Centre of mass of three particles of masses 1 kg, 2 kg and 3 kg lies at the point (1, 2, 3) and centre of mass of another system of particles 3 kg and 2 kg lies at the point (-1, 3, -2). Where should we put a particle of mass 5 kg so that the centre of mass of entire system lies at the centre of mass of first system?

Options

A. (0 , 0 , 0)
B. (1 , 3 ,2)
C. (-1 , 2 , 3)
D. (3 , 1 , 8)

Answer

D

Explanation

According to the definition of centre of mass , we can imagine one particle of mass (1 + 2+ 3)kg at (1 , 2, 3) , another particle of mass (2 + 3) kg at (-1 , 3 , -2)
Let the third particle of mass 5 kg put at `(x_(3) , y_(3) , z_(3))` i.e,
`m_(1) = 6 kg , (x_(1) , y_(1) , z_(1)) = (1 , 2 , 3)`
`m_(2) = 5 kg, (x_(2) , y_(2) , z_(2)) = (-1 , 3 , -2)`
`m_(3) = 5 kg, (x_(3) , y_(3) , z_(3)) = ?`
Given , `(X_(CM) , Y_(CM) , Z_(CM) ) = (1 , 2 , 3)`
`X_(CM) = (m_(1) x_(1) + m_(2) x_(2) + m_(3) x_(3))/(m_(1) + m_(2) + m_(3)) or 1 = (6 xx 1 + 5 xx (-1) xx 5x _(3))/(6 + 5 + 5)`
`5x_(3) =16 - 1 = 15 or x_(3) = 3`
Similarly `, y_(3) = 1` and `z_(3) = 8`