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  • one year ago

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.

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  1. Chealyn98
    • one year ago
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    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.

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