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### Center of mass reference frame

What we'll do is change to a moving coordinate system, which goes with the velocity of the center of mass! That's sounds weird. It makes the problem sound evem harder. But actually you'll see how simple things look in this frame.

Remember we said that if momentum is conserved, the center of mass velocity of the system is also. As the collision is taking place, it doesn't alter the motion of the center of mass a bit. It just plods along at a constant velocity. If we were coasting along on a bike at this center of mass velocity, watching the collision, what would we see?

Well in this reference frame, the center of mass velocity, by definition, is zero. And therefore by eqn. 1.37 the total momentum is also zero. I'll notate all variables in this new system the same as the old, but just to remind ourselves that we're in this new frame I'll also add " ' " to them. So for example, the initial momentum of thr first particle is denoted .

So let's play before and after again, but this time in the center of mass reference frame.

Before:
The total momentum . The total energy is .
After:
The total momentum . The total energy is .

This looks a lot simpler. The momentum equations say that the particles have equal and opposite momenta, and Using this, equating energy is almost as easy

Factoring the masses and cancelling gives . There are two solutions to this. One is kind of boring, . It means that before and after, nothing changes. This certainly obeys conservation of energy and momentum, but means that the particles haven't bounced off each other. So what's the other more interesting solution? It's . From conservation of momentum, that means . In terms of velocity this gives

This says that after the collision, the two balls have reversed their initial velocities. That's it. This satisfies both momentum and energy conservation.

Next: Back to the old Up: Elastic collisions Previous: elastic collisions in one

Joshua Deutsch
Fri Jan 17 12:19:41 PST 1997