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CGPET 2022 Solved Paper

Section: Physics

Question : 7 of 150

Marks: +1, -0

2\,\text{kg}, 4\,\text{kg}

4\,\text{kg}

(1,0,0),(1,1,0)

(0,1,0)

Solution:

7. (a) Coordinates by centre of mass are given by

\overline{x}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3}

=\frac{2\times 1+4\times 0+4\times 1}{2+4+4}=\frac{6}{10}=\frac{3}{5}

and

\overline{y}=\frac{m_1 y_1+m_2 y_2+m_3 y_3}{m_1+m_2+m_3}

=\frac{2\times 0+4\times 1+4\times 1}{2+4+4}=\frac{8}{10}=\frac{4}{5}

Also,

\overline{z}=\frac{m_1 z_1+m_2 z_2+m_3 z_3}{m_1+m_2+m_3}

=\frac{2\times 0+4\times 0+4\times 0}{2+4+4}=0

Position vector of Centre of mass is

r=\hat{xi} + \hat{yj} + \hat{zk} = \hat{35i} + \frac{4}{\hat{5j}}

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