1. The problem statement, all variables and given/known data A M=1.5kg horizontal, uniform tray is attached to a vertical ideal spring of force constant k=185 N/m and m=275g 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 15cm below its equilibrium position and released from rest. 2. Relevant equations How high above the release position will the tray be when the ball leave the tray? 3. The attempt at a solution Lets call distance from the release point to equilibrium point A. Imagine the tray pushing the ball from release point to equilibrium point, when this distance is reached the acceleration of the system instantaneously becomes zero and the spring is now about to pull the tray downwards. At this moment the ball is no longer pushes the tray down, however because the ball does not push the tray down the equilibrium position is no longer the same as before but is a little above. So the tray is going to keep up same speed with the ball as long it reaches its new equilibrium position. If my idea is right, then the extra distance kx = mg, and x = mg/k, and final answer should be A + x Is that right?