Let $\text{v}$ be the velocity of COM of the ring just after the impulse is applied and $\text{v'}$ is the velocity when pure rolling starts.

Angular velocity $\text{ω}$ of the ring at this instant will be $\mathrm{\omega }=\frac{\text{v'}}{\text{r}}$

  
 $\Rightarrow$Impulse = change in linear momentum
      we have,   $\text{J}=\text{mv}$
        or   $\text{v=J/m}$

   Between the two positions shown in figure force of
   friction on the ring acts backwards. Angular momentum
   of the ring about bottommost point will remain conserved

$\therefore {\text{L}}_{\text{i}}={\text{L}}_{\text{f}}$

or    $\text{mvr}=\text{mv'r}+\text{l}\omega$

       $=\text{mv' r +}\left({\text{mr}}^2\right)\frac{\text{v'}}{\text{r}}=2\text{mv' r}$

$\therefore$     $\text{v'}=\frac{\mathrm{v}}{2}=\text{J/2m}$