A block of mass $m$ is placed on a wedge of mass $M$ and angle $\theta$. The wedge is placed on a smooth horizontal surface. If the block is released from rest on the wedge, the acceleration of the wedge is given by:

Solution

1. Let $a$ be the acceleration of the wedge and $a^{\prime }$ be the acceleration of the block relative to the wedge.
2. The horizontal force on the wedge is $N\sin \theta =Ma$, where $N$ is the normal force.
3. The relative acceleration along the incline is $a^{\prime }=g\sin \theta +a\cos \theta$.
4. Using the constraint $N=m(g\cos \theta -a\sin \theta )$, we solve for $a$.
5. Hence the answer is (A).