Question

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

Answer 1

By the law of conservation of momentum, we have:

$$p_i=p_f,$$ $$m_{stone}v_{stone}+m_{man}v_{man}=0,$$ $$v_{man}=-\dfrac{m_{stone}v_{stone}}{m_{man}}=-\dfrac{0.045\ kg\cdot3.05\ \dfrac{m}{s}}{90.5\ kg}=-1.5\cdot10^{-3}\ \dfrac{m}{s}.$$

The sign minus means that the man moves in the opposite direction to the motion of the stone.

Answer:

$v_{man}=1.5\cdot10^{-3}\ \dfrac{m}{s},$ in the opposite direction to the motion of the stone.