Two blocks having masses m_1 and m_2 are connected by a spring of spring constant k and placed on a smooth horizontal surface. Each block is pulled by a constant force F in opposite directions. Find the maximum

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Correct answer: A. (2F)/(k), \; (2Fm_2)/(k(m_1 + m_2)), \; (2Fm_1)/(k(m_1 + m_2))

Let the acceleration of the centre of mass be

a = \dfrac{F - (-F)}{m_1 + m_2} = \dfrac{2F}{m_1 + m_2}

.

In the centre of mass frame, each block experiences an effective force stretching the spring. The relative acceleration between the two blocks is:

a_{rel} = \dfrac{F}{m_1} + \dfrac{F}{m_2} = F\left(\dfrac{1}{m_1} + \dfrac{1}{m_2}\right) = \dfrac{F(m_1 + m_2)}{m_1 m_2}

Using work-energy principle for relative motion:

\dfrac{1}{2} \mu v_{rel}^2 = \dfrac{1}{2} k x_\max^2

, where

\mu = \dfrac{m_1 m_2}{m_1 + m_2}

Also, effective force in relative motion is

2F

, so maximum extension occurs when:

k x_\max = 2F \Rightarrow x_\max = \dfrac{2F}{k}

Distances moved are in inverse ratio of masses:

x_1 = \dfrac{2F m_2}{k(m_1 + m_2)}, \quad x_2 = \dfrac{2F m_1}{k(m_1 + m_2)}

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