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(C) The initial momentum of the block is $P = Mu$,where $u$ is the initial velocity. Thus,$u = \frac{P}{M}$.Since the surface is rough,the frictional

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$\frac{P^{2}}{2 \mu M^{2} g}$

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(C) The initial momentum of the block is $P = Mu$,where $u$ is the initial velocity. Thus,$u = \frac{P}{M}$.Since the surface is rough,the frictional force $f = \mu N = \mu Mg$ acts on the block.The retardation (deceleration) $a$ is given by $a = \frac{f}{M} = \frac{\mu Mg}{M} = \mu g$.Using the kinematic equation $v^{2} = u^{2} - 2as$,where $v = 0$ (final velocity when it stops):$0 = u^{2} - 2as$$2as = u^{2}$$s = \frac{u^{2}}{2a}$Substituting $u = \frac{P}{M}$ and $a = \mu g$:$s = \frac{(\frac{P}{M})^{2}}{2 \mu g} = \frac{P^{2}}{M^{2} \cdot 2 \mu g} = \frac{P^{2}}{2 \mu M^{2} g}$.

$A$ block of mass $M$ is moving on a rough horizontal surface with momentum $P$. The coefficient of friction between the block and the surface is $\mu$. The distance covered by the block before it stops is $[g = \text{acceleration due to gravity}]$.

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