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A wooden block weighing 80 N rests on a rough horizontal plane, the coefficient of friction between the two being µ=0.4. If a bullet weighing 0.4 N is fired horizontally in to the block with muzzle velocity v= 500 m/s, how far will the block be displaced from its initial position? Assume that the bullet remains inside the block.
Asked by theavengers0203 | 24 May, 2023, 08:40: AM
Weight of bullet , ( m g  ) = 0.4 N ;

Velocity of bullet before impact, v = 500 m/s ;

Weight of wooden block , ( M g ) = 80 N ;

Let velocity of wooden block and bullet after impact = V

By conservation of momentum , total momentum of bullet embedded block after impact equals to
the initial momentum of bullet ( wooden block is under rest ) .

hence we have,

Kinetic energy K of bullet embedded wooden block  after impact is

Work done W against friction force is

where μ is friction coefficient and d is the distance moved by bullet embedded wooden block  after impact.

By work-energy theorem , K = W

Answered by Thiyagarajan K | 24 May, 2023, 10:37: PM
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