From Stick to Bounce — the Coefficient of Restitution
Two balls of the same size lie in a straight groove on a lab bench. Ball A slides in from the left, without spinning, towards ball B, which can wait at rest, slide towards A or run ahead of it. Five sliders set the two masses, the two velocities and the coefficient of restitution e — and with e, what the balls are made of: dull putty at e = 0, rubber in the middle, polished steel at e = 1. Arrows show each ball's velocity, a bracket in front of the groove shows how fast the gap between them closes and opens again, and a chart tracks how much kinetic energy each value of e loses. Press play and the collision runs once, eight times slower than real life and slower still with the playback chips; the timeline takes you back to any moment.
How to use this simulation
- Mass of A, mA — 0.5 to 3.0 kg, starting at 1.0 kg.
- Velocity of A, vA — +1.0 to +6.0 m/s (to the right), starting at +4.0 m/s.
- Mass of B, mB — 0.5 to 3.0 kg, starting at 1.5 kg.
- Velocity of B, vB — −4.0 to +2.0 m/s, starting at 0 (at rest); negative means moving left, towards A.
- Coefficient of restitution e — 0 (the balls stick) to 1 (perfectly elastic), starting at 0.25; the balls' material changes with it. A change keeps the timeline where it is.
- Play/pause, back to the start, a timeline you can drag to any moment, and ½×, ¼×, ⅛× slow motion. At 1× the collision already runs eight times slower than real life, so ⅛× shows the squash 64 times slower.
- Drag anywhere on the scene to look around.
What to look for
- At the starting values the gap closes at 4.0 m/s and opens at 1.0 m/s; A leaves at +1.0 m/s and B at +2.0 m/s, and 4.5 J of the 8.0 J is lost.
- At e = 0 the putty balls stick and move on together at 1.6 m/s, losing 4.8 J, the most this collision can lose.
- At e = 1 A bounces back at −0.80 m/s and B leaves at +3.2 m/s; all 8.0 J of kinetic energy is still there.
- Send B towards A at −2.0 m/s: they now close at 6.0 m/s, A bounces back at −0.50 m/s and 10.1 J of the 11 J is lost — yet the momentum is +1.0 kg m/s before and after.
- At ⅛× watch the squash: the dent left on each ball shrinks as e grows — putty keeps its dent, steel springs back perfectly round — and the point on the chart runs along the curve ½μ(vA − vB)²(1 − e²).
- Set vA below vB and the card reports no collision: A never catches B, and nothing is lost.
- Drag the timeline through the contact: the balls flatten while the gap is shut, and at the deepest squash they move at the same speed for an instant.
The physics behind it
Take rightward as positive. Conservation of momentum alone cannot give two unknown velocities, so a second equation is added — the restitution equation, v′A − v′B = −e(vA − vB): the balls close at vA − vB and part at e times that. At the starting values (1.0 kg at +4.0 m/s into 1.5 kg at rest, e = 0.25), solving it with 1.0 × 4.0 = 1.0 v′A + 1.5 v′B gives v′A = +1.0 m/s and v′B = +2.0 m/s. The kinetic energy falls from 8.0 J to 0.5 + 3.0 = 3.5 J, so 4.5 J is lost. In the model the squeezed balls store the whole relative kinetic energy, ½μ(vA − vB)² with μ = mA mB/(mA + mB), and give back only e² of it, so the loss is ½μ(vA − vB)²(1 − e²): 4.8 J at e = 0, when the putty balls stick and move on at 1.6 m/s, and nothing at e = 1, when A bounces back at 0.80 m/s and B leaves at 3.2 m/s. The masses and velocities decide how much energy there is to lose; e alone sets the fraction that is lost. The momentum is the same before and after at every setting, and if A is not faster than B the gap never closes and there is no collision.