Question

A ball is dropped on to a horizontal plate from a height $h=9m$ above it. If the coefficient of restitution is $e=$ $1/2,$ the total distance travelled before the ball comes to rest is

• A. 10 m
• B. 15 m
• C. 20 m
• D. 25 m

Answer 1

Answer: (B)

The impact between the ball and the plate is direct impact. If $u$ is the velocity of the ball before impact, its velocity after impact is $eu,$ where $e$ is the coefficient of restitution.
But $u=\sqrt{2gh}$ since the ball falls from a height $h$.
So the velocity of the ball at the first rebound
$=v_1=eu=e\sqrt{2gh}$
The height $h_1$ to which it rises after first rebound
$h_1=\frac{v_1^2}{2g}=\frac{e^2\cdot 2gh}{2g}=e^{2}h$
In general, the height $h_n$ to which the ball rises after $n^{\text{th }}$ rebound is given by $h_n=e^{2n}h$
Total distance travelled
$=h+2h_1+2h_2+2h_3+\dots ..+2h_n+\dots ..$ to $\infty$
$=h+2\left(e^{2}h+e^{4}h+e^{6}h+\dots ..\right)=h+2h{e}^2\left(1+e^2+e^4+\dots ..\right)$
$=h+\frac{2h{e}^2}{1-e^2}=h\left(1+\frac{2e^2}{1-e^2}\right)$
$=\frac{h\left(1+e^2\right)}{\left(1-e^2\right)}=\frac{9\left(1+{\left(\frac{1}{2}\right)}^2\right)}{\left(1-{\left(\frac{1}{2}\right)}^2\right)}$
$=15m$