Conservation of momentum for moving inclined plane

I'm confused about the conservation of momentum in problems involving inclined planes that moves. Consider an inclined plane free to move on the floor and a body on the edge, there is no friction at all. Calculate the accelerations of the body and the inclined plane with respect to the floor.

So I considered the fact that, on the $x$ axis momentum is conserved, which means

$m \dot{x} +M \dot {X}=0$

And then

$m \ddot{x} +M \ddot {X}=0$ (1)

Since there is relative motion is possible to write

$\ddot{x}= a_{relative_x} + \ddot {X}$ (2)

The problem is about $a_{relative_x}$, I think that is

$a_{relative_x}=-mg sin(\alpha) cos(\alpha)$

But using equation (1) and (2) I get

$\ddot{X}=\frac{g sin(\alpha) cos(\alpha) m }{M+m}$

While the answer is $\ddot{X}=\frac{g sin(\alpha) cos(\alpha) m }{M+m sin^2(\alpha)}$

Further more, as long as $\ddot{y}$ is concerned, I thought that is should not be influenced at all by the motion of the inclined plane, i.e.

$\ddot{y}=-mg sin(\alpha) sin(\alpha)$, but the answer is

$\ddot{y}=\frac{-(m+M) Sin^2(\alpha) g }{M+mSin^2(\alpha) }$

Is this the right way to use the conservation of momentum?

• draw fbd and use pseudo force for relative motion Mar 6, 2016 at 18:08

As a suggestion I would make the $z-$axis, or $y-$axis if you wish, to point downwards because you know that the acceleration is downwards and this will remove a number of negative signs from your equations.
Rather than using subscripts calling $X$ the acceleration of the wedge and $x$ and $z$ the acceleration of the block you can write down three Newton's second law equations for the block and the wedge.
The block keeps in contact with the wedge and slides down at an angle $\alpha$.
The acceleration of the block relative to the wedge is $z$ downwards and $x-X$ in the positive horizontal direction.
This will give you a connection between $\alpha$, $z$ and $x$.