If the centre of mass of three particles of masses of 1kg, 2 kg, 3 kg is at (2, 2, 2), then where should a fourth particle of mass 4 kg be placed so that the combined centre of mass may be at (0, 0, 0).

Explanation

Let `(x_(1), y_(1), z_(1)), (x_(2), y_(2), z_(2))` and `(x_(3), y_(3), z_(3))` be the position of masses 1 kg, 2 kg, 3 kg and let the co - ordinates of centre of mass of the three particles system is `(x_(cm), y_(cm), z_(cm))` respectively.
`x_(cm)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))`
`rArr 2=(1xx x_(1)+2xx x_(2)+3xx x_(3))/(1+2+3)`
(or) `x_(1)+2x_(3)+3x_(3)=12 " "` . ........(1)
Suppose the fourth particle of mass 4 kg is placed at `(x_(4), y_(4), z_(4))` so that centre of mass of new system shifts to (0, 0, 0). For X - co - ordinate of new centre of mass we have
`0=(1xx x_(1)+2xx x_(2)+3xx x_(3)+4xx x_(4))/(1+2+3+4)`
`rArr x_(1)+2x_(2)+3x_(3)+4x_(4)=0 " "` .......(2)
From equations (1) and (2)
`12+4x_(4)=0 rArr x_(4)=-3`
Similarly, `y_(4)=-3` and `z_(4)=-3`
Therefore 4 kg should be placed at `(-3, -3, -3)`.