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cherry_cat
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Homework Statement
We can use our results for head-on elastic collisions to analyze the recoil of the Earth when a ball bounces off a wall embedded in the Earth. Suppose a professional baseball pitcher hurls a baseball (m = 160 grams) with a speed (v1 = 44 m/s) at a wall, and the ball bounces back with little loss of kinetic energy. What is the recoil speed of the Earth (M = 6e24 kg)?
Homework Equations
v=sqrt(2gh)
p=mv
p(total)=m1v1+m2v2
v=p/m
The Attempt at a Solution
M(Earth)=6e24 kg
M(Ball)=0.16 kg
P(Ball) = 0.16*44 kg m/s = 7.04 kg.m/s
The momentum must be the same before and after the collision. Therefore, the momentum provided to the Earth when the ball hits is 6.82 kg.m/s
v(Earth) = P(total)/M(Earth+ball)
v(Earth)=7.04/ 6e24
v(Earth)=1.173e-24
I was sure all my working was correct, but the answer is wrong. Is there anything I have missed?
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