Two bodies meet. Whatever happens in the milliseconds between — noise, heat, bent steel — total momentum is the same before and after, provided nothing outside the pair pushes on them. That is the law: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$. Fix the convention once and the rest is bookkeeping. Subscript 1 is the first body and 2 is the second, both masses in kilograms. The letter u is a velocity BEFORE the collision and v is the same body's velocity AFTER — so $u_1$ is body 1 going in and $v_1$ is body 1 coming out. All four are velocities, not speeds: pick a positive direction, and anything travelling the other way carries a minus sign.

Kinetic energy is NOT protected. Some of it always goes into the coupler, the deformation, the bang. That is the difference between the two families, and choosing between them is the formula-pick of this lesson.

When the bodies stick together — a coupling, a wet clay ball, a bullet in a block — they leave with one shared velocity, and momentum alone gives it: $v = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$. Read it as a weighted average of the two incoming velocities. Note what sits in the denominator: the COMBINED mass, both bodies, because both of them are now going wherever they are going together. Forgetting the second mass down there is the classic error, and it always gives you an answer that is too fast.

When the bodies bounce apart, the missing information is how bouncy they were, and that is the coefficient of restitution: $e = \dfrac{v_2 - v_1}{u_1 - u_2}$ — separation speed over approach speed, a bare ratio with no units. $e = 1$ is perfectly elastic, all the relative motion returned, which no real material quite manages; $e = 0$ is the sticking case; everything real lies between. Hardened steel bearings reach about 0.95, a squash ball off a cold wall barely 0.3. If you ever compute an $e$ above 1, the collision would be handing out energy it never received — go back and look.