The block of mass m₁ shown in figure (12−E2) is fastened to the spring and the block of mass m₂ is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance (2/k) (m₁ + m₂)g sin θ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of blocks at the time of separation?

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Solution

(a) As it can be seen from the figure,
Restoring force = kx
Component of total weight of the two bodies acting vertically downwards = (m₁ + m₂) g sin θ

At equilibrium,
kx = (m₁ + m₂) g sin θ
$\Rightarrow x = \frac{(m_{1} + m_{2}) g \sin θ}{k}$

(b) It is given that:
Distance at which the spring is pushed, $x_{1} = \frac{2}{k} (m_{1} + m_{2}) g \sin θ$

As the system is released, it executes S.H.M.
where $ω = \sqrt{\frac{k}{m_{1} + m_{2}}}$

When the blocks lose contact, P becomes zero. (P is the force exerted by mass m₁on mass m₂)
$∴ m_{2} g \sin θ = m_{2} x_{2} ω^{2} = m_{2} x_{2} \times \frac{k}{m_{1} + m_{2}} \Rightarrow x_{2} = \frac{(m_{1} + m_{2}) g \sin θ}{k}$

Therefore, the blocks lose contact with each other when the spring attains its natural length.

(c) Let v be the common speed attained by both the blocks.

$Total compression = x_{1} + x_{2} \frac{1}{2} (m_{1} + m_{2}) v^{2} - 0 = \frac{1}{2} k {(x_{1} + x_{2})}^{2} - (m_{1} + m_{2}) g \sin θ (x + x_{1}) \Rightarrow \frac{1}{2} (m_{1} + m_{2}) v^{2} = \frac{1}{2} k (\frac{3}{k}) (m_{1} + m_{2}) g \sin θ - (m_{1} + m_{2}) g \sin θ (x_{1} + x_{2}) \Rightarrow \frac{1}{2} (m_{1} + m_{2}) v^{2} = \frac{1}{2} (m_{1} + m_{2}) g \sin θ \times (\frac{3}{k}) (m_{1} + m_{2}) g s in θ \Rightarrow v = \sqrt{\{\frac{3}{k} (m_{1} + m_{2})\}} g \sin θ$

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