A block of mass m_1 is fastened to a spring on an incline of angle...

Let the compression of the spring be $x$. The equilibrium position of the system of two blocks is given by balancing the forces along the incline:

$k x_{eq} = (m_1 + m_2) g \sin\theta \Rightarrow x_{eq} = \dfrac{(m_1 + m_2) g \sin\theta}{k}$

The blocks are pushed a further distance $A$ from equilibrium, which becomes the amplitude of their simple harmonic motion:

$A = \dfrac{2(m_1 + m_2) g \sin\theta}{k}$

The blocks move together up the incline. Let $N$ be the normal force between the blocks. The equation of motion for block $m_2$ is:

$N - m_2 g \sin\theta = m_2 a$

The blocks separate when the normal force becomes zero ($N = 0$), which gives the acceleration at the moment of separation:

$a = -g \sin\theta$

For the combined system, the equation of motion is:

$k x - (m_1 + m_2) g \sin\theta = (m_1 + m_2) a$

Substituting $a = -g \sin\theta$, we get:

$k x - (m_1 + m_2) g \sin\theta = -(m_1 + m_2) g \sin\theta \Rightarrow k x = 0 \Rightarrow x = 0$

Thus, the blocks separate when the spring reaches its natural length. The natural length is at a distance $x_{eq}$ from the equilibrium position. The speed of the blocks at this position can be found using the velocity formula for simple harmonic motion, $v = \omega \sqrt{A^2 - y^2}$, where $y = x_{eq}$ is the displacement from equilibrium:

$v = \sqrt{\dfrac{k}{m_1 + m_2}} \sqrt{\left(\dfrac{2(m_1 + m_2) g \sin\theta}{k}\right)^2 - \left(\dfrac{(m_1 + m_2) g \sin\theta}{k}\right)^2}$

$v = \sqrt{\dfrac{k}{m_1 + m_2}} \sqrt{\dfrac{3(m_1 + m_2)^2 g^2 \sin^2\theta}{k^2}}$

$v = \sqrt{\dfrac{3(m_1 + m_2)}{k}} g \sin\theta$

Answer: $\sqrt{\dfrac{3(m_1 + m_2)}{k}} g \sin\theta$