A) \[\sqrt{2gh}\]
B) \[\frac{m+m}{m}\sqrt{2gh}\]
C) \[\frac{m}{M+m}2gh\]
D) \[\frac{M+m}{M}\sqrt{2gh}\]
Correct Answer: A
Solution :
Initial KE of block when bullet strikes to it \[=\frac{1}{2}(m+M){{v}^{2}}\] Due to this KE of block will rise to a height h. Its potential energy \[=(m+M)gh\] By the law of conservation of energy \[\frac{1}{2}(m+M){{v}^{2}}=(m+M)gh\] \[\therefore \] \[v=\sqrt{2gh}\]You need to login to perform this action.
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