A particle of mass m is placed at rest on the top of a smooth wedge of mass M, which in turn is placed at rest on a smooth horizontal surface as shown in figure. Then the distance moved by the wedge as the particle reaches the foot of the wedge is :

Explanation

When block slides down the wedge, the wedge slides towards left. Displacement of the block relative to the wedge is L towards right in the horizontal direction. Letx be the displacement of wedge towards left, then net displacement of block will be (L - x) towards right.
If we include both the blocks into a single system then net force on it in the horizontal direction is zero hence acceleration of centre of mass in the horizontal direction is zero. Initially both the objects are at rest, hence initial velocity of the centre of mass is zero. We can conclude that centre of mass will remain at rest in the horizontal direction and for displacement of centre of mass to be zero we can write the following equation:
`Mx-m(K-x)`
`rArr x=(mL)/(m+M)`