Four particles of mass $m_{1} = 2 m, m_{2} = 4 m, m_{3} = m a n d m_{4}$ are placed at four corners of a square. What should be the value of $m_{4}$ so that the centre of mass of all the four particles are exactly at the centre of the square?

2 m

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8 m

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6 m

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Centre of mass of the four particles cannot be at centre of square.

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Solution

The correct option is D Centre of mass of the four particles cannot be at centre of square.
With reference to centre of square,
$X_{C M} = 0$
$\frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}} = 0$
$(2 m) (- a) + 4 m (a) + m (a) + m_{4} (- a) = 0$
$m_{4} = 3 m$

Similarly for $Y_{C M} = 0$
$(2 m) (- a) + 4 m (- a) + m (a) + m_{4} (a) = 0$
$m_{4} = 5 m$
Since, value of $m_{4}$ are different to satisfy both $x_{C M} = 0$ and $Y_{C M} = 0$ .
Hence, it is not possible.

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