A man of mass m_1 = 80 kg , standing on rough ground, can…
A man of mass $m_1 = 80\ \mathrm{kg}$, standing on rough ground, can throw a stone of mass $m_2 = 2\ \mathrm{kg}$ at $v_2 = 10\ \mathrm{m/s}$. (a) What speed can he give the stone standing on smooth ice? (b) What is then the stone's speed relative to the man? (c) Does he develop the same power in both cases?
Explanation of Method
Throwing a stone while standing on ice.
Step 1: (a) Same effort, same work
With the same arm motion and force, the man does the same work $W = \tfrac12 m_2 v_2^2 = 100\ \mathrm{J}$. On ice it is shared between the stone and his own recoil, with total momentum zero:
\[ m_2 v_2' = m_1 v_1, \]
\[ \tfrac12 m_2 v_2'^2 + \tfrac12 m_1 v_1^2 = \tfrac12 m_2 v_2^2. \]
\[ v_2' = v_2\sqrt{\frac{m_1}{m_1 + m_2}} = 10\sqrt{\frac{80}{82}}. \]
\[v_2' \approx 9.88\ \mathrm{m/s}, \quad v_1 \approx 0.25\ \mathrm{m/s}\]
Step 2: (b) Relative speed
\[v_{\text{rel}} = v_2\sqrt{\frac{m_1 + m_2}{m_1}} \approx 10.12\ \mathrm{m/s}\]
Step 3: (c) Power
The work is the same, but on ice the stone moves away from the hand faster (the man also moves back), so the same arm stroke takes a little less time. The average power is therefore slightly greater on ice, by about $\sqrt{1 + m_2/m_1} - 1 \approx 1\%$.
Result
\[\boxed{\,v_2' \approx 9.88\ \mathrm{m/s}, \quad v_1 \approx 0.25\ \mathrm{m/s}\,}\]
\[\boxed{\,v_{\text{rel}} = v_2\sqrt{\frac{m_1 + m_2}{m_1}} \approx 10.12\ \mathrm{m/s}\,}\]
Remark
Check. The original results for (a) and (c) are right. In (b) it rounded to 9.8 + 0.25 ≈ 10.05 m/s; the accurate value is 10.12 m/s. Its expression for $v_1$ is also more complicated than needed; $v_1 = m_2 v_2'/m_1$ follows directly from momentum.