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 is :

• 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 $t\left[\frac{1+e}{1-e}\right]$

Let $v$ represents the velocities ob the ball when it is about to reach the ground whereas $v^{\prime }$ represents the rebound velocities after the collision.

Also $t$ represents the time taken by ball to reach the ground whereas $t^{\prime }$ represents the time taken to reach the maximum height after the collision.

Initial height from where the ball is dropped be $h$

$\therefore$ $v_1=\sqrt{2gh}$

$h=0+\frac{1}{2}g{t}^2$ $⟹t=\sqrt{\frac{2h}{g}}$ ..............(1)

Rebound velocity after 1st collision $v_1^{\prime }=e{v}_1=e\sqrt{2gh}$

Rebound height after 1st collision $h_1=e^{2}h$

$0=v_1^{\prime }-g{t}_2^{\prime }$ $⟹t_2^{\prime }=\frac{v_1^{\prime }}{g}=e\sqrt{\frac{2h}{g}}$

Also $h_1=0+\frac{1}{2}g{t}_2^2$ $⟹t_2=\sqrt{\frac{2h_1}{g}}=e\sqrt{\frac{2h}{g}}$

Similarly, $t_3=t_3^{\prime }=e^2\sqrt{\frac{2h}{g}}$, $t_4=t_4^{\prime }=e^3\sqrt{\frac{2h}{g}}$, and so on

Total time for the ball to come to rest $T=t+t_2^{\prime }+t_2+t_3^{\prime }+t_3+t_4^{\prime }+t_4+.................$

$T=\sqrt{\frac{2h}{g}}$ $+2e\sqrt{\frac{2h}{g}}$ $+2e^2\sqrt{\frac{2h}{g}}$ $+2e^3\sqrt{\frac{2h}{g}}+............$

$T=t\times [1+2e(1+e+e^2+.......\left)\right]$

$T=t\times [1+2e\times \frac{1}{1-e}]$ $⟹T=t(\frac{1+e}{1-e})$