A thin spherical shell resting on a rough horizontal surface is...

Let the mass of the thin spherical shell be $m$ and its radius be $R$.

The moment of inertia of the shell about its center is $I = \dfrac{2}{3}mR^2$.

Initially, the shell has a linear velocity $v$ and zero angular velocity. The angular momentum of the shell about any point on the horizontal surface is conserved because the frictional force passes through the surface (exerting zero torque) and the torques due to the normal reaction and weight cancel each other.

Initial angular momentum about a point on the surface is:

$L_i = m v R$

Let $v'$ be the linear speed and $\omega$ be the angular speed when pure rolling commences. For pure rolling, $v' = \omega R$.

Final angular momentum about the same point is:

$L_f = I \omega + m v' R$

Substituting $I = \dfrac{2}{3}mR^2$ and $\omega = \dfrac{v'}{R}$:

$L_f = \left(\dfrac{2}{3}mR^2\right) \left(\dfrac{v'}{R}\right) + m v' R = \dfrac{5}{3}m v' R$

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

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

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