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? \[\Rightarrow \,\,\,\,\,\,\,\,\,\frac{3h}{4}=\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,t=\sqrt{\frac{3}{2}\frac{h}{g}}\]You need to login to perform this action.
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