Conserve Momentum Always; Conserve Energy Only If the Collision Is Elastic

A perfectly inelastic collision gives v = (m1 u1 + m² u2) / (m1 + m²) with velocities signed: a 10 g clay lump moving east at 10 cm/s that hits a 20 g lump moving west at 15 cm/s gives (100 - 300)/30 = -6.7 cm/s, about 6.7 cm/s toward the west. In a head-on elastic collision between identical masses the velocities simply interchange: a 6 m/s ball hitting an identical ball at rest stops, and the second moves off at 6 m/s; balls at +5 and -3 m/s leave with -3 and +5 m/s. Momentum is conserved in all collisions, but kinetic energy only in elastic ones.

If 60% of the energy is lost at a bounce, only 40% remains, so a ball dropped from 10 m rebounds to 4 m. The total distance travelled by a ball dropped from h before it stops bouncing is h + 2 e² h + 2 e⁴ h + ... = h (1 + e²) / (1 - e²).

At maximum compression of a spring between two moving masses both have the same velocity. For 2 kg at 10 m/s behind 5 kg at 3 m/s, the common velocity is 35/7 = 5 m/s and energy conservation, (1/2)(2)(100) + (1/2)(5)(9) = (1/2)(7)(25) + (1/2)(1120) x², gives x = 0.25 m. In two dimensions, conserve momentum along both axes: a 9 m/s ball striking an identical ball at rest, with both leaving at 30° to the original line, satisfies 9 = 2 v cos 30°, so v = 3 sqrt(3) m/s. If sphere B moving at 2v along x leaves at v along y, sphere A receives momentum components along x and minus y, so it moves at tan θ = -1/2 relative to the x-axis (or +1/2 if B had gone the other way).