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» 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,3 kg` and let the co-ordinates of centres 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=(1xxx_(1)+2xxx_(2)+3xxx_(3))/(1+2+3)`
Or `x_(1)+2x_(2)+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` coordinates of new centre of mass we have.
`0=(1xxx_(1)+2xxx_(2)+3xxx_(3)+4xxx_(4))/(1+2+3+4)`
`rArrx_(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 `z_(4)=-3`
Therefore `4 kg` should be placed at `(-3,-3,-3)`.


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