A hammer of mass m_1 = 100 kg falls from a height H = 3 m…

A hammer of mass $m_1 = 100\ \mathrm{kg}$ falls from a height $H = 3\ \mathrm{m}$ onto an anvil of mass $m_2 = 500\ \mathrm{kg}$ standing on an elastic (springy) base. After the impact the hammer and anvil move together. Find their speed just after the impact.

Explanation of Method

A hammer hitting an anvil on springs.

Step 1: Speed of the hammer

\[ v_1 = \sqrt{2gH} = \sqrt{58.9} \approx 7.67\ \mathrm{m/s}. \]

Step 2: Inelastic impact

The impact is very short, so the springs' force can be neglected during it and momentum is conserved: $m_1 v_1 = (m_1 + m_2)\,v_2$.

\[v_2 = \frac{m_1}{m_1 + m_2}\sqrt{2gH} \approx 1.28\ \mathrm{m/s}\]

$100\ \mathrm{kg}$ $500\ \mathrm{kg}$ $H = 3\ \mathrm{m}$ $v_2 = \tfrac{m_1}{m_1 + m_2}\sqrt{2gH} \approx 1.28\ \mathrm{m/s}$

The hammer (7.67 m/s at impact) and the anvil on its springy base move off together; momentum, not kinetic energy, is conserved in the impact.

Result

\[\boxed{\,v_2 = \frac{m_1}{m_1 + m_2}\sqrt{2gH} \approx 1.28\ \mathrm{m/s}\,}\]

Remark

Check. The original result is correct. As it notes, energy is not conserved in the impact: 5/6 of the hammer's 2943 J is lost.