A block of mass $m$ is placed at rest on a smooth wedge of mass $M$ placed at rest on a smooth horiz

Question

A block of mass $m$ is placed at rest on a smooth wedge of mass $M$ placed at rest on a smooth horizontal surface. As the system is released the center of mass of the system

Professor PC Thomas Classes · Answer

The net vertical downward force acting on the system is $(M+m) g-N$, where $N$ is the normal reaction from the horizontal surface. Therefore, the downward acceleration of the system is $ a=\frac{(M+m) g-N}{M+m}=g-\frac{N}{M+m}<g $