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.
Como usar esta simulação
- 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.
O que observar
- 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.
A física por trás
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.