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