Let the velocity of the object relative to the inclined plane at a certain moment be v and the velocity of the inclined plane at this moment be u, then the horizontal velocity of the object relative to the ground is (u-vcosθ).

If the momentum in the horizontal direction of the system is conserved, then:

mu+m(u-VCOθ)= 0

The solution is u= mvcosθ/(M+m)

Bilateral integration of time t

∫udt = M/(M+M)∫VCOθdt

∫ UDT-that is, the displacement x of the inclined plane relative to the ground.

∫VCOθdt- refers to the horizontal displacement of the object relative to the inclined plane, that is, h/tanθ.

Therefore, the displacement of the inclined plane x=mh/(M+m)tanθ.