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 + (1)/(2)at^2`

`5 = (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 * 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* 10^(24)(v_f - 0) = 140`

`v_f = 2.33 * 10^(-23) m/s`

so above is the speed of Earth