Question

If a ball is dropped from rest, it bounces from the floor repeatedly. The coefficient of restitution is $0.5$ and the speed just before the first bounce is $5m/s$. The total time (in $s$) taken by the ball to come to rest finally is

Answer 1

Hint: “Time to reach maximum height after every collision”

Formula used:

$t=\sqrt{\frac{2h}{g}}$

$e=\frac{\text{Velocity of seperation}}{\text{Velocity of approach}}$

Solution:

$v=0+gt\Rightarrow t=0.5\text{sec}$

After first collision:

Speed becomes $5\left(0.5\right)=2.5m/s$

$t_1=2\left(0.25\right)=0.5$

$t_2=2\left(0.125\right)=0.25$

$t_3=0.125$

[where $t_i$ is the time taken to complete the $i^{th}$ to and fro motion after collision]

Total time = $0.5+[0.5+0.25+0.125+...]$

(Since above is a G.P. with $a=0.5$ and $r=0.5$)
$=0.5+\frac{0.5}{1-0.5}$
$=0.5+1=1.5sec$

Final Answer : $1.5$.