= Elastic rough-sphere bounce
{title2=$K_t=-2(v+a\Omega)/(M^{-1}+a^2/I)$}
An ideal <elastic collision> between a sphere and a stationary plane reverses normal approach <velocity>. On the fully rough, tangentially elastic branch, the tangential <contact-point velocity> reverses too. If $K_t$ is the signed tangential <impulse>, translation and spin obey $M\Delta v=K_t$ and $I\Delta\Omega=aK_t$, with positive spin anticlockwise. Restoring tangential translational plus rotational <kinetic energy> gives
$$
0=K_t(v+a\Omega)+\frac12K_t^2(M^{-1}+a^2/I).
$$
The nonzero root is the displayed <impulse>; the other root is the smooth-contact branch. For $I=2Ma^2/5$, the map is $v'=(3v-4a\Omega)/7$ and $a\Omega'=-(10v+3a\Omega)/7$. It preserves <kinetic energy> and the <angular momentum> about the contact point, $I\Omega-Mav$. An unspecified rough surface need not realize this idealization: finite <friction> or a dissipative tangential contact law requires additional constitutive data. Setting the outgoing contact slip to zero instead loses $M(v+a\Omega)^2/7$ of <kinetic energy> for this uniform sphere.
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