Question

A particle is dropped from a height $h$ on a floor, if the coefficient of restitution is $e$. Find the time taken by ball before second bounce.

• A. $\sqrt{\frac{2h}{g}}(1+2e)$
• B. $\sqrt{\frac{2h}{g}}(1+\frac{2}{e})$
• C. $\sqrt{\frac{2h}{g}}(1+e)$
• D. $\sqrt{\frac{2h}{g}}(2+e)$

Answer 1

The correct option is A $\sqrt{\frac{2h}{g}}(1+2e)$


As the particle is dropped from height $h$. Thus, time taken to hit the floor is,

$t_1=\sqrt{\frac{2h}{g}}$

Velocity of the particle when hits the floor is, $v=\sqrt{2gh}$

Velocity with which rebound first time is, $v^{\prime }=ev=e\sqrt{2gh}$

The total time taken by the particle to hit the floor at the second time is,

$t_2=2\times \text{time taken to reach the maximum height}$

$t_2=2\times \frac{v^{{}^{\prime }}}{g}=\frac{2e\sqrt{2gh}}{g}$

$\therefore t_2=2e\sqrt{\frac{2h}{g}}$

Now, the time taken by the ball before the second bounce is,

$t=t_1+t_2$

$t=\sqrt{\frac{2h}{g}}+2e\sqrt{\frac{2h}{g}}=\sqrt{\frac{2h}{g}}(1+2e)$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.

Why this question ? Key concept: After, the rebound velocity changes by a factor of $e$. Thus, it does not rebounce back to original height. Hence, time taken also changes.