There is a simple shortcut to a problem like this. I can give you the proof if you like, but quite simply:Originally Posted bybabygirl

(Where , is the net external force on the system, and is the CM acceleration.) This is true even if the masses are not the same.

So N (Where and are the unit vectors in the x and y directions, respectively.)

Thus =

So the magnitude of is

and the direction is and is in Quadrant I.

Thus the acceleration of the center of mass is (to the correct number of sig. digs.) 6.0 m/s2 at an angle of 27 degrees above the positive x-axis.

-Dan