Placed on a smooth horizontal surface, two blocks of masses m_1 and...

Let the displacements of the blocks $m_1$ and $m_2$ be $x_1$ (to the left)$and x_2$ (to the right) respectively.

The total elongation of the spring is $x = x_1 + x_2$.

The equations of motion for the two blocks are:
$m_1 \ddot{x}_1 = F_1 - kx$
$m_2 \ddot{x}_2 = F_2 - kx$

Dividing by their respective masses, we get the accelerations:
$\ddot{x}_1 = \dfrac{F_1}{m_1} - \dfrac{kx}{m_1}$
$\ddot{x}_2 = \dfrac{F_2}{m_2} - \dfrac{kx}{m_2}$

The relative acceleration of the blocks (rate of change of elongation) is:
\ddot{x} = \ddot{x}_1 + \ddot{x}_2 = \left(\dfrac{F_1}{m_1} + \dfrac{F_2}{m_2} \right) -kx \left(\dfrac{1}{m_1} + \dfrac{1}{m_2} \right)

$\ddot{x} =$\dfrac{m_2 F_1 + m_1 F_2}{m_1 m_2} - \dfrac{k(m_1 + m_2)}{m_1 m_2}$x$

Let the reduced mass be $\mu =$\dfrac{m_1 m_2}{m_1 + m_2} and the effective force be $F_{\text{eff}} =$\dfrac{m_1 F_2 + m_2 F_1}{m_1 + m_2}.

Multiplying the relative acceleration equation by $\mu$, we get:
$\mu \ddot{x} = F_{\text{eff}} - kx$

To find the maximum elongation x_{\text\max}, we integrate the equation of motion. The maximum elongation occurs when the relative velocity is zero ($\dot{x} = 0$).

\int_{0}^{x_{\text\max}} (F_{\text{eff}} - kx) dx = 0

F_{\text{eff}} x_{\text\max} -\dfrac{1}{2}k x_{\text\max}^2 = 0

Since x_{\text\max} \neq 0, we have:
x_{\text\max} =\dfrac{2 F_{\text{eff}}}{k}

Substituting the value of $F_{\text{eff}}$:
x_{\text\max} =\dfrac{2(m_1 F_2 + m_2 F_1)}{k(m_1 + m_2)}

Answer: \dfrac{2(m_1 F_2 + m_2 F_1)}{k(m_1 + m_2)}