• anonymous
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.
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • schrodinger
I got my questions answered at in under 10 minutes. Go to now for free help!
  • Chealyn98
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.