(A) By the law of conservation of linear momentum,the initial momentum of the system equals the final momentum of the system.$mv + M(0) = (m + M)V$Whe

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$\frac{1}{2}mv^2 \times \frac{m}{m+M}$

(A) By the law of conservation of linear momentum,the initial momentum of the system equals the final momentum of the system.$mv + M(0) = (m + M)V$Where $V$ is the velocity of the composite block after the collision.$V = \frac{mv}{m + M}$The kinetic energy $(KE)$ of the composite block is given by $KE = \frac{1}{2}(m + M)V^2$.Substituting the value of $V$:$KE = \frac{1}{2}(m + M) \left( \frac{mv}{m + M} \right)^2$$KE = \frac{1}{2}(m + M) \frac{m^2 v^2}{(m + M)^2}$$KE = \frac{1}{2} m v^2 \left( \frac{m}{m + M} \right)$

Class 11 Physics Work, Energy, Power and Collision Perfectly Inelastic Collision

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$A$ bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be

A
$\frac{1}{2}mv^2 \times \frac{m}{m+M}$
B
$\frac{1}{2}mv^2 \times \frac{M}{m+M}$
C
$\frac{1}{2}mv^2 \times \frac{M+m}{M}$
D
$\frac{1}{2}Mv^2 \times \frac{m}{m+M}$

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Perfectly Inelastic Collision

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$A$ bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be

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