In a simple Atwood machine, two unequal masses m_1 and m_2 are...

Given $m_1 = 300 \text{ g}$ and $m_2 = 600 \text{ g}$.

The initial acceleration of the system is:
$a = \dfrac{m_2 - m_1}{m_2 + m_1} g = \dfrac{600 - 300}{600 + 300} g = \dfrac{300}{900} g = \dfrac{g}{3}$

The velocity of the system just before the larger mass is stopped ($t = 2.0 \text{ s}$) is:
$v = u + at = 0 + \left(\dfrac{g}{3}\right)(2) = \dfrac{2g}{3}$

When $m_2$ is momentarily stopped, its velocity becomes zero. However, $m_1$ is still moving upwards with velocity $v = \dfrac{2g}{3}$. At this point, the string becomes slack.

Both masses now move freely under gravity. Let $t$ be the time until the string becomes tight again.

For the string to become tight, the distance $m_1$ moves relative to the pulley must equal the distance $m_2$ falls. In other words, the upward displacement of $m_1$ must equal the downward displacement of $m_2$.

Displacement of $m_1$ (moving upwards against gravity):
$y_1 = vt - \dfrac{1}{2}gt^2$

Displacement of $m_2$ (falling downwards from rest):
$y_2 = \dfrac{1}{2}gt^2$

Equating the displacements for the string to regain tension:
$vt - \dfrac{1}{2}gt^2 = \dfrac{1}{2}gt^2$

$vt = gt^2$

Since $t \neq 0$, we have:
$t = \dfrac{v}{g}$

Substitute $v = \dfrac{2g}{3}$:
$t = \dfrac{\dfrac{2g}{3}}{g} = \dfrac{2}{3} \text{ s}$