A bag of mass M is suspended by a string. A b | vedclass

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(D) According to the law of conservation of linear momentum,the initial momentum of the bullet equals the final momentum of the system (bag + bullet).

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) According to the law of conservation of linear momentum,the initial momentum of the bullet equals the final momentum of the system (bag + bullet).Initial momentum $P_i = mv$.Final momentum $P_f = (M + m)V$,where $V$ is the common velocity.$mv = (M + m)V \implies V = \frac{mv}{M + m}$.The kinetic energy $K$ of the system after the collision is given by $K = \frac{1}{2}(M + m)V^2$.Substituting the value of $V$:$K = \frac{1}{2}(M + m) \left( \frac{mv}{M + m} \right)^2$.$K = \frac{1}{2}(M + m) \frac{m^2v^2}{(M + m)^2}$.$K = \frac{m^2v^2}{2(M + m)}$.

$A$ bag of mass $M$ is suspended by a string. $A$ bullet of mass $m$ moving with velocity $v$ enters the bag and gets embedded in it. Then,for the system:

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