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
$V_{\text{upward}}=e{V}_{\text{downward}}$
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

Answer: (D)

Given, steel ball bounces on a steel plate held horizontally, so
$V_{\text{upward }}=e{V}_{\text{downward }}$
According to the relation,
$h_n=e^{2n}h$
$\therefore t=\sqrt{\frac{2h}{g}}+2\sqrt{\frac{2h{e}^2}{g}}+2\sqrt{\frac{2h{e}^4}{g}}+$
$=\sqrt{\frac{2h}{g}}\left[1+2e+2e^2+\dots \right]$
$\therefore t=10sec$
$\therefore 10=\sqrt{\frac{2(0\cdot 4)}{10}}\left(\frac{1+e}{1-e}\right)$
Solving above relation
$\therefore e=\frac{25\sqrt{2}-1}{25\sqrt{2}+1}\approx \frac{17}{18}$