A mass, m1 = 9.23 kg, is in equilibrium while connected to a light spring of constant k = 106 N/m that is fastened to a wall.A second mass, m2 = 6.42 kg, is slowly pushed up against mass m1, compressing the spring by the amount A = 0.169 mThe system is then released, and both masses start moving to the right on the frictionless surface. When m1 reaches the equilibrium point, m2 loses contact with m1 and moves to the right with speed v. A) Determine the value of v.B)How far apart are the masses when the spring is fully stretched for the first time. I have solved part A by using : 1/2KA^2=1/2mv^2 v=0.439 m/s for part b i know i have to find a new amplitude for block m1 and i did so. 1/2KA^2=1/2mv^2 A=0.1295 m Here is where i am stuck, the amplitude i found tells me the furthest distance that block m1 travels. However everytime i try to solve for time and then distance i get the wrong answer. Can someone help me find time and distance thus solving this problem??