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(D) Initial kinetic energy of the bullet is $K_i = \frac{1}{2}mv^2$.After the perfectly inelastic collision,the bullet gets embedded in the sandbag,an

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$\frac{1}{2}m{v^2}\left( {\frac{M}{{M + m}}} \right)$

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(D) Initial kinetic energy of the bullet is $K_i = \frac{1}{2}mv^2$.After the perfectly inelastic collision,the bullet gets embedded in the sandbag,and the system moves with a common velocity $V$.By the principle of conservation of linear momentum:$mv + M(0) = (m + M)V$$V = \frac{mv}{m + M}$The final kinetic energy of the system is $K_f = \frac{1}{2}(m + M)V^2$.Substituting the value of $V$:$K_f = \frac{1}{2}(m + M) \left( \frac{mv}{m + M} \right)^2 = \frac{1}{2} \frac{m^2v^2}{m + M}$.The loss in kinetic energy is $\Delta K = K_i - K_f$.$\Delta K = \frac{1}{2}mv^2 - \frac{1}{2} \frac{m^2v^2}{m + M}$$\Delta K = \frac{1}{2}mv^2 \left( 1 - \frac{m}{m + M} \right)$$\Delta K = \frac{1}{2}mv^2 \left( \frac{m + M - m}{m + M} \right)$$\Delta K = \frac{1}{2}mv^2 \left( \frac{M}{m + M} \right)$.

$A$ bag of sand of mass $M$ is suspended by a string. $A$ bullet of mass $m$ is fired at it with velocity $v$ and gets embedded into it. The loss of kinetic energy in this process is

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