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Q.

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 a height h as a result of this collision, the velocity of the bullet before collision is

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a

υ=2gh

b

υ=2gh[1+mM]

c

υ=2gh[1+Mm]

d

υ=2gh[1−mM]

answer is C.

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Detailed Solution

Let  v' is the velocity of  (M+m) after collision. 

Then from conservation of linear momentum we have,

mv=(M+m)v' ∴  v'=(mM+m)v

Now,  v'=2gh

∴ (mM+m)v=2gh or  v=2gh(M+mm)

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