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Question

A system of two particles is having masses $${m}_{1}$$ and $${m}_{2}$$. If the particle of mass $${m}_{1}$$ is pushed towards the center of mass of particles through a distance, $$d$$, then determine by what distance the particle of mass $${m}_{2}$$ should be moved so as to keep the centre of mass of particles at the original position?


A
m1m2+m2d
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B
d
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C
m1m2d
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D
m2m1d
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Solution

The correct option is C $$\cfrac { { m }_{ 1 } }{ { m }_{ 2 } } d$$
Position of center of mass   $$x_{cm} = \dfrac{m_1x_1+m_2x_2}{m_1+m_2}$$

Shift in position of center of mass   $$\Delta x_{cm} = \dfrac{m_1\Delta x_1+m_2\Delta x_2}{m_1+m_2}$$

Given :  $$\Delta x_{cm} = 0$$       $$\Delta x_1 = d$$

$$\therefore$$   $$0= \dfrac{m_1d+m_2\Delta x_2}{m_1+m_2}$$

$$\implies \ \Delta x_2 = -\dfrac{m_1}{m_2}d$$

Thus $$m_2$$ must also be shifted towards the center of mass by a distance  $$\dfrac{m_1}{m_2}d$$.

Physics

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