A skater of mass m standing on ice throws a stone of mass M=2 textitm with a velocity of v=5 textm/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 )

Question

A skater of mass m standing on ice throws a stone of mass M=2 textitm with a velocity of v=5 textm/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 textm (B) 15 textm (C) 10 textm (D) 5 textm

Tardigrade Admin · Accepted Answer

By momentum conservation, 2m× 5=m× V' So, V'=10m/sec Let skater stops after travelling distance s. So, 02=102-2× μ × g× s s=10 m

Q. A skater of mass $m$ standing on ice throws a stone of mass $M = 2 m$ with a velocity of $v = 5 m/s$ 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 $g = 10 m s^{- 2}$ )

$20 m$

$15 m$

$10 m$

$5 m$

By momentum conservation, $2 m \times 5 = m \times V^{^{'}}$
So, $V^{^{'}} = 10 m / sec$
Let skater stops after travelling distance s.
So, $0^{2} = 1 0^{2} - 2 \times μ \times g \times s$
$s = 10 m$

Q. A skater of mass m standing on ice throws a stone of mass M=2m with a velocity of v=5m/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 )

20m

15m

10m

5m