Question

A small steel ball bounces on a steel plate held horizontally. On each bounce the speed of the ball arriving at the plate is reduced by a factor $e$ (coefficient of restitution) in the rebound, so that
$V_{upward}=e{V}_{downward}$
If the ball is initially dropped from a height of $0.4m$ above the plate and if $10$ seconds later the bouncing ceases, the value of $e$ is

• A. $\sqrt{\frac{2}{7}}$
• B. $\frac{3}{4}$
• C. $\frac{13}{18}$
• D. $\frac{17}{18}$

Answer 1

The correct option is D $\frac{17}{18}$

$h_n=e^{2n}.h$
$\therefore t=\sqrt{\frac{2h}{g}}+2\sqrt{\frac{2h{e}^2}{g}}+2\sqrt{\frac{2h{e}^4}{g}}+....=\sqrt{\frac{2h}{g}}(1+2e+2e^2+...)$
$=\sqrt{\frac{2h}{g}}\left[\frac{1+e}{1-e}\right]\therefore 10=\sqrt{\frac{2\left(0.4\right)}{10}}\left[\frac{1+e}{1-e}\right]\therefore e=\frac{25\sqrt{2}-1}{25\sqrt{2}+1}\approx \frac{17}{18}$