As shown in the figure, a solid sphere is set into motion on a...

Let the mass of the solid sphere be $m$ and its radius be $R$. The moment of inertia of the solid sphere about its center of mass is $I = \dfrac{2}{5}mR^2$.

Consider a fixed point on the horizontal surface along the line of motion. The net torque about this point is zero because the line of action of the frictional force passes through the surface, and the torques due to the normal reaction and gravitational force cancel each other out.

Therefore, the angular momentum of the sphere about any fixed point on the ground is conserved.

Taking the clockwise direction as positive, the initial angular momentum due to translation is $mvR$ (clockwise) and due to rotation is $I\omega$ (anticlockwise).

Total initial angular momentum is:
$L_i = mvR - I\omega$

Substituting $I = \dfrac{2}{5}mR^2$ and $\omega = \dfrac{v}{R}$:
$L_i = mvR - \left(\dfrac{2}{5}mR^2\right)\left(\dfrac{v}{R}\right) = mvR - \dfrac{2}{5}mvR = \dfrac{3}{5}mvR$

Let $v'$ be the linear speed of the sphere at the instant it stops rotating. At this instant, its final angular velocity $\omega'$ is zero.

Total final angular momentum is:
$L_f = mv'R + I(0) = mv'R$

By conservation of angular momentum ($L_i = L_f$):
$\dfrac{3}{5}mvR = mv'R$

$v' = \dfrac{3v}{5}$