# Homework Help: Applying conservation of momentum to find recoil of the Earth

1. Apr 9, 2013

### cherry_cat

1. The problem statement, all variables and given/known data

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)?

2. Relevant equations

v=sqrt(2gh)
p=mv
p(total)=m1v1+m2v2
v=p/m

3. 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?

Last edited: Apr 9, 2013
2. Apr 9, 2013

### Simon Bridge

My immediate thought was that the ball is bouncing off a wall some distance from the center of mass of the Earth ... so you are really interested in conservation of angular momentum.

However - looking at your relations, why did you add the mass of the ball to the mass of the Earth?

The way to do these problems is to sketch the "before" and "after" situations - so you know what you are describing. Then, write out the momentum calculations separately. Then put before=after.

From what you've done, your "before" should be a stationary Earth and a moving ball.
What would the "after" be?

3. Apr 9, 2013

### haruspex

It isn't the same - the ball is going the other way.
I have no idea where you get that number from.

4. Apr 9, 2013

### cherry_cat

The before section would be m1v1 = 0.16*44 = 7.04 kg.m/s as the ball is travelling towards the wall

After the ball hits the wall, it would have negative momentum but assuming no loss in kinetic energy, wouldn't it have a negative momentum of 7.04 kg.m/s and the Earth would have two times the momentum that the ball originally has. or -m1v1+2m1v1

5. Apr 9, 2013

### haruspex

Yes, but that would be 14.08 kg.m/s, not 6.82.

6. Apr 9, 2013

### cherry_cat

Yes the momentum would have to be constant before and after the collision, so it would be 7.04 kg.m/s, meaning 2*p would be 14.08 kg.m/s

Which means you could set the after section, of m1v1+m2v2=7.04
And because the ball lost no energy, it would be -7.04+m2v2=7.04
So m2v2=14.08
v2=14.08/6e24
v2=2.35e-24

Is that logical?

7. Apr 9, 2013

### cherry_cat

That's true. I must have made a foolish mistake somewhere. Thanks for pointing it out, I'm new to physics and sometimes I make mistakes and don't notice. I will try to be more careful.