Question

Two particles of equal mass m are at the vertices of the base of an equilateral triangle. The triangle’s center of mass is midway between the base and the third vertex. What’s the mass at the third vertex?

Answer 1

Answer:Twice of given mass

Explanation:

Given

Two Particles of Equal mass placed at the base of an equilateral Triangle

let mass of two equal masses be m and third mass be m'

Taking one of the masses at origin

Therefore co-ordinates of first mass be (0,0)

Co-ordinates of other equal mass is (a,0)

if a is the length of triangle

co-ordinates of final mass $(\frac{a}{2},\frac{\sqrt{3}a}{2})$

Given its center of mass is at midway between base and third vertex therefore

$x_{cm},y_{cm}=\frac{a}{2},\frac{\sqrt{3}a}{4}$

$y_{cm}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$

$\frac{\sqrt{3}a}{4}=\frac{m\cdot 0+m\cdot 0+m'\cdot \frac{\sqrt{3}a}{2}}{m+m+m'}$

$2m+m'=4\times (\frac{m'}{2})$

$2m+m'=2m'$

$m'=2m$