Question

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 $\mu $ , what will be the distance across which the skater will slide back ?

• A. $\frac{M^{2} v^{2}}{2 m \mu g}$
• B. $\frac{Mv^{2}}{2 m^{2} \mu g}$
• C. $\frac{M^{2} v^{2}}{2 m^{2} \mu g}$
• D. $\frac{M^{2} v^{2}}{2 m^{2} \mu ^{2} g}$

Answer 1

Answer: (C)

$P_{i}=0$ ...........(i)
$P_{f}=MV-mV_{1}$ ..............(ii)
$MV-mV_{1}=0\Rightarrow v_{1}=\frac{M}{m}V$
using $0^{2}=v_{1}^{2}-2ax$
$\Rightarrow v_{1}^{2}=2\mu gx$
$\Rightarrow \left(\frac{M V}{m}\right)^{2}=2\mu gx.\therefore x=\frac{M^{2} V^{2}}{2 m^{2} \mu g}$