Question

A ball is dropped on to a horizontal plate from a height $h=9 \,m$ 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 $e u,$ where $e$ is the coefficient of restitution.
But $u=\sqrt{2 g h}$ since the ball falls from a height $h$.
So the velocity of the ball at the first rebound
$=v_{1}=e u=e \sqrt{2 g h}$
The height $h_{1}$ to which it rises after first rebound
$h_{1}=\frac{v_{1}^{2}}{2 g}=\frac{e^{2} \cdot 2 g h}{2 g}=e^{2} h$
In general, the height $h_{n}$ to which the ball rises after $n^{\text {th }}$ rebound is given by $h_{n}=e^{2 n} h$
Total distance travelled
$=h+2 h_{1}+2 h_{2}+2 h_{3}+\ldots . .+2 h_{n}+\ldots . .$ to $\infty$
$=h+2\left(e^{2} h+e^{4} h+e^{6} h+\ldots . .\right)=h+2 h e^{2}\left(1+e^{2}+e^{4}+\ldots . .\right)$
$=h+\frac{2 h e^{2}}{1-e^{2}}=h\left(1+\frac{2 e^{2}}{1-e^{2}}\right)$
$=\frac{h\left(1+e^{2}\right)}{\left(1-e^{2}\right)}=\frac{9\left(1+\left(\frac{1}{2}\right)^{2}\right)}{\left(1-\left(\frac{1}{2}\right)^{2}\right)}$
$=15\, m$