How Does a Box's Speed Change When Sliding Down a Frictionless Ramp?

[Footballer010]

Homework Statement

A frictionless ramp of mass 3m is initially at rest on a horizontal frictionless floor. A small box of mass m is placed at the top of the ramp and then released from rest. After the box is released, it slides down the ramp and onto the horizontal floor, where it is measured to have a speed v, having fallen a total distance h.

What is the speed v of the box after it has left the ramp?

Homework Equations

U=mgh

The Attempt at a Solution

I have no idea.

 

[rl.bhat]

As the block slides down, wedge moves towards right. When the box slides on the horizontal floor with velocity v, the wedge moves with velocity V towards right. Since no external force is acting on the system, according to the conservation of linear momentum, we have
MV = mv...(1)
According to conservation of energy
mgh = 1/2*MV^2 + 1/2mvf^2...(2) where vf is the final velocity of the box when it reaches the bottom of the wedge.From the eq.1 we can write
1/2MV^2 = 1/2*(mv)^2/M...(3) Substitute this in eq.2. We get
mgh = 1/2*(mv)^2/M + 1/2*mvf^2. Now solve for v.
relation between v and vf is given as
v = vf*cosθ - V.