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Question

# A bullet of mass m hits a target of mass M hanging by a string and gets embedded in it. If the block rises to height h as a result of this collision, the velocity of the bullet before collision is :

A
v=2gh
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B
v=2gh[1+mM]
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C
v=2gh[1+Mm]
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D
v=2gh[1mM]
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Solution

## The correct option is C v=√2gh[1+Mm]Let v′ is the velocity of (M+m) after collision. Then from conservation of linear momentum in the horizontal direction,mv=(M+m)v′.......(1)Both the bullet and the block have the same velocity after collision since the bullet gets embedded in the blockSince the block+bullet rise to a height h, from conservation of energy,12(M+m)v′2=(M+m)gh∴v′=√2gh.....(2)∴ by putting v′ from eqn (2) in eqn (1), we get:v=(M+mm)√2gh=√2gh(1+Mm)

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