Question

please help

Answer 1

A 27 kg rock starting from rest free-falls through a distance of 18.0 m with no air resistance. Find the momentum change of the rock caused by its fall and the resulting change in the magnitude of the earth’s velocity. The earth’s mass is 6.0 × 10^24 kg. Show all your work, assuming the rock–earth system is closed.

Answer 2

To find the momentum change (the impulse) of the rock, we first need to find the final velocity. You can find the final velocity of the rock using the following equation:

\[v_f^2=v_i^2+2ad\]
Where v(f) is the final velocity, v(i) is the initial velocity, a is the acceleration, and d is the distance travelled.

Using the final velocity you obtain, you can then find the impulse of the rock:

\[J_{rock}=m_{rock}(v_{f_{rock}}-v_{i_{rock}})\]
Where J is the impulse, m is the mass, v(f) is the final velocity, and v(i) is the initial velocity.

By conservation of momentum, the impulse the Earth undergoes is equal to impulse the rock undergoes. In other words:

\[J_{Earth}=J_{rock}\]\[m_{Earth} \Delta v_{Earth}=J_{rock}\]
Solve for Δv and you have your answer! If you have any questions let me know!