question_answer

A particle of mass m is dropped from a height h above the ground. At the same time 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, the time taken for the combined mass to reach the ground, in units of \[\sqrt{\frac{h}{g}}\] is                                                                                                 [JEE MAIN Held on 08-01-2020 Evening]

A)   \[\sqrt{\frac{1}{2}}\]               

B)        \[\sqrt{\frac{3}{4}}\]

C)   \[\frac{1}{2}\]            

D)        \[\sqrt{\frac{3}{2}}\]

Correct Answer: D

Solution :

[d]

Time of collision

\[\Rightarrow {{t}_{0}}=\frac{h}{\sqrt{2gh}}=\sqrt{\frac{h}{2g}}\]

\[\therefore {{s}_{1}}=\frac{1}{2}gt_{0}^{2}=\frac{1}{2}g.\frac{h}{2g}=\frac{h}{4}\]

\[\therefore {{s}_{2}}=\frac{3h}{4}\]

Speed of [a] just before collision \[{{v}_{1}}\,\,\downarrow \]

\[=g{{t}_{0}}=\sqrt{\frac{gh}{2}}\]

And speed of [b] just before collision \[{{v}_{2}}\uparrow \]

\[=\sqrt{2gh}-\sqrt{\frac{gh}{2}}\]

After collision velocity of centres of mass

\[{{v}_{cm}}=\frac{m\left( \sqrt{2gh}-\sqrt{\frac{gh}{2}} \right)-m\sqrt{\frac{gh}{2}}}{2m}=0\]

So from there, time of fall ‘t’

\[\Rightarrow \,\,\,\,\,\,\,\,\,\frac{3h}{4}=\frac{1}{2}g{{t}^{2}}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,t=\sqrt{\frac{3}{2}\frac{h}{g}}\]