Bounce lab - 2D height-time trace

Each bounce returns a fixed fraction e of the impact speed, so peak heights shrink as e^2 per bounce and the ball stops after a finite time.

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Presets

Playback

Parameters

Model checks

The science

Coefficient of restitution

For a collision with a fixed floor the normal speed after impact is v' = e v, with 0 ≤ e ≤ 1. A ball dropped from height h0 hits at v0 = √(2 g h0) and rises to h1 = (e v0)2 / 2g = e2 h0. After n bounces hn = e2n h0, so e is recovered from any two successive peaks as e = √(hn+1 / hn).

Energy and time

Each bounce keeps the fraction e2 of the mechanical energy and dissipates 1 - e2 as heat, sound and deformation. Flight times form a geometric series: the first fall takes t0 = √(2 h0/g) and the n-th flight takes 2 en t0, so the ball comes to rest after the finite time T = t0 (1 + e)/(1 - e) although it bounces infinitely often (Zeno behaviour).

Checks

A fine-step numerical integration with collision detection is compared with the closed forms for the fifth peak, the total rest time, and the restitution recovered from heights.

Honest simplifications: e is constant, whereas real balls have an e that falls with impact speed; no air drag, no spin or friction (the 3D page models spin), and the ball is a point mass. The last bounces are cut off at a height of 0.1 mm.

What to try
  • Press Tennis ball and then Steel ball: a high e makes many bounces and a long rest time.
  • Select the Moon preset: the peaks are the same shape but everything takes 2.5 times longer.
  • Set e to 0.5 and read the third bar: it should be h0 times 0.5 to the sixth power.
  • Compare the energy graph with the peak graph: they are the same curve in different units.
Height against time
Peak height of each bounce (bars: simulated, line: h0 e^(2n))
Mechanical energy kept after n bounces (e^(2n))