- #1

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We start with a light mass moving at velocity v, having linear momentum p=mv and kinetic energy (mv^2)/2, and a heavy mass at rest, with linear momentum p=M(0)= 0 and kinetic energy (m(0^2))/2= 0. The total momentum of the system is 0 + mv = mv, and the total kinetic energy is 0 + (mv^2)/2 = (mv^2)/2.

Suppose the light mass collides with the heavy mass in such a way that it has a new velocity nearly equal to its original velocity but opposite in direction. It's new momentum is m(-v)= -mv, and its new kinetic energy is (m(-v)^2)/2= (mv^2), which is the same kinetic energy it began with. Since the initial momentum of the system was p = mv, we have

mv = x -mv, where x is the final momentum of the heavy mass. So x = 2mv. Since the heavy mass has a momentum, it must have kinetic energy. But the light mass has the same kinetic energy that it had at the start, so adding any energy to that would give us more kinetic energy than we started with.

What did I miss?