Three particles, each of mass 'm', are placed at the corners of a right angled triangle as shown in Fig. If OA = a and OB = b, the position vector of the centre of mass is (here i and j are unit vectors along x and y axes respectively).

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Solution

The correct option is A

The (x, y) co-ordinates of the masses at 0, A and B respectively are (refer to fig. 5.4)
$(x_{1} = 0, y_{1} = 0), (x_{2} = a, y_{2} = 0) a n d (x_{3} = 0, y_{3} = b)$
The (x,y) co - ordinates of the centre of the centre of mass are $x_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{m \times 0 + m \times a + m \times 0}{m + m + m} = \frac{a}{3}$
$y_{C M} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}} = \frac{m \times 0 + m \times 0 + m \times b}{m + m + m} = \frac{b}{3}$
The position vector of the centre of mass is $x_{C M} i + y_{C M} j$
$= \frac{a}{3} i + \frac{b}{3} j = \frac{1}{3} (a i + b j)$ , which is choice (a).

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