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 center of mass at the same position?

Options

A. `(m_1)/(m_1 + m_2) d`
B. `(m_1)/(m_2) d`
C. d
D. `(m_2)/(m_1) d`

Answer

B

Explanation

(b) The system of two given particles of masses `m_1` and `m_2` are shown in the figure.
Initially the centre of mass
`r_(CM) = (m_1 r_1 + m_2 r_2)/(m_1 + m_2)`
When mass `m_1` moves towards centre of mass by a distance `d`, then let mass `m_2` move a distance `d'` away from `CM` to keep the `CM` in its initial position.
So `r_(CM) = (m_1 r_1 + m_2 r_2)/(m_1 + r_2)`...(ii)
Equating Eqs. (i) and (ii), we get
`(m_1 r_1 + m_2 r_2)/(m_1 + m_2) =(m_1(r_1 -d)+ m_2(r_2 + d))/(m_1+ m_2)`
`rArr -m_1 d + m_2 d' = 0`
`rArr d' = (m_1)/(m_2) d`.
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