In the figure provided, the masses are m_1 = 5 kg and m_2 = 2 kg ,...

Let $a$ be the common acceleration of the blocks and $T$ be the tension in the string.

For block $m_1$ (moving downward), the equation of motion is:
$m_1 g + F - T = m_1 a$

For block $m_2$ (moving upward), the equation of motion is:
$T - m_2 g - F = m_2 a$

Adding the two equations, we get:
$m_1 g - m_2 g = (m_1 + m_2) a$

$a = \dfrac{m_1 - m_2}{m_1 + m_2} g$

Substituting $m_1 = 5 \text{ kg}$, $m_2 = 2 \text{ kg}$, and taking $g = 10 \text{ m/s}^2$:
$a = \dfrac{5 - 2}{5 + 2} \times 10 = \dfrac{30}{7} \approx 4.3 \text{ m/s}^2$

When the string breaks, the tension $T$ becomes zero. The forces acting on $m_1$ are its weight $m_1 g$ and the applied force $F$, both directed downwards.

The net downward force on $m_1$ is:
$F_{\text{net}} = m_1 g + F$

The new acceleration $a'$ of $m_1$ is:
$a' = \dfrac{m_1 g + F}{m_1} = g + \dfrac{F}{m_1}$

Substituting $F = 1 \text{ N}$ and $m_1 = 5 \text{ kg}$:
$a' = g + \dfrac{1}{5} = g + 0.2 \text{ m/s}^2$

Thus, $m_1$ moves downward with an acceleration of $g + 0.2 \text{ m/s}^2$.