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Question

# A bullet of mass m moving with a velocity v, strikes a wooden block of mass M and gets embedded in the block which was initially at rest. The final speed of the system is :

A
Mm+Mv
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B
mM+mv
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C
(mm+M)v
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D
(M+mM)v
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Solution

## The correct option is C (mm+M)vLets, m= mass of the bulletM= mass of the wooden block which is at rest initiallyv= velocity of the bulletv′= velocity of the bullet-block systemfrom the law of conservation of momentum,mv=(m+M)v′v′=mm+MvThe corrrect option is C.

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