A bullet of mass m travelling with a speed v | vedclass

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(D) Mass of bullet $= m$, Initial speed of bullet $= v$. Mass of block $= M$, Initial speed of block $= 0$. Let the common velocity of the system afte

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

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(D) Mass of bullet $= m$, Initial speed of bullet $= v$. Mass of block $= M$, Initial speed of block $= 0$. Let the common velocity of the system after collision be $V$. According to the law of conservation of linear momentum: $m v + M(0) = (m + M)V$ $V = \frac{m v}{m + M}$ Heat generated is equal to the loss in kinetic energy $(KE)$. $\Delta KE = KE_{initial} - KE_{final}$ $\Delta KE = \frac{1}{2} m v^{2} - \frac{1}{2} (m + M) V^{2}$ Substituting the value of $V$: $\Delta KE = \frac{1}{2} m v^{2} - \frac{1}{2} (m + M) \left( \frac{m v}{m + M} \right)^{2}$ $\Delta KE = \frac{1}{2} m v^{2} - \frac{1}{2} (m + M) \frac{m^{2} v^{2}}{(m + M)^{2}}$ $\Delta KE = \frac{1}{2} m v^{2} \left( 1 - \frac{m}{m + M} \right)$ $\Delta KE = \frac{1}{2} m v^{2} \left( \frac{m + M - m}{m + M} \right)$ $\Delta KE = \frac{1}{2} \frac{m M v^{2}}{m + M}$

$A$ bullet of mass $m$ travelling with a speed $v$ hits a block of mass $M$ initially at rest and gets embedded in it. The combined system is free to move and there is no other force acting on the system. The heat generated in the process will be

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