Consider a two particle system with the particles having masses `m_1 and m_2`. If the first particles pushed towards the centre of mass through a distance d, by what distance should the second particle be moved so as the keep the centre of mass at the same position?

Options

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

Answer

C

Explanation

To keep the centre of mass at the same position, velocity of centre of mass is zero, so
`(m_1v_1 + m_2v_2)/(m_1 + m_2) = 0`
Where, `v_1 and v_2` are the velocities of particles 1 and 2, respectively.
`implies m_1(dr_1)/(dt) + m_2(dr_2)/(dt) = 0 " "[ :' v_1 = (dr_1)/(dt) and v_2 = (dr_2)/(dt)]`
`implies m_1 dr_1 + m_1dr_2 = 0`
`dr_1 and dr_2` represent the change in displacement of particles.
Let `2^(nd) particle has been displaced by distance x. `implies` m_1(d) + m_2(x) = 0 " or " x = - (m_1d)/(m_2)` .

`X_(CM) = (m_1x_1 + m_2x_2)/(m_1 + m_2) " ".....(i)`
After moving `m_1` through a distance towards CM and to keep the position of CM unchanged, let be the shift of `m_2`.
`X_(CM) = (m_1(x_1 - d) + m_2(x_2 + d'))/(m_1 + m_2) " ".....(iii)`
From Eqs. (i) and (ii), we get
`(m_1x_1 + m_2x_2)/(m_1 + m_2) = (m_1(x_1 -d)+m_2(x_2 + d))/(m_1 + m_2) implies -m_1d + m_2d' = 0 " ":. " "d' = (m_1)/(m_2)d` .