Two objects are connected by a light string passing over a light, frictionless pulley as shown in figure. The object of mass `5.00kg` is released from rest. Using the isolated system model, (a) determine the speed of the `3.00kg` object just as the `5.00-kg` object hits the ground, (b) find the maximum height to which the `3.00-kg` object rises.

Explanation

Using conservation of energy for the system of Earth and the two objects,
a. `(5.00kg)g(4.00m)=(3.00kg)g(4.00m)+1/2(5.00+3.00)v^2`

`v=sqrt(20)=2sqrt5ms^-1`
b. Now we apply conservation of energy for the system of the `3.00kg` object and Earth during the time interval between teh instant when the string goes slack and the instant at which the `3.00kg` object reaches its highest position in its free fall.
`1/2(3.00)v^2=mgDeltay=3.00gDeltay`
`Deltay=1.00m`
`y_(max)=4.00m+Deltay=5.00m`