A plastic ball falls from a height h_1 = 1 m onto a…

A plastic ball falls from a height $h_1 = 1\ \mathrm{m}$ onto a horizontal plate and bounces. The time from release to the second impact is $t = 1\ \mathrm{s}$. Find the coefficient of restitution $k$ (ratio of the rebound speed to the impact speed).

Explanation of Method

Coefficient of restitution from the bounce time.

Step 1: Times

First fall: $t_1 = \sqrt{2h_1/g} = 0.4515\ \mathrm{s}$. The ball rebounds at $k$ times its impact speed, so the bounce lasts $k$ times as long as a fall from $h_1$ up and down: $2kt_1$.

\[ t = t_1 + 2kt_1 = t_1\,(1 + 2k). \]

Step 2: Coefficient

\[ k = \frac{1}{2}\left(\frac{t}{t_1} - 1\right) = \frac{1}{2}\left(\frac{1}{0.4515} - 1\right). \]

\[k \approx 0.61\]

The ball rises to $k^2h_1 \approx 0.37\ \mathrm{m}$ after the first bounce.

$y\ (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $1$ $0.37$ $1$ $0.45$ $h_2 = k^2h_1$

Height against time. The first fall takes 0.45 s; the bounce rises to $k^2h_1 \approx 0.37\ \mathrm{m}$ and comes back down at exactly 1 s.

Result

\[\boxed{\,k \approx 0.61\,}\]

Remark

Check. The original result is correct.