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Concave or convex doesn't matter. The point about circles is the surface is always perpendicular to the vector to the CM, so there is no torque. The basic idea is to resolve the collision force into vectors through the CM of each body and perpendicular to the vector to the CM. The force through the CM accelerates the body. The transverse force changes the angular velocity around the CM. It sounds like in your case the bodies are rigid. In that case you can think in terms of impulse (which is the integral of force by time). Newton's $f=\frac{d(mv)}{dt}$ becomes $i_{\text{parallel}}=\Delta v/m$ and $i_{\text{perpindicular}}=\Delta \omega/I$

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:Can you point me to a more specific explination, or tell me what to search for to get the right results? Also some things don't make much sense... =p (is "perpendicular to the vector to the CM" a typo?) Thanks for replying! –  Griffin Sep 24 '11 at 5:15

I know, love it when people just give theorems and not the actual case. Words are hard to translate into pure mathematics. I do have a PDF I found online that is very useful in circle to circle elastic collisions. http://www.imada.sdu.dk/~rolf/Edu/DM815/E10/2dcollisions.pdf read all of it. its very straight forward.

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