A hammer of mass M strikes a nail of mass m with velocity of ums−1 and drives it a metres into a fixed block of wood. The average resistance of the wood to the penetration of the nail is:
A
(MM+m)u22a
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B
(M(M+m)2)u22a
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C
(M+mM)u22a
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D
(M2M+m)u22a
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Solution
The correct option is D(M2M+m)u22a Using conservation of momentum we have v=MuM+m , Thus we get FS=12(M+m)v2 Fa=12(M+m)v2 F=12a(M+m)(MuM+m)2=(M2M+m)u22a