Question

A skater of mass m standing on ice throws a stone of mass M with a velocity of V in a horizontal direction. The distance over which the skater will move back (the coefficient of friction between the skater and the ice is $\mu$) :

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

Answer 1

Answer: (C)

$\frac{M^2{V}^2}{2m^2\mu g}$

$P_i=0$....(i)

$P_f=MV-m{V}_1$....(ii)

$MV-m{V}_1=0\Rightarrow V_1=\frac{M}{m}V.$

using $0^2=V_1^2-2ax$

$\Rightarrow V_1^2=2\mu gx$

$\Rightarrow (\frac{MV}{m}{)}^2=2\mu gx.$

$\therefore x=\frac{M^2{V}^2}{2m^2\mu g}$