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Let the mass of small block be $m$ and that of wedge be $M$ Now when the block slides down to the lowest point the change in its Gravitational potential energy is $mgh$.

Suppose it has velocity $u$ along the incline(w.r.t. wedge). Let $v$ be velocity of wedge.

So, its final $KE = \frac{1}{2}m(v^2+u^2+2uv\sin\beta)$ where $\beta$ is the angle of wedge.

Now in this situation how do we apply energy conservation i.e. do we conserve energy only of the block

$$mgh + 0 = 0+ \frac{1}{2}\left(m(v^2+u^2+2uv\sin\beta\right)$$


that of the wedge+block system

$$mgh + 0 + PE_{wedge} +0 = 0 + \frac{1}{2}m\left(v^2+u^2+2uv\sin\beta\right) + PE_{wedge} + \frac{1}{2}Mv^2$$


Assuming that energy is conserved in the first place, you would conserve energy for them both. Notice that the potential energy of the wedge is unlikely to change so you can just take it to be zero, as only energy differences are well-defined: actual energies are ambiguous up to an additive constant.

Depending on whether the wedge is stuck to the ground or not, it will either move or not and thus may or may not have a kinetic energy term. But taking this as 0 gives the right expressions when the wedge is fixed in-place and so conserving energy gives the right answer either way.


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