Question

A 90.5kg man standing on a surface of negligible friction kicks forward a 45gm stone lying at his feet so that it acquires a speed of 30.5m/s. What velocity does the man acquire as a result?

Answer 1

Apply conservation of momentum:

$$p_m=p_s,\\ v_mm_m=v_sm_s,\\\space\\ v_m=\frac{v_sm_s}{m_m}=\frac{30.5·0.045}{90.5}=0.0152\text{ m/s}.$$