Centre of mass is at point X (1,1,1) when system consists of particles of masses 2, 3, 4 and 5 kg. If the centre of mass shifts to point Y (2, 2, 2) on removal of the mass of 5 kg. What was its position ?

Explanation

Let `vec(r )_(1), vec(r )_(2), vec(r )_(3)`, are the position vectors of the particles of mass `m_(1)=2kg, m_(2)=3 kg, m_(3)=4 kg` and `m_(4)=5 kg` respectively.
Initially, the centre of mass was at the point `xx(1,1,1)`, then
`hat(i)+hat(j)+hat(k)=(m_(1)vec(r )_(1)+m_(2)vec(r )_(2)+m_(3)vec(r )_(3)+m_(4)vec(r )_(4))/(m_(1)+m_(2)+m_(3)+m_(4))`
`rArr (2vec(r )_(1)+3vec(r )_(2)+4vec(r )_(3)+5vec(r )_(4))/(2+3+4+5)=hat(i)+hat(j)+hat(k)`
`rArr 2vec(r )_(1)+3vec(r )_(2)+4vec(r )_(3)+5vec(r )_(4)=14hat(i)+14hat(j)+14hat(k)` ....(1)
On removing the mass `m_(4)=5 kg`, the centre of mass shifts to the point Y (2, 2, 2), then
`2hat(i)+2hat(j)+2hat(k)=(m_(1)vec(r )_(1)+m_(2)vec(r )_(2)+m_(3)vec(r )_(3))/(m_(1)+m_(2)+m_(3))`
`rArr 2vec(r )_(1)+3vec(r )_(2)+4vec(r )_(3)=18hat(i)+18hat(j)+18hat(k)` ....(2)
By subtracting the equation (2) from (1).
`vec(r )_(4)=-(4)/(5)(hat(i)+hat(j)+hat(k))`