The balls A, B and C of masses 50 g, 100 g and 150 g, respectively are placed at the vertices of an equilateral triangle. The length of each side is 1 m. If A is placed at (0, 0) and B is placed at (1, 0) m, find the coordinates (x, y) for the centre of mass of this system of the balls

Options

A. `(7/12, sqrt(3/4)) m`
B. `(5/18, sqrt(1/4))m`
C. `(7/12, sqrt(3/2))m`
D. `(5/18, sqrt(3/4)) m`

Answer

C

Explanation

Given,
mass of ball `A, m_(1)=50` g
mass of ball `B, m_(2)=100` g
mass of ball `C, m_(3)=150` g
and length of each side of triangle `=1 m`
According to the question,

`:' y_(3)=sqrt(AC^(2)-AM^(2))`
`=sqrt(1-(0.5)^(2))=sqrt(0.75)=sqrt(3)/2" "[ :' AM =x_(3)=0.5]`
Coordinates of Centre of mass is (x, y), then
`x=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))`
`=(50xx0+100xx1+150xx0.5)/(50+100+150)=175/300=7/12`
and `y=(m_(1)y_(1)+m_(2)y_(2)+m_(3)y_(3))/(m_(1)+m_(2)+m_(3))`
`=(50xx0+100xx0+150xx sqrt(3)/2)/(50+100+150)=(75 sqrt(3))/300=sqrt(3)/4`
The coordinates (x, y) for the centre of the mass of this system of balls is `(7/12 m, sqrt(3)/4 m)`. So, no option given is correct.