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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)v

Let 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 ms1
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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