The block of mass $m_{1}$ shown in figure (12-E2) is fastened to the spring and the block of mass $m_{2}$ is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance $(2 / k) (m_{1} + m_{2}) g s i n θ$ 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) At the equilibrium condition,

$k x = (m_{1} + m_{2}) g s i n θ$

$\Rightarrow x = \frac{(m_{1} + m_{2}) g s i n θ}{k}$

$(b) x_{1} = \frac{2}{k} (m_{1} + m_{2}) g s i n θ$ (Given)

when the system is released, it will start to make SHM.

where, $ω = \sqrt{\frac{k}{m_{1} + m_{2}}}$

when the blocks lose contact,

$p = 0$

So, $m_{2} g s i n θ = m_{2} x_{2} \times \frac{k}{m_{1} + m_{2}}$

$\Rightarrow x_{2} = \frac{(m_{1} + m_{2}) g s i n θ}{k}$

So the blocks will lose contact with each other when the springs attain its natural length.

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

$\frac{1}{2} (m_{1} + M_{2}) v^{2} - 0$

$= \frac{1}{2} k (x_{1} + x_{2})^{2} -) (m_{1} + m_{2}) g s i n θ (x + x_{1})$

[ $x + x_{1} =$ total compression ]

$\Rightarrow \frac{1}{2} (m_{1} + m_{2}) v^{2}$

$= \frac{1}{2} k (\frac{3}{k}) (m_{1} + m_{2}) g s i n θ - (m_{1} + m_{2}) g s i n θ - (x_{1} + x_{2})$

$\Rightarrow \frac{1}{2} (m_{1} + m_{2}) v^{2} = \frac{1}{2} (m_{1} + m_{2}) g s i n θ \times (\frac{3}{k}) (m_{1} + m_{2}) s i n θ$

$\Rightarrow v = \sqrt{{\frac{3}{k} (m_{1} + m_{2})}} g s i n θ$

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