Two objects, m1 = 5.00 kg and m2 = 3.00 kg, are connected by a light string passing over a light frictionless pulley as shown in the figure below. The object of mass 5.00 kg is released from rest, h = 4.00 m above the ground.
Using the isolated system model, determine the speed of the 3.00 kg object just as the 5.00 kg object hits the ground.
Kf + Uf = Ki + Ui
The Attempt at a Solution
So I thought that initially, energy was only in the form of gravitational potential energy for m1 and at the final state there is kinetic energy for m2 and gravitational potential energy for m2.
When I isolate for final velocity of m2, I end up getting:
v = (2*g*h(m1-m2)/m2)^0.5
But I am doing something wrong. Could someone point me in the right direction please?