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A system of two particles is having masses m₁ and m₂. If the particle of mass m₁ is pushed towards the center of mass of particles through a distance d, by what distance the particle of mass m₂ should be moved so as to keep the centre of mass of particles at the original position?

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

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

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

answer is C.

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Detailed Solution

Position of center of mass is = $x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$

Center of Mass Position Shift =

$∆ x_{cm} = \frac{m_{1} ∆ x_{1} + m_{2} ∆ x_{2}}{m_{1} + m_{2}}$

$∆ x_{cm} = 0$

$∆ x_{1} = d$

$0 = \frac{m_{1} d + m_{2} ∆ x_{2}}{m_{1} + m_{2}}$

$∆ x_{2} = - \frac{m_{1}}{m_{2}} d$

Hence the correct answer is $\frac{m_{1}}{m_{2}} d$ .

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