A bullet of mass m and velocity v when fired | vedclass

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(D) Mass of the bullet $= m$. Speed of the bullet $= v$. According to the problem,the bullet gets embedded into the bag,so they move together with a c

Vedclass · Accepted Answer

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

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(D) Mass of the bullet $= m$. Speed of the bullet $= v$. According to the problem,the bullet gets embedded into the bag,so they move together with a common velocity $v_1$. This is a case of a perfectly inelastic collision. Initial kinetic energy of the bullet,$K_i = \frac{1}{2} m v^2$. By the law of conservation of linear momentum: $m v = (M + m) v_1 \Rightarrow v_1 = \frac{m v}{M + m}$. Final kinetic energy of the system,$K_f = \frac{1}{2} (M + m) v_1^2 = \frac{1}{2} (M + m) \left( \frac{m v}{M + m} \right)^2 = \frac{m^2 v^2}{2(M + m)}$. Loss in kinetic energy $= K_i - K_f = \frac{1}{2} m v^2 - \frac{m^2 v^2}{2(M + m)}$. $= \frac{1}{2} m v^2 \left( 1 - \frac{m}{M + m} \right) = \frac{1}{2} m v^2 \left( \frac{M + m - m}{M + m} \right) = \frac{m M v^2}{2(M + m)}$.

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

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