Center of Mass!

There are two phases in the motion, due to the wall to the left of the groove. In the first phase, the ball will slide to the bottom gaining speed and the groove will not move. The ball velocity at the bottom is $v_1$ , where from conservation of energy, we have

$\dfrac{1}{2} m v_1^2 = m g r$

In the second phase, the ball is still moving to the right, but the groove starts moving to the right as well, until their velocities are the same. Let's call this velocity $v_2$ , at this point, the ball makes an angle of $\theta$ with the vertical, and from conservation of energy, and equal velocities, we have,

$\dfrac {1}{2} (3 m) v_2^2 + m g r (1 - \cos \theta ) = m g r$

so that,

$\dfrac {1}{2} (3 m) v_2^2 = m g r \cos \theta$

which implies that,

$\cos \theta = \dfrac{\dfrac {1}{2} (3 m) v_2^2}{mgr} = \dfrac{\dfrac {1}{2} (3 m) v_2^2}{\dfrac{1}{2} m v_1^2} = 3 {(\dfrac{v_2}{v_1})}^2$

We also have from conservation of linear momentum,

$m v_1 = (3 m) v_2$

hence,

$\dfrac{v_2}{v_1} = \dfrac{1}{3}$

Substituting this above, we get

$\cos \theta = 3 (\dfrac{1}{3})^2 = \dfrac{1}{3}$

making the answer $1 + 3 = \boxed{4}$ .