A) \[\frac{3h}{4}\]
B) \[\frac{2h}{3}\]
C) \[\frac{h}{2}\]
D) \[\frac{h}{4}\]
Correct Answer: D
Solution :
When bob A strikes the bob B, then \[mu=(m+m)v'\] \[\therefore \] \[v'=\frac{\sqrt{2gh}}{2}=\sqrt{\frac{gh}{2}}\] ...(i) The potential energy of A at height \[h\]converts into kinetic energy of this mass, at point 0. ie, \[mgh=\frac{1}{2}m{{u}^{2}}\]or \[u=\sqrt{2gh}\] \[\therefore \] \[v'=\frac{\sqrt{2gh}}{2}=\sqrt{\frac{gh}{2}}\] Let combined mass moves to a height\[h',\]then \[2mgh'=\frac{1}{2}(2m)v{{'}^{2}}\] Or \[gh'=\frac{gh}{4}\] Or \[h'=\frac{h}{4}\]You need to login to perform this action.
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