Two particles of masses 1 kg and 3 kg have the position vectors \left(\right. 2 \textrm{ } \hat{i} + 5 \textrm{ } \hat{j} + 13 \textrm{ } \hat{k} \left.\right) , \left(\right. - 6 \textrm{ } \hat{i} + 4 \textrm{ } \hat{j} - 2 \textrm{ } \hat{k} \left.\right) and velocities \left(\right. 10 \hat{i} - \hat{j} + 3 \hat{k} \left.\right) , \left(\right. 7 \hat{i} - 9 \hat{j} + 6 \hat{k} \left.\right) respectively. Find the position vector of the center of mass after 1 sec.

Engineering

Physics

Center of Mass

Question

Two particles of masses 1 kg and 3 kg have the position vectors $(2 \hat{i} + 5 \hat{j} + 13 \hat{k}), (- 6 \hat{i} + 4 \hat{j} - 2 \hat{k})$ and velocities $(10 \hat{i} - \hat{j} + 3 \hat{k}), (7 \hat{i} - 9 \hat{j} + 6 \hat{k})$ respectively. Find the position vector of the center of mass after 1 sec.

$\frac{15 \hat{i}}{4} - \frac{11 \hat{j}}{4} + 7 \hat{k}$

$\frac{31 \hat{i}}{4} - 7 \hat{j} + \frac{21}{4} \hat{k}$

$- 4 \hat{i} + \frac{17 \hat{j}}{4} + \frac{7}{4} \hat{k}$

$\frac{21}{4} \hat{i} - \frac{25}{4} \hat{j} + 3 \hat{k}$

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${\vec{r}}_{c} = \frac{1 \times (2 \hat{i} + 5 \hat{j} + 13 \hat{k}) + 3 \times (- 6 \hat{i} + 4 \hat{j} - 2 \hat{k})}{1 + 3} = \frac{- 16 \hat{i} + 17 \hat{j} + 7 \hat{k}}{4}$

${\vec{V}}_{c} = \frac{1 \times (10 \hat{i} - \hat{j} + 3 \hat{k}) + 3 \times (7 \hat{i} - 9 \hat{j} + 6 \hat{k})}{4} = \frac{31 \hat{i} - 28 \hat{j} + 21 \hat{k}}{4}$

${\vec{r}}_{c}^{'} = {\vec{r}}_{c} + {\vec{V}}_{c} t = \frac{- 16 \hat{i} + 31 \hat{j} + 17 \hat{i} - 28 \hat{j} + 7 \hat{k} + 21 \hat{k}}{4}$ .

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