Distance moved till rolling.

Let the ball be given angular velocity $\omega$ in *clock wise * direction.

Consider $k$ in the diagram as $\mu$ Let us fix our frame of reference with the moving wedge. As it is a non-inertial frame so we have to apply pseudo force on the sphere. As the acceleration of the plank is $\mu g$ towards left so pseudo force acting on the ball is $\mu mg$ in right ward direction.

Acceleration of the CoM of theball $(a)=2\mu g$

Let us assume that the ball starts pure rolling motion after time $t$ .

So velocity of CoM of the ball after timr $t$ is $2\mu gt$

Let the angular velocity of the ball be $\omega_{0}$ when it starts pure rolling motion.

So $R\omega_{0}=2\mu gt$ .................(1)

Now torque acting on the ball the CoM is $\mu mgR$

So

$I \alpha=\mu mgR$

Note that the directions of the angular velocity and the angular acceleration are opposite.

Putting $I=\frac{2}{5} mR^{2}$

we get $\alpha=\frac{5\mu g}{2R}$

Now

$\omega_{0}=\omega+(-\alpha t)$

Putting the values in this equation we get

$t=\frac{2R\omega}{9\mu g}$

Now distance moved by plank $(S)$ is $0.5at^2$

So $S=\frac{2R^2 \omega^2}{81\mu g}$

Let's consider the forces acting on the sphere and the block individually. First of all, we note that as the sphere rolls, the bottom most point is travelling to the left, so friction on the sphere will be acting to the right. Clearly, an equal and opposite force will be acting to the left on the plank. We can set up the equations of motion for both systems:

$I \alpha = F_f R = mg \mu R$ $ma = F_f = mg \mu$ $mA = -F_f = -mg \mu$

where $a$ is the translational acceleration of the sphere, $A$ is the acceleration of the plank, and $\alpha$ is the angular acceleration of the sphere.

We are interested in the time at which the sphere rolls without slipping with respect to the plank. If we let the translational velocity of the sphere be $v$ and the velocity of the plank be $V$ , then the velocity of the sphere with respect to the plank is $v - V$ , so we wish to compute the time at which:

$\omega = \frac{v - V}{R}$

where $\omega$ is the angular velocity of the sphere. Since the acceleration is constant, we then will have:

$\omega = \omega_0 - \alpha t = \omega_0 - \frac{ mg \mu R t}{I}$ $V = At = -mg \mu t$ $v = at = mg \mu t$

which gives us:

$\omega_0 - \frac{ mg \mu R t}{I} = \frac{2 mg \mu t}{R} \ \Rightarrow \ \Big( \frac{2 g \mu}{R} + \frac{mg R \mu}{I} \Big) t = \omega_0 \ \Rightarrow \ t = \frac{2 R \omega_0}{9 g \mu}$

as a sphere's moment of inertia is given by $I = \frac{2}{5} mR^2$ . Finally, in the case of constant, acceleration, the distance that the plank travels is given by:

$D = \frac{1}{2} |a| t^2 = \frac{1}{2} g \mu \Big( \frac{2 R \omega_0}{9 g \mu} \Big)^2 = \frac{2 R^2 \omega_0^2}{81 g \mu}$

So we will have:

$a = 2, \ b = 81 \ \Rightarrow \ \boxed{a + b = 2 + 81 = 83}$