A bullet of mass m moving with velocity v str | vedclass

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Question

(A) According to the law of conservation of linear momentum,the initial momentum of the bullet equals the final momentum of the combined system:$mv =

Vedclass · Accepted Answer

$\frac{1}{2}mv^2 	imes \frac{m}{(m + M)}$

Vedclass · Answer

(A) According to the law of conservation of linear momentum,the initial momentum of the bullet equals the final momentum of the combined system:$mv = (m + M)V$where $V$ is the final velocity of the combined system.$V = \frac{mv}{m + M}$The kinetic energy $K$ of the combined system 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{1}{2} \frac{m^2v^2}{m + M}$$K = \left( \frac{1}{2}mv^2 \right) \left( \frac{m}{m + M} \right)$

$A$ bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ and gets embedded in it. The kinetic energy of the system is:

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