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Question

Two identical blocks A and B of mass M each are kept on each other on a smooth horizontal plane. There exists friction between A and B. If a bullet of mass m hits the lower block with a horizontal velocity v and gets embedded into it, the work done by friction between A and B is


  1. - Mm2v22(m+2M)(m+M)

  2. -  Mm2v2(m+2M)(m+2M)

  3. - Mm2v22(m+2M)(m+M)

  4. - Mm2v22(m+2M)(2m+M)


Solution

The correct option is C

- Mm2v22(m+2M)(m+M)


Let the velocity of lower block just after collision be V1

Applying conservation of liner momentum between bullet and lower block as impluse due to friction is negligible.

mv = (M + m )V1  V1 = ( mM+m) V

After collision frictional force acts on both and will act till velocities of both blocks become equal.

(M + m )V1 = (2M + m )V2  V2 ( m2M+m) V

Applying work energy theorem ,WFriction = Kf - Ki = 12 [2 M + m] V22

 KfrictionMm2v22(m+2M)(m+M)

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