 Relative speed retention in an oblique collision

For a two-particle collision conserving total momentum, the outgoing speeds in the centre-of-momentum frame are multiplied by the same $q$. The relative kinetic energy is multiplied by $q^2$. For a smooth oblique collision with normal coefficient of restitution $e_n$ and incident relative velocity inclined by $\chi$ to the contact normal, unchanged tangential relative motion gives $q^2=e_n^2{\cos }^2\chi +{\sin }^2\chi$. Thus $q=e_n$ is not a general oblique-impact identity: $e_n=1/2$ and $\chi =\pi /4$ give $q=\sqrt{5/8}$. Sticking requires all relative motion to vanish, not merely its normal component.