The block of mass m₁ shown in figure is fastened to the spring and the block of mass m₂ is placed against it. The blocks are pushed a further distance $(2 / k) (m_{1} + m_{2}) g \sin θ$ from mean position against the spring and released. The common speed of blocks at the time of separation is

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$\sqrt{\frac{2}{k} (m_{1} + m_{2})}) \cdot g \sin θ$

$\sqrt{\frac{3}{k} (m_{1} + m_{2})} \cdot g \sin θ$

$\sqrt{(\frac{m_{1} + m_{2}}{k})} \cdot g \sin θ$

$\sqrt{\frac{5}{k} (m_{1} + m_{2})} \cdot g \sin θ$

answer is B.

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Detailed Solution

At the equilibrium condition

$\begin{array}{l} x = \frac{(m_{1} + m_{2}) g \sin θ}{k} \to (1) \\ Let x_{1} = \frac{2}{k} (m_{1} + m_{2}) g \sin θ \to (2) \end{array}$

Total compression in spring = x + x₁
Let x₂ be the displacement of m₁ at the time of loosing the contact with m₂ from mean position F.B.D. at the time of loosing contact.

$\begin{array}{l} ∴ m_{1} a = m_{1} g \sin θ - k (x - x_{2}) \\ \Rightarrow m_{1} ω^{2} x_{2} = m_{1} g \sin θ - k x + k x_{2} \\ \Rightarrow \frac{m_{1} \cdot k}{(m_{1} + m_{2})} x_{2} - k_{2} + k \cdot \frac{(m_{1} + m_{2}) g \sin θ}{k} = m_{1} g \sin θ \\ (∵ ω = \sqrt{\frac{k}{m_{1} + m_{2}}}) \Rightarrow \frac{- m_{2} k}{(m_{1} + m_{2})} x_{2} = - m_{2} g \sin θ \\ \Rightarrow x_{2} = \frac{(m_{1} + m_{2}) g \sin θ}{k} = x \to (3) \end{array}$

$∴$ At the tine of loosing contact, no compression in spring $\Rightarrow$ from law of conservation of energy, E_initial = E_final

$\begin{array}{l} \Rightarrow \frac{1}{2} k {(x_{1} + x)}^{2} - (m_{1} + m_{2}) g \sin θ \cdot (x + x_{1}) = \frac{1}{2} (m_{1} + m_{2}) v^{2} \\ v = \sqrt{\frac{3}{k} (m_{1} + m_{2})} \cdot g \sin θ \end{array}$