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Question

# A bullet of mass ‘m′, moving with a speed ‘v′ strikes a wooden block of mass M kept at rest and gets embedded in it. The speed of this embedded block will be:

A
(MM+m)v
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B
(mM+m)v
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C
(mM+m)v
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D
(m+MMm)v
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Solution

## The correct option is C (mM+m)vLet the velocity of the embedded block be V. Given: the mass of bullet is m, and mass of the wooden block is M, initial velocity of the bullet be v. Using law conservation of momentum: m1u1+m2u2=m1v1+m2v2 Initially the block is at rest. So u1=0 ms−1 ∴ Total momentum before impact = mv Finally the bullet is embedded in the block ∴Total momentum after impact = (M+m)V Now, applying conservation of momentum, mv = (M+m)V ∴V=(mM+m)v

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