If the mass falls freely then it obeys the equation $\Delta y = \tfrac{1}{2}gt^2$. That means the wedge must accelerate sideways to make the height of the wedge decrease at the same rate.
I would guess we're assuming there is no friction between the block and the wedge. The idea is that the wedge moves sideways at such an acceleration that the block falls straight down i.e. moves downwards in a straight line.
@JohnRennie ....In this question:i.sstatic.net/4oMa2.jpg In the (a) why is the potential difference across C2 Va-Vb?....It should be Vb-Va right?....could you explain?
@Abcd and the question says the block falls freely so it accelerates downwards at an acceleration of $g$.
In effect it's like you are pulling out the wedge from under the block at a rate that matches the free fall of the block.
So if you consider the point on the wedge directly under the block then that point must accelerate downwards at $g$ so it doesn't impede the block's fall.
