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 \[u=\sqrt{2gh}\] \[\therefore \] \[v=\frac{\sqrt{2gh}}{2}=\sqrt{\frac{gh}{2}}\] 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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