Hi Andrew

Yeah, so the basic center-of-mass-frame argument was set kinetic energies equal to spring energy:

m1v1^2 + m2v2^2 = kx^2



But this also works in any arbitrary frame. Call the two velocities (vcm + v1) and (vcm + v2), corresponding to the center of mass velocity and their relative velocities in the center of mass frame (so that they add up to the total velocity in any arbitrary frame). Then the energy equation looks like:

m1(vcm + v1)^2 + m2(vcm + v2)^2 = kx^2 + (m1 + m2)vcm^2

Expanding the LHS, you get:

[m1vcm^2 + m2vcm^2] + 2vcm * (m1v1 + m2v2) + m1v1^2 + m2v2^2 = kx^2 + [(m1 + m2)vcm^2]

The first two terms on the left hand side (LHS; bracketed) cancel w/ the last term on the RHS (bracketed). The second term, in the parentheses, *vanishes* by definition of the center of mass: m1v1 + m2v2 = 0 in this frame! That's the crux of the argument :)

Energy is a bit weird to work through sometimes, since it's not linear in the velocities, but with proper bookkeeping it should always work out :)