Question

A particle of mass $m$ is dropped from a height $h$ above the ground. Simultaneously another particle of the same mass is thrown vertically upwards from the ground with a speed of $\sqrt{2gh}$ . If they collide head-on completely inelastically, then the time taken for the combined mass to reach the ground is

• A. $\sqrt{\frac{3h}{4g}}$
• B. $\sqrt{\frac{3h}{2g}}$
• C. $\sqrt{\frac{h}{2g}}$
• D. $\sqrt{\frac{h}{4g}}$

Answer 1

Answer: (B)

The time for collision
$t_1=\frac{h}{\sqrt{2gh}}$
After $t_1$ ,
$v_A=0-g{t}_1=-\sqrt{gh/2}$
$V_B=\sqrt{2gh}-g{t}_1=\sqrt{gh}\left[\sqrt{2}-\frac{1}{\sqrt{2}}\right]$
At the time of collision
$$\begin{gathered} {\overrightarrow{P}}_i={\overrightarrow{P}}_f \\ \Rightarrow m{\overrightarrow{V}}_A+m{\overrightarrow{V}}_B=2m{\overrightarrow{V}}_f \\ \Rightarrow -\sqrt{\frac{gh}{2}}+\sqrt{gh}\left[\sqrt{2}-\frac{1}{\sqrt{2}}\right]=2{\overrightarrow{V}}_f \\ {V}_f=0 \end{gathered}$$
and height from ground
$=h-\frac{1}{2}g{t}_1^2=h-\frac{h}{4}=\frac{3h}{4}$
so time $=\sqrt{2\times \frac{\left(\frac{3h}{4}\right)}{g}}=\sqrt{\frac{3h}{2g}}$