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

If the centre of mass of three particles of masses of 1kg, 2kg, 3kg is

1.	If the centre of mass of three particles of masses of 1kg, 2kg, 3kg is at (2,2,2), then where should a fourth particle of mass 4kg be placed so that the combined centre of mass may be at (0,0,0).
Answer» Solution :Let `(x_(1),y_(1),z_(1)),(x_(2),y_(2),z_(2))` and `(x_(3),y_(3),z_(3))` be the positions of masses `1kg`, `2kg`, `3kg` and let the co-ordinates of centre of mass of the three particle 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))` `implies2=(1xx x_(1)+2xx x_(2)+3XX x_(3))/(1+2+3)` (or) `x_(1)+2x_(2)+3x_(3)=12`.............`(1)` Suppose the FOURTH particle of mass `4kg` is placed at `(x_(1),y_(1),z_(1))` 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)` `impliesx_(1)+2x_(2)+3x_(3)+4x_(4)=0`...........`(2)` From equations `(1)` and `(2)` `12+4x_(4)=0impliesx_(4)=-3` Similarly `y_(4)=-3` and `x_(4)=-3` Therefore 4kg should be placed at `(-3,-3,-3)`.