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 1st 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 the centre of mass we can imagine one particle of mass (1 + 2 + 3)kg located at (1, 2, 3) and another particle of mass (3 + 2)kg located at (-1, +3, -2). Assume that the 3rd particle mass 5 kg is placed at `(x_(3), y_(3), z_(3))`. Hence
`m_(1)=6kg, (x_(1), y_(1), z_(1))=(1, 2, 3), m_(2)=5kg, (x_(2), y_(2), z_(2))=(-1, 3, -2)`
`m_(3)=5kg, (x_(3), y_(3), z_(3))=?`
Given that `(X_(CM), Y_(CM), Z_(CM))=1, 2, 3," Now, "X_(CM)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))`
On substituting values: `x_(3)=3`
`Y_(CM)=(m_(1)y_(1)+m_(2)y_(2)+m_(3)y_(3))/(m_(1)+m_(2)+m_(3))`
On substituting values, `y_(3)=1`
`Z_(CM)=(m_(1)z_(1)+m_(2)z_(2)+m_(3)z_(3))/(m_(1)+m_(2)+m_(3))" or "z_(3)=8" "therefore (x_(3), y_(3), z_(3))=(3, 1, 8)`