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Question

# A bullet of mass m is fired at speed v_0 into a wooden block of mass M. The bullet instantaneously comes to rest in the block. The block with the embedded bullet slides along a horizontal surface with a coefficient of kinetic friction μ. Which one of the following expressions determines how far the block slides before it comes to rest (the magnitude of displacement Δx in the figure)?

A

Δx=mm+M(v20μg)

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B

Δx=mmM(v20μg)

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C

Δx=(v20μg)

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D

Δx=12μg(mM+mV0)2

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Solution

## The correct option is D Δx=12μg(mM+mV0)2 p0=mv0;pf=(M+m)vf p0=pf⇒vf=mM+mv0 So, retardation would be due to kinetic friction, and would be equal to −μg (constant). So, from equations of motion we can obtain the distance travelled by the block embedded with bullet before coming to rest. So, using v2=u2+2as,vf=mM+mv0,0=(mM+mv0)2+2(−μg)s,s=12μg(mM+mv0)2

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