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(C) Let the speed of the bullet be $v$. Let the speed of the combined system (block + bullet) immediately after the collision be $V$. By the law of co

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$\sqrt {2\mu gx} \left( {\frac{{M + m}}{m}} \right)$

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(C) Let the speed of the bullet be $v$. Let the speed of the combined system (block + bullet) immediately after the collision be $V$. By the law of conservation of linear momentum: $mv = (m + M)V$,which gives $V = \frac{mv}{m + M}$. The kinetic energy of the combined system immediately after the collision is $K.E. = \frac{1}{2}(m + M)V^2 = \frac{1}{2}(m + M) \left( \frac{mv}{m + M} \right)^2 = \frac{m^2v^2}{2(m + M)}$. This kinetic energy is dissipated by the work done against the force of friction $f = \mu N = \mu(m + M)g$. The work done by friction over distance $x$ is $W = f \cdot x = \mu(m + M)gx$. Equating the initial kinetic energy to the work done against friction: $\frac{m^2v^2}{2(m + M)} = \mu(m + M)gx$. Solving for $v^2$: $v^2 = 2\mu gx \left( \frac{m + M}{m} \right)^2$. Taking the square root,we get $v = \sqrt{2\mu gx} \left( \frac{m + M}{m} \right)$.

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