The balls A, B and C of masses 50 g, 100 g | vedclass

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Question

(A) Given masses: $m_1 = 50 \ g$,$m_2 = 100 \ g$,$m_3 = 150 \ g$. Coordinates of the vertices: $A = (0, 0)$ $B = (1, 0)$ Since it is an equilateral tr

Vedclass · Accepted Answer

$\left(\frac{7}{12}, \frac{\sqrt{3}}{4}\right) \ m$

Vedclass · Answer

(A) Given masses: $m_1 = 50 \ g$,$m_2 = 100 \ g$,$m_3 = 150 \ g$. Coordinates of the vertices: $A = (0, 0)$ $B = (1, 0)$ Since it is an equilateral triangle with side length $1 \ m$,the $x$-coordinate of $C$ is the midpoint of $AB$,i.e.,$x_3 = 0.5 \ m$. The $y$-coordinate of $C$ is $h = \sqrt{1^2 - 0.5^2} = \sqrt{0.75} = \frac{\sqrt{3}}{2} \ m$. So,$C = (0.5, \frac{\sqrt{3}}{2})$. The $x$-coordinate of the centre of mass is: $x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{50(0) + 100(1) + 150(0.5)}{50 + 100 + 150} = \frac{100 + 75}{300} = \frac{175}{300} = \frac{7}{12} \ m$. The $y$-coordinate of the centre of mass is: $y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \frac{50(0) + 100(0) + 150(\frac{\sqrt{3}}{2})}{50 + 100 + 150} = \frac{75\sqrt{3}}{300} = \frac{\sqrt{3}}{4} \ m$. Thus,the centre of mass is $\left(\frac{7}{12}, \frac{\sqrt{3}}{4}\right) \ m$.

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