Q.Define the coefficient of restitution for a collision between two bodies, in terms of their velocities of approach and separation, and explain why it equals exactly $1$ for a perfectly elastic collision and exactly $0$ for a perfectly inelastic collision.

The coefficient of restitution is defined as the ratio of the relative velocity of separation (measured just after the collision) to the relative velocity of approach (measured just before it), both along the line of impact: $e = \frac{v_2'-v_1'}{u_1-u_2}$

Why $e=1$ for a perfectly elastic collision. In §5.10, combining momentum conservation with kinetic-energy conservation for an elastic collision led to the relation $u_1-u_2 = -(v_1'-v_2') = v_2'-v_1'$ -- the relative velocity of approach and the relative velocity of separation are equal in magnitude (the bodies separate exactly as fast, relatively, as they approached, just with the sense reversed from "closing" to "opening"). Substituting this directly into the definition of $e$ gives $e = (u_1-u_2)/(u_1-u_2) = 1$.

Why $e=0$ for a perfectly inelastic collision. By definition, in a perfectly inelastic collision the two bodies stick together and move off with one single, common final velocity, so $v_1' = v_2'$ exactly. The relative velocity of separation is then $v_2'-v_1' = 0$, and since the relative velocity of approach $u_1-u_2$ is (in any real collision that actually occurs) nonzero, $e = 0/(u_1-u_2) = 0$. …

State the definition of $e$, then substitute the elastic-collision relation (approach speed equals separation speed) and the perfectly-inelastic condition (co …

• Writing the ratio upside down (approach speed over separation speed) or with the velocities in the wrong order, flipping the sign. …