Question

A 90.5kg man standing on a surface of negligible friction kicks forward a 45gm stone lying at his feet so that it acquire a speed of 3.05m/s. What velocity does the man acquire as a result?

Answer 1

Given:

$M=90.5\:\rm kg$

$m=0.045\:\rm kg$

$u=3.05\:\rm m/s$

The law of conservation of momentum says

$$0=Mv+mu$$

Hence, the velocity of the man

$$v=-\frac{mu}{M}=-\frac{0.045*3.05}{90.5}=-0.0015\:\rm m/s$$