Three particle each of 1kg mass are placed at corners of a right angled triangle AOB, 'O' being the origin of co-ordinate system (OA and OB along the +ve x-direction and +ve y direction ). If OA = OB = 1m, the position vector of center of mass (in meters) is

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$i - j$

$\frac{i + j}{3}$

$\frac{i - j}{3}$

$\frac{2 (i + j)}{3}$

answer is A.

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Detailed Solution

${\bar{x}}_{cm} = \frac{m_{1} {\bar{x}}_{1} + m_{2} {\bar{x}}_{2} + m_{3} {\bar{x}}_{3}}{m_{1} + m_{2} + m_{3}}$

Here m = 1kg

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

$= (\frac{1 (0) + 1 (0) + 1 (1)}{3}) \hat{i} + (\frac{1 (1) + 1 (0) + 1 (0)}{3}) \hat{j}$

$= \frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} = \frac{\hat{i} + \hat{j}}{3}$

$∴$ The position vector of centre of mass is $= \frac{\hat{i} + \hat{j}}{3}$

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