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 $\sqrt{2 g h}$ . If they collide head-on completely inelastically, then the time taken for the combined mass to reach the ground is

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

Answer 1

Answer: (B)

The time for collision
$t_{1}=\frac{h}{\sqrt{2 g h}}$
After $t_{1}$ ,
$v_{A}=0-gt_{1}=-\sqrt{g h / 2}$
$V_{B}=\sqrt{2 g h}-gt_{1}=\sqrt{g h}\left[\sqrt{2} - \frac{1}{\sqrt{2}}\right]$
At the time of collision
$\overset{ \rightarrow }{P}_{i}=\overset{ \rightarrow }{P}_{f} \\ \Rightarrow m\overset{ \rightarrow }{V}_{A}+m\overset{ \rightarrow }{V}_{B}=2m\overset{ \rightarrow }{V}_{f} \\ \Rightarrow -\sqrt{\frac{g h}{2}}+\sqrt{g h}\left[\sqrt{2} - \frac{1}{\sqrt{2}}\right]=2\overset{ \rightarrow }{V}_{f} \\ V_{f}=0$
and height from ground
$=h-\frac{1}{2}gt_{1}^{2}=h-\frac{h}{4}=\frac{3 h}{4}$
so time $=\sqrt{2 \times \frac{\left(\frac{3 h}{4}\right)}{g}}=\sqrt{\frac{3 h}{2 g}}$