If the centre of mass of three particles of masses of 1 kg, 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

Total mass of system of given three particles is `m_(1)` = 1+ 2 +3 = 6 kg
All this mass can be assumed to be existing as a point mass at `(x_(1),y_(1),z_(1))=(2,2,2)`.
Let second mass `m_2 = 4` kg be located at a position with co-ordinates `(x_(2),y_(2),z_(2))`
CM of the system will be at (0,0,0)
`X_(cm) =(m_(1)x_(1) + m_(2)x_(2))/(m_(1) + m_(2)), 0= (6(2) + 4(x_(2)))/(6+4) rArr x_(2)=-3`
Similarly `y_(2)=-3, z_(2)=-3`
`therefore` 4 kg mass should be placed at `(-3,-3,-3)`