Two piloted satellites approach each other head on at a relative speed of 0.250 m/s but collide elastically instead of docking. Using the general one-dimensional elastic collision equations, what is their final relative speed?

1. Momentum conservation gives m1(v1 minus v1 prime) equal to m2(v2 prime minus v2). 2. Internal kinetic energy conservation gives m1(v1 squared minus v1 prime squared) equal to m2(v2 prime squared minus v2 squared), which factors using the difference of squares. 3. Dividing the factored energy equation by the momentum equation cancels the masses and leaves v1 plus v1 prime equal to v2 prime plus v2. 4. Rearranging that result shows v1 minus v2 equal to negative (v1 prime minus v2 prime), meaning the relative velocity reverses sign but keeps the same size. 5. Since the two satellites approached at a relative speed of 0.250 m/s, they must separate at that same 0.250 m/s, just moving apart instead of coming together. 6. This result depends only on both conservation laws holding at once, not on the actual masses of either satellite. _Source: OpenStax College Physics (CC BY 4.0), Ch 8 "Linear Momentum and Collisions", section 8.4 Elastic Collisions in One Dimension_