Three point particles of mass 1 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The centre of mass of the system is at the point:

0.9 cm right and 2.0 cm above 1 kg mass

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2.0 cm right and 0.9 cm above 1 kg mass

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1.5 cm right and 1.2 cm above 1 kg mass

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0.6 cm right and 2.0 cm above 1 kg mass

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Solution

The correct option is A 0.9 cm right and 2.0 cm above 1 kg mass
Taking 1 kg as the origin
$\begin{matrix} x_{c o m} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} x_{c o m} = \frac{1 \times 0 + 1.5 \times 3 + 2.5 \times 0}{5} x_{c o m} = 0.9 y_{c o m} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}} y_{c o m} = \frac{1 \times 0 + 1.5 \times 0 + 2.5 \times 4}{5} y_{c o m} = 2 \end{matrix}$
Centre of mass is at $(0.9, 2)$

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Three point particles of mass 1 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The centre of mass of the system is at the point:

The correct option is A 0.9 cm right and 2.0 cm above 1 kg massTaking 1 kg as the origin xcom=m1x1+m2x2+m3x3m1+m2+m3xcom=1×0+1.5×3+2.5×05xcom=0.9ycom=m1y1+m2y2+m3y3m1+m2+m3ycom=1×0+1.5×0+2.5×45ycom=2 Centre of mass is at (0.9,2)