A uniform solid sphere of mass m and radius R rolls without slipping with speed v on a rough horizontal floor towards a vertical wall. The wall is…

The wall is smooth and the collision elastic, so just after it the centre moves with $-v$ while the spin is unchanged ($\omega=v/R$ in the original sense). Now the sphere slips on the floor. Friction acts at the floor, so angular momentum about any point on the floor line is conserved during slipping. Take the original sense of rotation as positive: just after the wall, $L=\tfrac25 mR^2\omega - mRv=\tfrac25 mRv-mRv=-\tfrac35 mRv$. Finally, rolling with centre velocity $v_f$ (spin sense fixed by $v_f=\omega_f R$): $L=\tfrac25 mRv_f+mRv_f=\tfrac75 mRv_f$. Equating: $v_f=-\tfrac37 v$, so the sphere finally rolls away from the wall with speed $3v/7$. Initial KE is $\tfrac12mv^2(1+\tfrac25)=\tfrac7{10}mv^2$ and final KE is $\tfrac7{10}m\cdot\tfrac9{49}v^2$. Fraction lost $=1-\tfrac9{49}=\tfrac{40}{49}$. The distractor $9/49$ is the fraction remaining; the others come from ignoring the retained spin.