Find Block's Acceleration w/ 45 Degree Wedge & Constant A

[Radarithm]

Homework Statement

A 45 degree wedge is pushed along a table with constant acceleration A. A block of mass m slides without friction on the wedge. Find the block's acceleration. Gravity is directed down.

Homework Equations

Constraint Equation:
$$\tan\theta=\frac{\ddot{y}}{\ddot{x}-\ddot{X}}$$

Equations (p is for parallel and p² is for perpendicular, eg. x and y directions):

$$F_{pb}=N\sin{\theta}=m\ddot{x}$$
$$F_{p^2b}=N\cos{\theta}-mg=m\ddot{y}$$
$$F_{pw}=F-N'\sin{\theta}=MA$$

The Attempt at a Solution

It's embarrassing how this problem seems difficult for me; I'm probably making some fundamental error somewhere.

Newton's 3rd Law: $N=N'$
$$N=\frac{m\ddot{x}}{\sin{\theta}}=\frac{m(g+\ddot{y})}{\cos{\theta}}$$

Because the angle is 45 degrees, the tangent of theta is just 1. Solving for the x acceleration, we get:
$$\ddot{x}=(g+\ddot{y})$$

Now, according to the constraint equation, the acceleration in the y direction is equal to the tangent of theta times the x acceleration of the block minus the x acceleration of the wedge itself. Following the constraint equation, I get:
$$\ddot{y}=(g+\ddot{y})-A$$

(tangent of 45 is 1). There is no way to solve for the y acceleration here. If, back when I solved for the normal force, I included the tangent of theta, I would've been able to solve for the acceleration but I would have to divide by zero, which means there is no y acceleration. The hint, however, states otherwise: If A = 3g then the y acceleration is g. Where am I going wrong? I'm going to try and solve for the y acceleration instead of the x acceleration first and see where that takes me.

 

[Radarithm]

I've reached a rather strange conclusion; I solved for the y acceleration in from the normal force equation (2 of them):
$$\frac{m\ddot{x}}{\cos{\theta}}=\frac{m(g+\ddot{y})}{\sin{\theta}}$$
$$\ddot{y}=\ddot{x}-g$$

From the constraint equation:
$$(\ddot{x}-\ddot{X})\tan{\theta}=\ddot{y}=\ddot{x}-g$$

Solving for x, we get:
$$\ddot{x}=\frac{-\ddot{X}\tan{\theta}}{(1-\tan{\theta})}-g$$

If A is 3g, then the y acceleration must be g; this equation shows that, but it is for the x direction. When I plug it into the y equation, things don't make any sense.

 

[Radarithm]

I feel like I'm making a mistake that deals with inertial reference frames; bump?