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 to point A, which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point A will the tray be when the metal ball leaves the tray? (?Hint: This does not occur when the ball and tray reach their maximum speeds.) (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?

Solution 71P Step 1: The ball tray-spring system is released under the equilibrium at point A. The ball lives the tray at some point between the equilibrium and the top of oscillation. Step 2: A) he equation for this is amax = A 2 x = Acos(t + ) v = Asin(t + ) So, by equating the acceleration of the ball and the string g = x2 Step 3: The period of oscillation to be T = 2( )m k 1.50 T = 2( 185 )=0.616sec To get equilibrium at point A t=T/4 = 0.415sec