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Question
The block of mass m1 shown in figure 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
\[\Rightarrow x = \frac{\left( m_1 + m_2 \right) g \sin \theta}{k}\]
(b) It is given that:
Distance at which the spring is pushed,
\[x_1 = \frac{2}{k}\left( m_1 + m_2 \right)g \sin \theta\]
As the system is released, it executes S.H.M.
where \[\omega = \sqrt{\frac{k}{m_1 + m_2}}\]
When the blocks lose contact, P becomes zero. (P is the force exerted by mass m1 on mass m2)
\[\therefore m_2 g \sin \theta = m_2 x_2 \omega^2 = m_2 x_2 \times \frac{k}{m_1 + m_2}\] \[ \Rightarrow x_2 = \frac{\left( m_1 + m_2 \right) g \sin \theta}{k}\]
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.
\[\text { Total compression } = x_1 + x_2 \]
\[\frac{1}{2} \left( m_1 + m_2 \right) v^2 - 0 = \frac{1}{2}k \left( x_1 + x_2 \right)^2 - \left( m_1 + m_2 \right) g \sin \theta \left( x + x_1 \right) \]
\[ \Rightarrow \frac{1}{2}\left( m_1 + m_2 \right) v^2 = \frac{1}{2}k\left( \frac{3}{k} \right) \left( m_1 + m_2 \right) g \sin \theta - \left( m_1 + m_2 \right) g \sin \theta \left( x_1 + x_2 \right)\]
\[ \Rightarrow \frac{1}{2}\left( m_1 + m_2 \right) v^2 = \frac{1}{2} \left( m_1 + m_2 \right) g \sin \theta \times \left( \frac{3}{k} \right) \left( m_1 + m_2 \right) g sin \theta\]
\[ \Rightarrow v = \sqrt{\left\{ \frac{3}{k} \left( m_1 + m_2 \right) \right\}}g \sin \theta\]
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