Colliding cylinders

Let the initial velocity of the CoM (of cylinder 1) of the cylinder and initial angular velocity of the cylinder be $v$ and $\omega$ respectively. Let $\omega$ be in clock wise direction initially.

Let the final velocity of the CoM of (of cylinder 1) the cylinder and final angular velocity of the cylinder be $v_{0}$ and $\omega_{0}$ respectively. Let $\omega_{0}$ be in clock wise direction.

Let the final velocity of the CoM (of cylinder 2) be $v_{1}$ and angular velocity be $\omega_{1}$ in anti-clock wise direction.

As net impulse is equal to change in momentum and net angular momentum is equal to the change in angular momentum so

$J=m(v_{0}+v)$ .....................(1) (Let $v$ and $v_{0}$ be in opposite direction)

$\mu JR=I(\omega_{0}-\omega)$ .........................(2)

For cylinder 2

As the cylinder don't jump so $\mu J=J_{N}$ (I have neglected mg as it is very small as compared to the impulses)...........................(3)

As net impulse is equal to change in momentum and net angular momentum is equal to the change in angular momentum so

$J-\mu J_{N}=mv_{1}$ ...................(4)

$I\omega_{1}=(\mu J-\mu J_{N})R$ .......(5)

Now as the collision is elastic so we can equate velocity of approach and separation in common normal direction so

$v=v_{0}+v_{1}$ .................(6)

On solving these equations we get

${ v }_{ 1 }=\frac { 6\times 17.5 }{ 7 }$

and ${ \omega }_{ 1 }=\frac { 4\times 17.5 }{ 7 }$

Now we can find final angular angular velocity by conserving angular momentum about point of contact.

So

$mv_{1}R-I\omega_{1}=mv_{f}R+I\omega_{f}$

when it starts pure rolling then $v_{f}=R\omega_{f}$

On solving this we get

${ \omega }_{ final }=\frac { 20 }{ 3 }$ . So $a+b=23$ :)