Posted by Anonymous on .
A mass, m1 = 9.36 kg, is in equilibrium while connected to a light spring of constant k = 112 N/m that is fastened to a wall.
A second mass, m2 = 6.71 kg, is slowly pushed up against mass m1, compressing the spring by the amount A = 0.201 m
The 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. Determine the value of v.
At the spring's equilbrium point, all of the stored potential energy of the compressed spring becomes kinetic energy of both masses. No further force is applied by the spring at that instant, and then the masses separate as the spring force reverses direction.
The V (of both masses) at that instant is given by
(1/2) k A^2 = (1/2)(m1 + m2) V^2
Solve for V