Question

A 18 kg rock starting from rest free falls through a distance of 7.0 m with no air resistance. Find the momentum change of the rock caused by its fall and the resulting change in the magnitude of earth’s velocity. Earth’s mass is 6.0 × 1024 kg. Show all your work, assuming the rock-earth system is closed.

Answer 1

Find the final velocity of the rock first:

The total energy of the rock at rest:
$\frac{1}{2} mv_1^2 + mgh_1$

The total energy of the rock after 7m:
$\frac{1}{2} mv_2^2 + mgh_2$

Energy must be conserved:
$\frac{1}{2} mv_1^2 + mgh_1 = \frac{1}{2} mv_2^2 + mgh_2 \\ \\ \frac{1}{2} m(v_2^2 - v_1^2) = mg(h_1 - h_2) \\ \\ \frac{1}{2} (v_2^2 - v_1^2) = g(h_1 - h_2) \\ \\ v_1 = 0, \delta h = 7m \\ \\ v_2^2 = 2g(\delta h) \\ \\ v_2 = \sqrt{2g(\delta h)}$

Momentum must be conserved also.
Momentum with the ball at rest:
$p = m_1v_1 + m_2v_2 \\ m_1 = 18kg \\ m_2 = 6 * 10^{24}kg \\ v_1 = v_2 = 0 \\ \\ p = 0$

The total momentum is zero with the ball at rest and must remain zero.

Momentum with the ball at 7m:
$p = m_1v_1 + m_2v_2 = m_1\sqrt{2g(\delta h)} + m_2v_2 = 0$

Solving for the earth's velocity v₂:
$v_2 = - \frac{m_1}{m_2} \sqrt{2g(\delta h)}$