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Center of Mass as an Average Point

The position ...

Question

The position vectors of two identical particles with respect to the origin in three-dimensional coordinate system are $r_{1}$ and $r_{2}$ . The position vector of centre of mass of the system is given by:

A

r_{1} + r_{2}

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B

2 (r_{1} + r_{2})

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C

r_{1} - r_{2}

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D

\frac{r_{1} + r_{2}}{2}

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E

\frac{r_{1} + r_{2}}{3}

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Solution

The correct option is D $\frac{r_{1} + r_{2}}{2}$
Given, position vector of first particle $= r_{1}$
Position vector of second particle $= r_{2}$
$m_{1} = m_{2} = m$
We know that,
Position of centre of mass of system
$R = \frac{m_{1} r_{1} + m_{2} r_{2}}{m_{1} + m_{2}}$
$R = \frac{m r_{1} + m r_{2}}{m + m}$
$R = \frac{m (r_{1} + r_{2})}{2 m} = \frac{r_{1} + r_{2}}{2}$ .

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