A bag (mass M) hangs by a long thread and a b | vedclass

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(D) Initial momentum of the system = $mv$.Final momentum of the system = $(m + M)V$,where $V$ is the common velocity of the system after collision.By

Vedclass · Accepted Answer

Kinetic energy is $\frac{m^2v^2}{2(M + m)}$

Vedclass · Answer

(D) Initial momentum of the system = $mv$.Final momentum of the system = $(m + M)V$,where $V$ is the common velocity of the system after collision.By the law of conservation of linear momentum,$mv = (m + M)V$.Therefore,the common velocity $V = \frac{mv}{M + m}$.The kinetic energy of the combined system is $K = \frac{1}{2}(m + M)V^2$.Substituting the value of $V$,we get $K = \frac{1}{2}(m + M) \left( \frac{mv}{M + m} \right)^2$.$K = \frac{1}{2}(m + M) \frac{m^2v^2}{(M + m)^2} = \frac{m^2v^2}{2(M + m)}$.Thus,option $D$ is correct.

$A$ bag (mass $M$) hangs by a long thread and a bullet (mass $m$) comes horizontally with velocity $v$ and gets caught in the bag. Then for the combined (bag $+$ bullet) system

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