A wedge having a vertical slot in it is placed on smooth horizontal surface as shown in the figure. Two blocks are arranged as shown in the figure. The system is released from rest calculate the speed of the wedge when block 1 comes down a distance `h`.

Explanation

We define system wedge `+` block 1 and block 2
Observation: No external force acts on the system in horizontal direction:
Conclusion: Both the linear momentum direction.
Analysis of Problem: If block 1 moves down a distance h with will be conserved for the system. respect to wedge, the block 2 will also slide towards left by the same distance with respect to wedge.
As wedge is not moving in vertical direction, the displacement of block 1 with respect to wedge will be same as with respect to the ground.
The linear momentum of the system in horizontal direction is conserved. Initial linear momentum of the system is zero. As block 2 moves towards left, to conserve linear momentum the wedge will start moving towards right.
As block 1 is moving the slot in the wedge, the velocity of block 1 horizontal direction will be same as the velocity of wedge.
Block 1 will also move in vertical direction with respect to wedge as well as with respect to ground.
The velocity of block 2 with respect to wedge will be same as velocity of block I in vertical direction as both are connected with same string.
Final step: Let velocity of block 1 in vertical direction be `v`. Hence velocity of block 2 with respect to wedge will be `v` towards left. Let velocity of wedge towards right be `V`.
Hence velocity of block 2 towards left (with respect to ground) will be (`v - V`).
Now using conservation of linear momentum for the system in horizontal direction.
`(M+m)V-m(v-V)=0`
as `m=3mimpliesv=5V`..............i
Now using conservation of mechanical energy for the system `/_\K+/_\U=0`
`[(1/2(M+m)V^(2)+1/2m(v-V)^(2)+1/2mv^(2))-0]+[-mg]=0`
`=1/2(3m+m)V^(2)+1/2m(5V-V)^(2)+1/2m(5V)^(2)=mgh`
`V=sqrt((2gh)/45)`