~ v²), so the impact speed is proportional to √h and the rebound speed to √h'. The coefficient of restitution is the ratio of speeds, not energies: e = √h'/√h = √(h'/h). Equivalently, e² = h'/h. A ball with e = 0.9 rebounds to h' = e²h = 0.81h, not 0.9h. This is a common error that confuses the ratio of heights (which gives e²) with the ratio of speeds (which gives e).
Question 4 True / False
The coefficient of restitution is a property of a single material — a rubber ball has the same e regardless of what surface it bounces against.
TTrue
FFalse
Answer: False
The coefficient of restitution characterizes the pair of materials in contact, not a single material alone. A rubber ball bouncing on hardwood, concrete, carpet, or another rubber ball will have different e values because the energy loss during impact depends on how both surfaces deform and interact. Specifying e requires specifying both surfaces. This is why engineering specifications for e always name both materials, and why a ball may bounce very differently on different surfaces even though the ball itself hasn't changed.
Question 5 Short Answer
Why is the coefficient of restitution useful specifically for solving inelastic collision problems, when energy conservation cannot be used?
Think about your answer, then reveal below.
Model answer: In any collision that is not perfectly elastic, kinetic energy is lost — converted to heat, sound, and deformation. The amount lost is not known in advance, so energy conservation cannot be written as a useful equation (it would introduce the unknown energy loss as a third variable). The coefficient of restitution sidesteps this problem: e is a known material property, and the equation e = (relative separation speed)/(relative approach speed) provides a second independent equation alongside momentum conservation. Two equations for two unknown final velocities makes the system exactly solvable without needing to know how much energy was lost.
This is the conceptual core of why e is introduced as a concept. For perfectly elastic collisions, energy conservation provides the second equation. For perfectly inelastic collisions, the objects stick together, so there is only one unknown final velocity and momentum alone suffices. For all intermediate cases — the vast majority of real collisions — neither energy conservation nor the sticking condition applies, and e fills the gap.