Question

# Two particles having mass ratio n : 1 are interconnected by a light in extensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is

A
(n1)2g
B
(n+1n1)2g
C
(n1n+1)2g
D
(n+1n1)9

Solution

## The correct option is C $$\left( \frac { n - 1 } { n + 1 } \right) ^ { 2 } g$$Given$$,$$$$\frac{{{m_1}}}{{{m_2}}} = \frac{n}{1} = n$$Each mass have the acceleration $$a = \frac{{\left( {{m_1} - {m_2}} \right)}}{{{m_1} + {m_2}}}$$however $${{m_1}}$$ which is heavier will have the will have acceleration $${{a_1}}$$ vertically down while the lighter mass $${{m_2}}$$ will have acceleration $${{a_2}}$$ vertically up $$\to {a_2} = - {a_1}$$The acceleration or the centre of mass of the system$$,$$ $${a_{cm}} = \frac{{{m_1}{a_1} + {m_2}{a_2}}}{{{m_1} + {m_2}}}$$given that $${a_2} = - {a_1} \to {a_{cm}} = \frac{{\left( {{m_1} - {m_2}} \right){a_1}}}{{{m_1} + {m_2}}} = \frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}} \times \frac{{\left( {{m_1} - {m_2}} \right)g}}{{{m_1} + {m_2}}} = \frac{{{{\left( {{m_1} - {m_2}} \right)}^2}g}}{{{m_1} + {m_2}}}$$Since $$\frac{{{m_1}}}{{{m_2}}} = n$$ diving by $${{m_2}}$$ and simplifying $$\Rightarrow {a_{cm}} = {\left( {\frac{{n - 1}}{{n + 1}}} \right)^2}g$$Hence,option $$(C)$$ is correct answer.Physics

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