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

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$\frac{2}{3} (ai + bj)$

$\frac{2}{3} (ai - bj)$

$\frac{1}{3} (ai + bj)$

$\frac{1}{3} (ai - bj)$

answer is A.

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Detailed Solution

$x_{cm} = (\frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}) i + (\frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}}) j$

$= (\frac{m (a)}{3 m}) i + (\frac{m (b)}{3 m}) j$

$= \frac{1}{3} (ai + bj)$

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