Answer to Question #208365 in Physics for Klemzy

Question #208365

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?

Expert's answer

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}.

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