Question

A glass marble dropped from a certain height above the horizontal surface reaches the surface in time $t$ and then continues to bounce up and down. The time in which the marble finally comes to rest will be (coefficient of restitution is $e$):

• A. $e^{n}t$
• B. $e^{2}t$
• C. $t\left[\frac{1+e}{1-e}\right]$
• D. $t\left[\frac{1-e}{1+e}\right]$

Answer 1

The correct option is C $t\left[\frac{1+e}{1-e}\right]$

The time taken by the body to fall from the height $h$ is

$t=\sqrt{\frac{2h}{g}}$....................................(1)

After the first rebound, time taken to rise to height $h_1$ and coming back is

$t_1=2\sqrt{\frac{2h_1}{g}}$

$t_1=2\sqrt{\frac{2e^{2}h}{g}}$

$t_1=2e\sqrt{\frac{2h}{g}}$

Time taken in rising to height $h_1$ and coming back is

$t_2=2e^2\sqrt{\frac{2h}{g}}$

Time taken by the body in coming to rest

$t_3=\sqrt{\frac{2h}{g}}+2e\sqrt{\frac{2h}{g}}+2e^2\sqrt{\frac{2h}{g}}$

$t_3=\sqrt{\frac{2h}{g}}+2e\sqrt{\frac{2h}{g}}[1+e+e^2+----------.]$

$t_3=\sqrt{\frac{2h}{g}}+2e\sqrt{\frac{2h}{g}[\frac{1}{1-e}}]$

$t_3=\sqrt{\frac{2h}{g}}[1+\frac{2e}{1-e}]$

$t_3=\sqrt{\frac{2h}{g}}\left[\frac{1-e+2e}{1-e}\right]$

$t_3=\sqrt{\frac{2h}{g}}\left[\frac{1+e}{1-e}\right]$

$t_3=t\left[\frac{1+e}{1-e}\right]$........................from (1)