The centre of mass of three particles of masses 10 kg, 20 kg and 30 kg is on origin (0, 0, 0). Now where should one place a mass of 40 kg, so that centre of mass of system is equal to (3, 3, 3)?

Options

A. `(0,0,0)`
B. `(7.5,7.5,7.5)`
C. `(1,2,3)`
D. `(4,4,4)`

Answer

B

Explanation

`m_(1)(x_(1),y_(1),z_(1))+m_(2)(x_(2),y_(2),z_(2))+`
`(x,y,z)=(m_(3)(x_(3),y_(3),z_(3))+m_(4)(x_(4),y_(4),z_(4)))/(m_(1)+m_(2)+m_(3)+m_(4))`
`therefore (x,y,z)=(0+m_(4)(x_(4),y_(y),z_(4)))/(10+20+30+40)`
`therefore (3,3,3)=(40(x_(4),y_(4),z_(4)))/(100)`
`therefore ((300)/(40),(300)/(40),(300)/(40))=(x_(4),y_(4),z_(4))`
`therefore (7.5,7.5,7.5)=(x_(4),y_(4),z_(4))`
`therefore (x_(4),y_(4),z_(4))=(7.5,7.5,7.5)` unit