Tackling a Tough Momentum Problem: Solving the Mystery

[EebamXela]

I had a test today that was really tough.

We just went over conservation of momentum and angular momentum and that stuff.

The hardest problem on the test looks like this:

http://alexmabee.googlepages.com/problem.jpg/problem-full.jpg

I can't for the life of me figure out how to solve it. I asked the teacher if there was missing information and she said that everything you need is there.

The problem I have with it is that the location of the collision between A and B is not given. I think it can be found, but i have no idea how. The collisions are perfectly elastic so the coefficient of restitution equals one, so you know that the velocities of A and B after the collide will be at 90 degrees to each other.

I drew the problem to scale in AutoCAD with exact angles and such because AutoCAD is sweet and can calculate it all. But how the hell does one go about doing that by hand on paper?

I tried using conservation of Linear momentum and also using conservation of angular momentum ( r x mv ) taking the center of rotation to be the lower left corner of the pool table and came out with a bunch of unknowns and not enough equations.


HELP!

(if this is the wrong forum for topics like this, sorry)

 

[TVP45]

You may not have wasted enough of your life in poolrooms to see it, although the problem is not completely realistic. Can you work out what the angle is between A1-A2 and A2-A3?

 

[EebamXela]

You may not have wasted enough of your life in poolrooms to see it, although the problem is not completely realistic. Can you work out what the angle is between A1-A2 and A2-A3?

See that's the hard part. No I can't figure it out. I know the problem isn't realistic, but no problems on a physics test are ever realistic. The problem states that it can be assumed to be frictionless and perfectly elastic.

 

[TVP45]

So have you figured it out yet? Look at the angle between B1-B2 and A1-A2. Do you know anything about that?

The lack of realism I mentioned is that ball A picks up some angular momentum at the collision with B (side English) and the collision with C will look a little different. A pool player would go "huh?" when looking at the diagram. However, for a Physics problem, that really doesn't matter - we can just ignore it.

 

[EebamXela]

In order to have English on a ball you need friction. This simplified problem has no friction. The balls themselves can be considered point masses.

If you use computer aided drafting software like AutoCAD you can calculate the angle in question. It turns out to be 31.717 degrees. You know that the exit velocities of A and B have to be perpendicular because it is a perfectly elastic collision, and you also know where A goes from there and where B goes from there.

 

[TVP45]

I missed the part where the balls are point masses; you're right that there can be no English under that condition. BTW, most pool simulations (and even games) always use that assumption.

So, you know that A and B depart perpendicularly. Then, you know the tan of one line is the cot of the other. Since you know the vertical travel of each, you can equate the horizontal travel of one in terms of the other. Then, when you subtract one horizontal from the other horizontal, you get how much?