A bullet of mass m moving horizontally with a velocity v strikes a block of wood of mass M and gets embedded in the block. The block is suspended from the ceiling by a massless string. The height to which block rises is:
A
v22g(mM+m)2
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B
v22g(M+mm)2
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C
v22g(mM)2
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D
v22g(Mm)2
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Solution
The correct option is Av22g(mM+m)2 The situation is as shown in the figure. Let V be velocity of the block-bullet system just after collision. Then by the law of conservation of linear momentum, we get mv=(m+M)V V=mvm+M Let the block rises to a height h. According to law of conservation of mechanical energy, we get 12(m+M)V2=(m+M)gh h=V22g=v22g(mm+M)2.