# Conservation of Momentum

1. Dec 10, 2008

### reallybanana

I'm having a little bit of trouble trying to figure out how to solve this problem. I know it has to do with conservation of momentum, but I can't seem to get my first step on the problem.

The problem is as follows:

A bowling ball with a mass of 5.5kg hangs from the end of a 5.0m cable, initially such that the came is vertical. You push horizontally on the ball with a varying force to move it 2.0m to the side. During the ball's displacement how much work do you do on the ball? Assume that the ball is motionless before and after you push on it.

I have an idea, but I'm unsure if it's towards the right direction. Because we have a distance covered, we can some how figure out a velocity and then figure out our work or energy placed into the ball. I know the ball has no energy at all whatsoever because it is also assumed that the ball is on rest on the ground (asked my professor about this as well). Because there is no energy initially, we have to give it energy, and since we are assume that no energy is lost in the form of heat or what ever, our end energy is equal to the same energy we put in and vice versa. My only problem is how to get there in this problem.

I was thinking that since the ball is on a dedicated 5.0m cable, that if I push the ball 2.0m, the ball has to travel in an arc, however my professor told me that's wrong too. I'm quite stuck on this problem. If anyone could not give me the answer, but give me a hint as what to do instead, that would be great.

2. Dec 11, 2008

### reallybanana

Working at the problem again, I was looking at different arcs how it relates to the 2.0m mentioned in the problem.

Doing some work, we know that I pushed the ball 2.0m in the x direction, and not the ball a total of 2m. Going from my initial idea dealing with arcs, 2.0m in the x direction would give us a change in height of about sqrt(21) or .417. Now placing this as the distance for potential energy mgy would give me about 22.5 J of work. Does this seem plasuable?

3. Dec 11, 2008

### PhanthomJay

Yes, excellent 'work'. Since there is no change in the kinetic energy of the system, and since the tension force does no work, then the work done by you must equal the negative of the work done by gravity, since the net work done by all forces must be zero, per the work-energy theorem.