Two balls each of mass 'm' are placed at the vertices A and B of an equilateral triangle of side 1 m. A third ball of mass 2 m is kept at vertex C. Taking vertex A as the origin, find the coordinates of the centre-of-mass of the given system.

Options

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

Answer

C

Explanation

Here centre of mass is given by
`bar(x)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3)), bar(x)=m times 0+m times 1+2m times 1//2)/(m+m+2m)`
`bar(x)=(2m)/(4m)=1/2, " "bar(y)=(m_(1)y_(1)+m_(2)y_(2)+m_(3)y_(3))/(m_(1)+m_(2)+m_(3))`
`bar(y)=(m times 0+m times 0+2m times sqrt(3)//2)/(m+m+2m)=sqrt(3)/4`
`therefore` Centre of mass is `[1/2m, sqrt(3)/4m]`