Consider a two particle system with particles having masses $m_{1}$ and $m_{2}$ . If the first particle is pushed towards the centre of mass through a distance $d$ by what distance should the second particle is moved, so as to keep the centre of mass at the same position?

\frac{m_{1}}{m_{1} + m_{2}} d

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\frac{m_{2}}{m_{1}} d

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\frac{m_{1}}{m_{2}} d

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d

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Solution

The correct option is C $\frac{m_{1}}{m_{2}} d$
Given:
Displacement of first particle towards center of mass $= d$

Initially $0 = \frac{m_{1} (- x_{1}) + m_{2} x_{2}}{m_{1} + m_{2}} \Rightarrow m_{1} x_{1} = m_{2} x_{2} . . . . (1)$

Let second particles is displaced through distance $d^{'}$ towards center of mass

$∴ 0 = \frac{- m_{1} (x_{1} - d) + m_{2} (x_{2} - d^{'})}{m_{1} + m_{2}}$

$\Rightarrow 0 = - m_{1} x_{1} + m_{1} d + m_{2} x_{2} - m_{2} d$

$\Rightarrow d^{'} = \frac{m_{1}}{m_{2}} d$

Hence, option $(C)$ is correct.

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Q. Consider a two particle system with particles having masses

m_{1}

and

m_{2}

. If the first particle is pushed towards the centre of mass through a distance

d

, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?

Consider a two-particle system with the particles having masses $m_{1}$ and $m_{2}$ . If the first particle is pushed towards the centre of mass by a distance 'd', by what distance should the second particle be moved so as to keep the centre of mass at the same position?