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

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1/4(-16i + 17j + 7k) m

1/4(-8i + 17j + 7k) m

1/4(-6i + 17j + 7k) m

1/4(-6i + 17j + 5k) m

answer is A.

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Detailed Solution

m₁ = 1kg, m₂ = 3kg

${\bar{r}}_{1} = 2 \hat{i} + 5 \hat{j} + 13 \hat{k}, {\bar{r}}_{2} = - 6 \hat{i} + 4 \hat{j} - 2 \hat{k}$

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

$= \frac{1 (2 \hat{i} + 5 \hat{j} + 13 \hat{k}) + 3 (- 6 \hat{i} + 4 \hat{j} - 2 \hat{k})}{4}$

${\bar{r}}_{cm} = \frac{1}{4} (- 16 \hat{i} + 17 \hat{j} + 7 \hat{k}) m$

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