Question

A 14 kg rock starting from rest free falls through a distance of 5.0 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 × 10^24 kg. Show all your work, assuming the rock-earth system is closed.

Answer 1

for the rock:

P0 = mv0
= (14)(0)
= 0

Pf = 14vf, so we need to find vf

vf = v0 + at
= 0 + (-9.8)t, we need to find t

xf = x0 + v0t + (1/2)at^2
-5 = 0 + 0t + (1/2)(-9.8)t^2
1.0102 = t

so vf = (-9.8)(1.1012)
= -9.89996

Pf = 14(-9.89996)
= -138.6 kg m/s

Change in momentum is -138.6 kg m/s

In a closed system, momentum is conserved, so a change in momentum for one body must be accompanied by a corresponding change in another body. So the earth's momentum must change in the same magnitude but opposite direction to the rock's change in momentum.

P = mv
138.6 = (6.0 x10^24)v
138.6/(6.0 x 10^24) = v
2.31 X10^ -23 m/s = v

So, as the rock falls toward the Earth, the Earth moves up toward the rock.