= Solution
Choose positive horizontal <velocity> to the right and positive spin <angular velocity> anticlockwise, as in the source drawing. At the bottom of the sphere, the rotational <velocity> is then to the right: the initial tangential <contact-point velocity> is
$$
w_1=v_1+a\Omega_1.
$$
For the illustrated positive $v_1,\Omega_1$, the contact slips to the right, so the tangential contact <force> acts leftward. The upward <normal force> acts through the sphere centre and produces no spin <torque>. The tangential contact <force> acts at lever arm $a$ and produces a clockwise <torque>, reducing the positive spin. The weight acts vertically downward through the <centre of mass>.
During the short impact the contact <impulses> dominate the weight's <impulse>. In an ideal elastic rough contact, the tangential deformation can store and return <energy>; its contact <force> need not be a dissipative sliding-friction law throughout the collision. For other signed incoming states, reverse the initial tangential <force> if $w_1<0$; if $w_1=0$, no tangential <impulse> is required.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2017-superball-contact-forces.png]
{title=Contact forces and spin convention for the rough elastic sphere}
{description=The arrows depict the initial contact tendency for $v_1+a\Omega_1>0$. The normal contact force has zero spin torque, whereas a leftward tangential contact force produces clockwise torque about the centre.}
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