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 total change in momentum of ball when it stops rebounding is (Here, $e$ is coefficient of restitution between the ball and the floor)

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

Answer 1

The correct option is D $m\sqrt{2gh}\left(\frac{1+e}{1-e}\right)$

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, $v_3=e^3{v}_0$

Change in momentum of ball after $1^{\text{st}}$ collision,
$\Delta p_1=m{v}_1-(-m{v}_0)=m{v}_1+m{v}_0$
Change in momentum of ball after $2^{\text{nd}}$ collision,
$\Delta p_2=m{v}_2-(-m{v}_1)=m{v}_2+m{v}_1$
$\Delta p_3=m{v}_3-(-m{v}_2)=m{v}_3+m{v}_2$
Similarly change in momentum of ball after $n^{\text{th}}$ collision,
$\Delta p_n=m{v}_n-(-m{v}_{n-1})=m{v}_n+m{v}_{n-1}$

Therefore, change in momentum of ball:
$\Delta p=\Delta p_1+\Delta p_2+\Delta p_3+......+\Delta p_n...\left(i\right)$
$=m{v}_0+2m{v}_1+2m{v}_2+.....+2m{v}_{n-1}+m{v}_n$
$=m{v}_0+2m\left(e{v}_0\right)+2m\left(e^2{v}_0\right)+.....+2m\left(e^{n-1}{v}_0\right)+m\left(e^n{v}_0\right)$
where number of collisions with floor $n\to \infty$
$\therefore \Delta p=m{v}_0+2me{v}_0(1+e+e^2+e^3+...\infty )...\left(ii\right)$
Applying formula for sum of infinite GP, where $a=1,r=e$
$\sum =\frac{a}{1-r}=\frac{1}{1-e}...\left(iii\right)$

From Eq.$\left(ii\right)\&\left(iii\right)$:
$\Delta p=m{v}_0[1+2e\left(\frac{1}{1-e}\right)]$
$\Delta p=m{v}_0\left(\frac{1+e}{1-e}\right)$
$\therefore \Delta p=m\sqrt{2gh}\left(\frac{1+e}{1-e}\right)$