What Determines When a Metal Ball Leaves an Oscillating Tray?

[iwonde]

Homework Statement

A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up-and-down. The tray is then pushed down 15.0 cm below its equilibrium point (call this point A) and released from rest. (a) How high above point A will the tray be when the metal ball leaves the tray? (b) How much time elapses between releasing the system at point A and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?

Homework Equations

ΣF = ma_y
-mg-ky = m((d^2y)/(dt^2)) + (k/m)y + g = 0

The Attempt at a Solution

I'm not sure of how to approach this problem, but I'm thinking of solving for y in the equation above.

 

[DavidWhitbeck]

I'll provide a hint:

When the ball reaches the position at which it leaves the tray, it must lose contact with the tray, that implies that the condition for leaving the tray is that the normal force vanishes. You can solve for the normal force on the ball using Newton's 2nd Law.

The first thing you should then do is draw a free body diagram, identify all forces acting on the ball and in which direction they act.

After that you can write down the form of Newton's 2nd Law that holds for the ball.

After that apply my hint to solve for the position at which the ball leaves the tray, and then do the usual song and dance with your periodic motion equations, and I'm sure by that point you'll have no trouble proceeding.