A skater of mass m standing on ice throws a stone of mass M with a velocity of v in a horizontal direction. Assuming coefficient of friction between the skater and the ice be $μ,$ what will be the distance across which the skater will slide back ?

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$\frac{M^{2} v^{2}}{2 m μ g}$

$\frac{{Mv}^{2}}{2 m^{2} μ g}$

$\frac{M^{2} v^{2}}{2 m^{2} μ g}$

$\frac{M^{2} v^{2}}{2 m^{2} μ^{2} g}$

answer is C.

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Detailed Solution

$\begin{array}{l} P_{i} = 0 \\ P_{f} = M V - m V_{1} \dots \dots \dots \dots . . (i i) \\ M V - m V_{1} = 0 \Rightarrow v_{1} = \frac{M}{m} V \\ using 0^{2} = v_{1}^{2} - 2 a x \\ \Rightarrow v_{1}^{2} = 2 μ g x \\ \Rightarrow {(\frac{M V}{m})}^{2} = 2 μ g x . ∴ x = \frac{M^{z} V^{2}}{2 m^{2} μ g} \end{array}$