A solid sphere rolling on a rough horizontal surface with a linear...

Before the collision, the solid sphere is in pure rolling. Its linear velocity is $v$ and its angular velocity is $\omega = \dfrac{v}{R}$.

During the elastic collision with the smooth vertical wall, the linear velocity of the sphere reverses its direction, becoming $v$ in the backward direction. Since the wall is smooth, it exerts no torque about the center of mass, meaning the angular velocity remains unchanged in magnitude and direction.

Immediately after the collision, the sphere moves backward with speed $v$ but continues to rotate in the original direction with angular speed $\omega = \dfrac{v}{R}$.

Considering a point on the horizontal surface along the line of motion, the torque due to friction about this point is zero. Thus, the angular momentum of the sphere about this point is conserved.

Taking the backward direction as positive for linear velocity and the corresponding pure rolling rotation as positive for angular velocity, the initial angular momentum about the point on the ground is:

$L_i = m v R - I \omega$

Substituting $I = \dfrac{2}{5}mR^2$ and $\omega = \dfrac{v}{R}$:

$L_i = m v R - \left(\dfrac{2}{5}mR^2\right)\left(\dfrac{v}{R}\right) = \dfrac{3}{5}m v R$

Let $v_f$ be the final speed of the sphere when it commences pure rolling in the backward direction. The final angular velocity will be $\omega_f = \dfrac{v_f}{R}$. The final angular momentum about the same point is:

$L_f = m v_f R + I \omega_f = m v_f R + \left(\dfrac{2}{5}mR^2\right)\left(\dfrac{v_f}{R}\right) = \dfrac{7}{5}m v_f R$

By conservation of angular momentum, $L_i = L_f$:

$\dfrac{3}{5}m v R = \dfrac{7}{5}m v_f R$

$v_f = \dfrac{3v}{7}$