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 \[mu=(m+m)v\] \[\Rightarrow \] \[v=\frac{u}{2}\] ?(i) The potential energy of A at height h converts into kinetic energy of this mass, at point O. ie, \[mgh=\frac{1}{2}m{{u}^{2}}\]or \[u=\sqrt{2gh}\] \[\therefore \] \[v=\sqrt{\frac{2gh}{2}}=\sqrt{\frac{gh}{2}}\] 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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