Question

PLEASE HELP ASAP AND SHOW WORK!!! 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 earths velocity. earth mass is 6.0 * 10^24 kg. Show your work assuming the rock earth system is closed.

Answer 1

Time taken to fall the rock from rest

here we know that

$\Delta y = 5 m$

$v_i = 0$

$a = 10 m/s^2$

now we have

$\Delta y = v_i t + \frac{1}{2}at^2$

$5 = \frac{1}{2}10t^2$

$t = 1 s$

Now we know that

$F\Delta t = \Delta P$

here Force (F) is due to gravity

so we have

$F = mg = 14(10) = 140 N$

now we can plug in all data

$140 \times 1 = \Delta P$

$\Delta P = 140 kg m/s$

So change in momentum of rock will be 140 kg m/s

Now we know from Newton's III law

$F_{rock} = F_{earth}$

so we have

$\Delta P_{earth} = \Delta P_{rock} = 140 Ns$

now we have

$m(v_f - v_i) = 140$

$6\times 10^{24}(v_f - 0) = 140$

$v_f = 2.33 \times 10^{-23} m/s$

so above is the speed of Earth