Two particles bearing mass ratio n : 1 are connected by a light inextensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is :

{(n - 1)}^{2}

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{(\frac{n - 1}{n + 1})}^{2}

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{(\frac{n + 1}{n - 1})}^{2}

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(\frac{n + 1}{n - 1})

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Solution

The correct option is B ${(\frac{n - 1}{n + 1})}^{2}$ g
Given,

$\frac{m_{1}}{m_{2}} = \frac{n}{1} = 1$

Each mass will have the acceleration, $a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}$

However $m_{1}$ which is heavier will have acceleration $a_{1}$ vertically down while the lighter mass $m_{2}$ will have acceleration $a_{2}$ vertically up

$⟹ a_{2} = - a_{1}$

The acceleration of the center of mass of the system,

$a_{c m} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}}$

Given that,

$a_{2} = - a_{1} ⟹ a_{c m} = \frac{(m_{1} - m_{2}) a_{1}}{m_{1} + m_{2}}$

$\frac{m_{1} - m_{2}}{m_{1} + m_{2}} \times \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}} = \frac{(m_{1} - m_{2})^{2} g}{(m_{1} + m_{2})^{2}}$

Since $\frac{m_{1}}{m_{2}} = n$ , dividing by $m_{2}$ and simplifying

$⟹ a_{c m} = (\frac{n - 1}{n + 1})^{2} g$

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