Answer
After the containers are released, the center of mass moves down with an acceleration magnitude of $~~0.0157~m/s^2$
Work Step by Step
The horizontal location of the center of mass will not move since the containers do not move horizontally.
However, the vertical location of the center of mass will move after the containers are released.
We can find the magnitude of acceleration of the system of containers:
$\sum F = Ma$
$(0.520~kg)(9.8~m/s^2)-(0.480~kg)(9.8~m/s^2) = (0.520~kg+0.480~kg)~a$
$a = \frac{(0.520~kg)(9.8~m/s^2)-(0.480~kg)(9.8~m/s^2)}{1.0~kg}$
$a = 0.392~m/s^2$
Container 2 will move down with this magnitude of acceleration while container 1 will move up with this magnitude of acceleration.
Let "up" be the positive direction.
We can find the acceleration of the center of mass:
$a_{com} = \frac{(0.520~kg)(-0.392~m/s^2)+(0.480~kg)(0.392~m/s^2)}{0.520~kg+0.480~kg}$
$a_{com} = -0.0157~m/s^2$
After the containers are released, the center of mass moves down with an acceleration magnitude of $~~0.0157~m/s^2$
