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Sandbox Physics

M050 · Every bounce tells a story

Bouncing Ball & Restitution

Where does the missing height go? It is lost in each impact. The coefficient of restitution is the rebound speed divided by the impact speed, and the height a ball reaches grows with its speed squared. Each new peak is the previous one times the coefficient squared, so a ball with a coefficient of 0.9 keeps 81 percent of its height on every bounce.

Drop a ball, watch the impact in slow motion and compare elastic, lossy and sticking collisions. Record apex heights to estimate the coefficient of restitution, and see how the ideal model fits infinitely many bounces into a finite time.

Interactive modelBouncing Ball & Restitution
Motion stateReady\text{Ready}
Ball-bottom heightReady\text{Ready}
Upward velocityReady\text{Ready}
Energy transferred at impactsReady\text{Ready}
Ideal collapse timeReady\text{Ready}
Restitution from recorded heightsReady\text{Ready}
Energy balance residualReady\text{Ready}
Recorded apex heightsReady\text{Ready}

Physics tutorial

Bouncing Ball & Restitution

BackgroundA fixed floor exchanges momentum with a ball at each impact.

Why it mattersEach bounce is lower than the last, and the lost height tells you how much mechanical energy left the ball.

Start with the essentials

Focus question
Where does the missing height go?
One-sentence intuition
The ratio of successive bounce heights is the square of the speed ratio.

Core mathematical model

Instantaneous impact

v+=−ev−,J=m(v+−v−)v^+=-e v^-,\qquad J=m(v^+-v^-)

The floor takes the opposite impulse; the ball alone does not conserve momentum.

Apex height and energy after each bounce

Hn=e2nH0,En=mgHnH_n=e^{2n}H_0,\qquad E_n=mgH_n

Uniform gravity and no drag are assumed between instantaneous impacts.

Infinitely many bounces in a finite time

t∗=2H0g+2e2gH0g(1−e)t_*=\sqrt{\frac{2H_0}{g}}+\frac{2e\sqrt{2gH_0}}{g(1-e)}

With zero restitution the first impact stops the ball. With a restitution of one, the bounces never die out.

Restitution from recorded heights

e^=∑nHnHn+1∑nHn2\widehat e=\sqrt{\frac{\sum_n H_nH_{n+1}}{\sum_n H_n^2}}

The fit is a line through the origin, using resolved pairs of successive recorded heights. Noise in both heights of a pair can bias the estimate.

Common difficulties

The floor pushes from outside

Typical misconceptionThe ball conserves momentum when it hits a fixed floor.

Better mental modelThe floor gives the ball an external impulse, so the ball's momentum changes. Count the energy transferred at impact as a separate term.

Run the experiment

  1. 01

    Set the height

    Drag the gold handle or use its keyboard slider.

    What to observe: The scene and predicted motion update together.
  2. 02

    Follow the event

    Play, pause and jump to the next event.

    What to observe: Height changes smoothly, but an impact can change the velocity instantly.
  3. 03

    Track the energy

    Drag the time between turning points.

    What to observe: Stored, kinetic, gravitational and transferred energy always add up to the same total.
  4. 04

    Measure a run

    Record apex heights, export the CSV and compare the inferred parameter.

    What to observe: Replaying preserves the same measurements; changing physical parameters clears them.
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