homework and exercises - Kinematics question - Newton's Law of Motion

Question: Find the mass M of the hanging block in the following figure which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.

What I did: Considering that M-mass block is moving downwards, I assumed that the string would transmit force $Mg$ to the inclined plane. Thus the inclined plane would move forward with acceleration $\frac{Mg}{M^-}$ towards the right, while the m-mass block would move backward, in a horizontal direction, with the same acceleration. Calculating components, the acceleration up the incline would be $\frac{Mg}{M^-}\cos \alpha$. The acceleration of the block on its own would be $g\sin \alpha$. Equating the two, I got $M = M^-\tan \alpha$

New approach (considering horizontal component of normal force exerted by small block) Considering the fact that the normal force exerted by the wedge on the block to be $mg\cos \alpha$, the block exerts an equal force on the wedge. The horizontal component of this normal force will be $mg\cos \alpha\sin \alpha$. Thus the acceleration of the wedge will be

$$\frac{Mg - m\cos \alpha\sin \alpha}{M^-+m}$$ This should be equal to the downward acceleration of the small block: $g\sin \alpha$. Equation the two and solving for $M$, I got: $$M = \sin \alpha(M^- + m) + m\cos \alpha\sin \alpha$$ Still not the correct answer: $\frac{M^- + m}{\cot \alpha -1}$