A uniform rod of mass M and length L is place | vedclass

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(B) According to the law of conservation of linear momentum,the total momentum of the system before the collision is equal to the total momentum of th

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$\frac{mv}{M}$

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(B) According to the law of conservation of linear momentum,the total momentum of the system before the collision is equal to the total momentum of the system after the collision.Let $v_{cm}$ be the velocity of the centre of mass of the rod after the collision.Before the collision,the momentum of the particle is $p_i = mv$ and the rod is at rest $(p_{rod} = 0)$.After the collision,the particle comes to rest,so its momentum is $0$. The rod moves with velocity $v_{cm}$.Applying conservation of linear momentum:$mv + 0 = 0 + Mv_{cm}$Solving for $v_{cm}$:$v_{cm} = \frac{mv}{M}$

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ particle of mass $m$ moving with velocity $v$ strikes the rod at one end perpendicularly and comes to rest. The velocity of the centre of mass of the rod after the collision is:

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