Consider a system of two particles having masses m₁ and m₂. If the particle of mass m₁is pushed towards the mass centre of particles through a distance d, by what distance would the particles of mass m₂ move so as to keep the centre of mass of particles at the original position ?

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

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

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

answer is B.

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Detailed Solution

The system of two given particles of masses m₁ and m₂ are shown in figure. Initially the centre of mass

$r_{C M} = \frac{m_{1} r_{1} + m_{2} r_{2}}{m_{1} + m_{2}}$ -------- (i)

When mass m₁ moves towards centre of mass by adistanced, then let mass m₂ moves adistance d’ away from CM to keep the CM in its initial position.
So, $r_{C M} = \frac{m_{1} (r_{1} - d) + m_{2} (r_{2} + d^{'})}{m_{1} + m_{2}} ... (i i)$
Equating Eqs, (i) and (ii), we get
$\frac{m_{1} r_{1 + m_{2} r_{2}}}{m_{1} + m_{2}} = \frac{m_{1} (r_{1} - d) + m_{2 (r_{2} + d)}}{m_{1} + m_{2}}$