Question

A skater of mass $m$ standing on ice throws a stone of mass $M=2 \, \textit{m}$ with a velocity of $v=5 \, \text{m/s}$ in horizontal direction. If the coefficient of friction between the skater and the ice is $0.5$ , find the distance over which the skater will move. (take $g=10ms^{- 2}$ )

• A. $20 \, \text{m}$
• B. $15 \, \text{m}$
• C. $10 \, \text{m}$
• D. $5 \, \text{m}$

Answer 1

Answer: (C)

By momentum conservation, $2m\times 5=m\times V^{'}$
So, $V^{'}=10m/sec$
Let skater stops after travelling distance s.
So, $0^{2}=10^{2}-2\times \mu \times g\times s$
$s=10 \, m$