A solid sphere is set into motion on a rough horizontal surface,...

Let the mass of the solid sphere be $m$ and its radius be $R$.

The initial linear velocity of the sphere is $v$ (forward) and its initial angular velocity is $\omega_0 = \dfrac{v}{R}$ (anticlockwise).

Taking the clockwise direction as positive, the initial angular momentum of the sphere about a fixed point on the ground is:

$L_i = m v R - I \omega_0$

Since the moment of inertia of a solid sphere is $I = \dfrac{2}{5} m R^2$, we have:

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

Let the final linear velocity of the sphere when pure rolling begins be $v_f$. For pure rolling, the final angular velocity will be $\omega_f = \dfrac{v_f}{R}$ (clockwise).

The final angular momentum about the same point on the ground is:

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

The frictional force acts at the point of contact, so its torque about any point on the ground is zero. The torques due to the normal force and gravity cancel each other. Thus, the net torque about the point on the ground is zero, and angular momentum is conserved:

$L_i = L_f$

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

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