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
$V_\text{upward} = eV_\text{downward}$
If the ball is initially dropped from a height of $0.4\, m$ 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^{2 n} h $
$\therefore t=\sqrt{\frac{2 h}{g}}+2 \sqrt{\frac{2 h e^{2}}{g}}+2 \sqrt{\frac{2 h e^{4}}{g}}+$
$=\sqrt{\frac{2 h}{g}}\left[1+2 e+2 e^{2}+\ldots\right]$
$\therefore t=10\, sec $
$\therefore 10=\sqrt{\frac{2(0 \cdot 4)}{10}}\left(\frac{1+e}{1-e}\right)$
Solving above relation
$\therefore e=\frac{25 \sqrt{2}-1}{25 \sqrt{2}+1} \approx \frac{17}{18}$