Visualizing the trajectory of a mass on top of a wedge on a frictionless tabletop A wedge with mass M rests on a frictionless, horizontal tabletop. A block with mass m is placed on the wedge (Fig. 1). There is no friction between the block and the wedge. The system is released from rest.
Question: As seen by a stationary observer, what is the shape of the trajectory of the block?

The answer key states the shape is a spiral. I'm trying to visualize how this is possible, but I can't seem to wrap my head around it. I don't understand the shape or how it's meant to look across a timeline. Thus my main issue is how do I create an equation to represent the x-position of the block, assuming our reference point was at x = 0?

In an effort to graph the trajectory, I created the above figure (Fig. 2). I calculated an equation for the acceleration of the block in both the x and y-direction.
$$a_x=\frac{gM}{\left(M+m\right)\tan\left(\alpha \right)+\left(\frac{M}{\tan\left(\alpha \right)}\right)}$$
$$a_y=\frac{-g\left(M+m\right)\tan\left(\alpha \right)}{\left(M+m\right)\tan\left(\alpha \right)+\left(\frac{M}{\tan\left(\alpha \right)}\right)}$$
Then I calculated the acceleration of the wedge in the horizontal direction, in order to ensure my center of mass was standstill. It was.
$$A=\frac{-gm}{\left(M+m\right)\tan\left(\alpha \right)+\left(\frac{M}{\tan\left(\alpha \right)}\right)}$$
At this point, I graphed the kinematic displacement equations using acceleration from mock values to receive the above graph. Unfortunately, it does not look like a spiral, making me even more confused, wondering if the equation I used for horizontal displacement is showing an incorrect graph.
The variables used for the graph above were, M = 1 kg, m = 10 kg, angle = 30°, initial block height = 200 m.
 A: Welcome to physics SE! :) Your calculations seem about right, and regardless of the details of the calculation, the acceleration of the block will be constant as long as it has not left the wedge. The way to see this is to realize that the free body diagram of the system doesn't depend on where the block is on the wedge (again, as long as it is on the wedge) and thus, since none of the forces are velocity dependent or position-dependent (again, beyond their dependence on the fact that the block is on the wedge) your equations for acceleration won't depend on where the block is on the edge or on time.  And thus, the motion of the block would be a constant acceleration motion. There are only two options for the trajectory of a constant acceleration motion in a plane:

*

*It will be linear iff the object's initial velocity is zero or if its initial velocity is along the direction of acceleration (either parallel or anti-parallel).

*It will be parabolic iff the object's initial velocity is non-zero and not aligned with the direction of acceleration.

So, there is no way for the motion to be spiral.
