Particles of masses 1kg and 3kg are at $(2 i + 5 j + 13 k) m$ and $(- 6 i + 4 j - 2 k) m$ then instantaneous position of their center of mass is

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$\frac{1}{4} (- 8 i + 17 j + 7 k) m$

$\frac{1}{4} (- 16 i + 17 j + 7 k) m$

$\frac{1}{4} (- 6 i + 17 j + 7 k) m$

$\frac{1}{4} (- 16 i + 17 j + 5 k) m$

answer is A.

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Detailed Solution

$m_{1} = 1 kg, m_{2} = 3 kg$

$r_{1} = 2 i + 5 j + 13 k$

$r_{2} = - 6 i + 4 j - 2 k$

$\bar{r_{c m}} = \frac{m_{1} \bar{r_{1}} + m_{2} \bar{r_{2}}}{m_{1} + m_{2}}$

$= \frac{(2 i + 5 j + 13 k) + 3 (- 6 i + 4 j - 2 k)}{4}$

$= \frac{2 i + 5 j + 13 k - 18 i + 12 j - 6 k}{4} = \frac{- 16 i + 17 j + 7 k}{4} = \frac{1}{4} (- 16 i + 17 j + 7 k)$

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