Minimum acceleration so that mass $m$ falls freely A block of mass $m$ rests on a frictionless wedge of angle $\theta$ as in the figure. The wedge is given an acceleration $a$. What is the minimum value of $a$ so that the mass $m$ falls freely?

I have solved a similar problem, where the acceleration was in the opposite direction as in this question, and the condition was to prevent the block from sliding down. The accleration in that question came to be equal to $(g\;tan\theta)$. What exactly must be done here so that the block falls freely? I don't need an actual answer. Just how are we to find it.
 A: A displacement of $x$ of the block moves the point under the center of inertia of the mass by the distance $x\tan\theta$. Therefore the acceleration of this point (the ground for the mass $m$) is $a\tan\theta$. For the block to experience a free fall, this acceleration must be larger than the gravity's acceleration $g$.
So you must have
$$a\geq\frac g{\tan \theta}.$$
A: In addition to the correct answer, a little strategy to solve problems involving constraints that are unilateral, that is, given by a condition$$f(x,y,z)\geq 0,$$
as in this case: $$y\geq \tan \theta (x+\frac {1}{2} at^2).$$
You can imagine that the constraint is bilateral  (that is, given by an equality) and look at the constraint force as a function of the parameters. When this changes sign, it means that the constraint force is attracting the point in order to satisfy the constraint. So the simplification of your unilateral constraint as a bilateral one is no longer valid and, in a physical situation, the point would lose contact with the surface.
