= Solution
Write the sphere's <mass> as $M$ and use $I=2Ma^2/5$ from <8/b>. Let $K_n$ and $K_t$ be the upward and rightward contact <impulses>. The initial vertical <velocity> is $-u_1$ and the outgoing vertical <velocity> is $u_2$. Perfect normal elasticity gives
$$
u_2=u_1,\qquad K_n=M(u_2+u_1)=2Mu_1.
$$
The tangential and angular <impulse> equations are
$$
M(v_2-v_1)=K_t,\qquad I(\Omega_2-\Omega_1)=aK_t.
$$
For the ideal <elastic rough-sphere bounce>, the tangential translational plus rotational <kinetic energy> is restored on separation. Substituting these two equations into that requirement gives
$$
0=K_t(v_1+a\Omega_1)
+\frac{K_t^2}{2}\left(\frac1M+\frac{a^2}{I}\right).
$$
The interacting rough-contact branch, rather than the smooth branch $K_t=0$, therefore has
$$
K_t=-\frac{2(v_1+a\Omega_1)}{1/M+a^2/I}
=-\frac47M(v_1+a\Omega_1).
$$
Consequently, with all tangential quantities signed in the convention of <8/c>,
$$
\boxed{u_2=u_1,\qquad
v_2=\frac{3v_1-4a\Omega_1}{7},\qquad
\Omega_2=-\frac{10v_1+3a\Omega_1}{7a}.}
$$
In particular, $v_2+a\Omega_2=-(v_1+a\Omega_1)$: the tangential <contact-point velocity> reverses. The outgoing $v_2$ may be negative; the sphere rebounds backward when $a\Omega_1>3v_1/4$. Substitution also checks both total <kinetic energy> and <angular momentum> about the contact point, $I\Omega-Mav$.
This result uses the conventional perfectly elastic, fully rough contact idealization. The word \i[rough] alone, with an unspecified finite <coefficient of friction>, would not determine the outgoing tangential state. For example, imposing sticking at separation would instead give $K_t=-2M(v_1+a\Omega_1)/7$ and dissipate $M(v_1+a\Omega_1)^2/7$, contradicting total elasticity unless $w_1=0$. The displayed elastic branch requires enough tangential <impulse>; a Coulomb bound $|K_t|\leq\mu K_n$ would at least require $\mu\geq2|v_1+a\Omega_1|/(7u_1)$.
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