Q155 / 155 · WBJEE 2017 · Physics
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 $V_{upward}=eV_{downward}$
If the ball is initially dropped from a height of $0.4 \mathrm{m}$ 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$\frac{13}{18}$
- Correct answerD$\frac{17}{18}$
Solution
Given, steel ball bounces on a steel ple held horizontally, so $$
V_{\text {upward }}=e V_{\text {downward }}
$$
According to the relation,
$h_{n}=e^{2 n} h$
$\begin{aligned} \therefore \quad t &=\sqrt{\frac{2 h}{g}}+2 \sqrt{\frac{2 h e^{2}}{g}}+2 \sqrt{\frac{2 h e^{4}}{g}}+\ldots \\ &=\sqrt{\frac{2 h}{g}}\left[1+2 e+2 e^{2}+\ldots\right] \\ &=\sqrt{\frac{2 h}{g}}\left[\frac{1+e}{1-e}\right] \\ \therefore \quad t &=10 \mathrm{sec} \\ \therefore 10 &=\sqrt{\frac{2(0 \cdot 4)}{10}}\left(\frac{1+e}{1-e}\right) \end{aligned}$
Solving above relation $\therefore e=\frac{25 \sqrt{2}-1}{25 \sqrt{2}+1}=\frac{17}{18}$
Hence, the correct answer is option: 'D'.