Answer
$a_{com} = (2.35~m/s^2)~\hat{i}-(1.57~m/s^2)~\hat{j}$
Work Step by Step
We can find the magnitude of the acceleration of the system:
$F = (m_1+m_2)~a$
$m_2~g = (m_1+m_2)~a$
$a = \frac{m_2~g}{(m_1+m_2)}$
$a = \frac{(0.400~kg)(9.8~m/s^2)}{(0.600~kg+0.400~kg)}$
$a = 3.92~m/s^2$
We can find the horizontal component of the acceleration of the center of mass:
$a_{com,x} = \frac{(0.600~kg)(3.92~m/s^2)+(0.400~kg)(0)}{0.600~kg+0.400~kg}$
$a_{com,x} = 2.35~m/s^2$
We can find the vertical component of the acceleration of the center of mass:
$a_{com,y} = \frac{(0.600~kg)(0)+(0.400~kg)(-3.92~m/s^2)}{0.600~kg+0.400~kg}$
$a_{com,y} = -1.57~m/s^2$
We can express the acceleration of the center of mass in unit-vector notation:
$a_{com} = (2.35~m/s^2)~\hat{i}-(1.57~m/s^2)~\hat{j}$
