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 :

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 just before collision \[{{v}_{1}}\,\,\downarrow \] \[=g{{t}_{0}}=\sqrt{\frac{gh}{2}}\] And speed of 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? \[\Rightarrow \,\,\,\,\,\,\,\,\,\frac{3h}{4}=\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,t=\sqrt{\frac{3}{2}\frac{h}{g}}\]