Question

A ball is dropped from height $h$ on a floor. If the coefficient of restitution is $e$, find the total distance covered by the ball before coming to rest on the ground.

• A. $\left(\frac{1-e}{1+e}\right)h$
• B. $\left(\frac{1+e}{1-e}\right)h$
• C. $\left(\frac{1-e^2}{1+e^2}\right)h$
• D. $\left(\frac{1+e^2}{1-e^2}\right)h$

Answer 1

The correct option is D $\left(\frac{1+e^2}{1-e^2}\right)h$



Using Newton's law of impact,

$v_2-v_1=e(u_1-u_2)......\left(1\right)$

For, reaching height $h_1$

$v_2=0;u_2=0;u_1=-\sqrt{2gh}$

$\left(\text{-ve sign indicates the downward direction}\right)$

Putting this values in (1) we get,

$-v_1=e{u}_1$

$\Rightarrow \sqrt{2g{h}_1}=e\sqrt{2gh}$

$\Rightarrow h_1=e^{2}h$

Similarly, for reaching $h_2$, $h_2=e^{4}h$ and so on,

So, the total distance travelled by the ball is,

$d=h+2h_1+2h_2+....+$

$\therefore d=h+2e^{2}h+2e^{4}h+2e^{6}h+...$

$\Rightarrow d=h+2e^{2}h(1+e^2+e^4+...)$

$\Rightarrow d=h+2e^{2}h\left(\frac{1}{1-e^2}\right)$

$\Rightarrow d=\frac{h(1-e^2)+2e^{2}h}{1-e^2}$

$\therefore d=h\left(\frac{1+e^2}{1-e^2}\right)$

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