a 7.00 kg bowling ball is dropped from rest at an initial height of 3.00 m. What

Question

a 7.00 kg bowling ball is dropped from rest at an initial height of 3.00 m. What is the speed of the Earth coming up to meet the ball just before the blal hits the ground? Use 5.98 x 10^24 kg as the mass of the Earth. I found some help with this and thought I was on the right track, but the answer is wrong.

m1=7 kg m2=5.98x10^24 kg h=3m
pball =mv
to find velocity
v^2f = v^2i + 2ad
v^2f = 0 + 2 (9.8)(3)
V^2f +58.8
v=7.67 m/s for the ball
conservation of momentum
mE vE = 7.67 m/s ball
vE =7.67/5.98x1024 = 1.28x10-24
the correct answer is 9.0x10-24 m/s
Thanks for the assist!

Explanation / Answer

You forgot the mass of the bowling ball while applying conservation on momentum. Multiply your answer by 7 you will get your answer :)

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