A uniform solid ball of mass 2.0 kg and radius 0.10 m is launched along a rough horizontal floor…

Step 1. Friction at the start. The bottom of the ball moves forwards at $v_0$ because of the centre's motion, and forwards again at $\omega_0 R = 9.0\ \text{m/s}$ because of the backspin. So it slides forwards at $15.0\ \text{m/s}$, and friction acts backwards, with $\mu mg = 4.0\ \text{N}$. It slows the centre at $2.0\ \text{m/s}^2$ and reduces the backspin at $\dfrac{fR}{I} = 50\ \text{rad/s}^2$.

Step 2. Angular momentum about the floor. Take a fixed point on the floor. Friction acts along the floor, so it has no torque about that point; the weight and the normal force cancel along one vertical line. So the angular momentum about that point is conserved. Taking forward rolling as positive:

When the ball rolls, $\omega = v/R$ and $L = mvR + I\dfrac{v}{R} = \tfrac75 mRv = 0.28\ \text{kg m} \times v$. So $v = \dfrac{0.48}{0.28}\ \text{m/s} = 1.71\ \text{m/s}$, forwards.

Step 3. Time. The centre slows at $2.0\ \text{m/s}^2$ from $6.0\ \text{m/s}$ to $1.714\ \text{m/s}$: $t = \dfrac{4.286\ \text{m/s}}{2.0\ \text{m/s}^2} = 2.14\ \text{s}$. In that time the spin goes from $-90\ \text{rad/s}$ to $-90 + 50 \times 2.143 = 17.14\ \text{rad/s}$, which is $v/R$, as rolling needs.

Step 4. Heat. At the start: $\tfrac12 mv_0^2 + \tfrac12 I\omega_0^2 = 36.0\ \text{J} + 32.4\ \text{J} = 68.4\ \text{J}$. When rolling: $\tfrac{7}{10}mv^2 = 4.11\ \text{J}$. So $64.3\ \text{J}$ has turned into heat.

Step 5. Coming back. The final velocity points backwards if $L < 0$, that is if $I\omega_0 > mv_0R$: $\omega_0 > \dfrac{5v_0}{2R} = 150\ \text{rad/s}$.