Question

A ball is dropped from a height $h$ on to a floor. If in each collision its speed becomes $e$ times of its striking value, then time taken by ball to stop rebounding is (Here, $e$ is coefficient of restitution between the ball and the floor)

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

Answer 1

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

When the ball is dropped $(u=0)$ from height $h$, time taken to reach the floor will be,
$t_0=\sqrt{\frac{2h}{g}}$
Its speed while striking floor is
$v_0=\sqrt{2gh}$
After $1^{\text{st}}$collision with floor, its rebound speed will be:
$v_1=e{v}_0$
Total time $\left(t_1\right)$ between $1^{\text{st}}\&2^{\text{nd}}$ collision is equal to time of flight of ball, after $1^{\text{st}}$ collision.
$\therefore t_1=\frac{2v_1}{g}$
Similarly, total time $\left(t_2\right)$ between $2^{\text{nd}}\&3^{\text{rd}}$ collision is equal to time of flight of ball, after $2^{\text{nd}}$ collision.
$t_2=\frac{2v_2}{g}$, where $v_2=e{v}_1$
(since ball will hit the ground at same velocity with which its starts)
Now, $v_2=e{v}_1=e\left(e{v}_0\right)=e^2{v}_0$
Similarly, $v_3=e^3{v}_0$

Thus, total time after which ball will stop rebounding is given by,
$\Delta T=t_0+t_1+t_2+t_3+.....\infty$
$\Delta T=t_0+\frac{2v_1}{g}+\frac{2v_2}{g}+\frac{2v_3}{g}+....\infty$
$\Delta T=t_0+\frac{2e{v}_0}{g}+\frac{2e^2{v}_0}{g}+\frac{2e^3{v}_0}{g}+....\infty$
Putting value of $v_0$ in above equation,
$\Delta T=\sqrt{\frac{2h}{g}}+\frac{2e{v}_0}{g}(1+e+e^2+e^3+.....\infty )$
$\Delta T=\sqrt{\frac{2h}{g}}+2e\sqrt{\frac{2h}{g}}(1+e+e^2+e^3+.....\infty )$
$\Delta T=\sqrt{\frac{2h}{g}}[1+2e(1+e+e^2+e^3+.....\infty )]....\left(i\right)$
Applying formula for sum of infinite GP, where $a=1,r=e$
$\sum =\frac{a}{1-r}=\frac{1}{1-e}...\left(ii\right)$
From Eq $\left(i\right)\&\left(ii\right)$:

$\Delta T=\sqrt{\frac{2h}{g}}[1+2e\left(\frac{1}{1-e}\right)]$
$\therefore \Delta T=\sqrt{\frac{2h}{g}}\left(\frac{1+e}{1-e}\right)$