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Question

# A bullet of mass m moving with a horizontal velocity v strikes a stationary block of mass M suspended by a string of length L. The bullet gets embedded in the block. What is the maximum angle made by the string after impact ?

A
θ=cos1(Mm)
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B
θ=cos1(mM)
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C
θ=cos1(mVM+m)
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D
θ=cos1(112gL(mvm+M)2)
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Solution

## The correct option is D θ=cos−1(1−12gL(mvm+M)2)If V is the velocity of the block- bullet systemjust after collision, then by conservation of linear momentum mv=(M+m)Vor v=[M+mm]v.......(1)So the KE of the block- bullet system just aftercollision is 12(m+M)V2 (which is less than 12mv2 as collision is inelastic).Now due to this remaining KE if the system rises upto a height h, conservation of ME for this part of problem yields,12(m+M)V2=(m+M)gh,i.e V=√2gh....(2)Substituting this valueof V from Eqn. (2) in(1),we get.v=(M+mm)√2gh.........(3)Since h=L−L cosθcosθ=1−hLcosθ=1−12gL(mvM+m)2

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