Two particles whose masses are 10,kg and 30,k | vedclass

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Question

(A) The formula for the position vector of the center of mass is given by: $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$.Given: $m_1

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$-\frac{(\hat{i} + \hat{j} + \hat{k})}{2}$

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(A) The formula for the position vector of the center of mass is given by: $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$.Given: $m_1 = 10\,kg$,$\vec{r}_1 = \hat{i} + \hat{j} + \hat{k}$.Given: $m_2 = 30\,kg$,$\vec{r}_2 = -\hat{i} - \hat{j} - \hat{k}$.Substituting the values:$\vec{r}_{cm} = \frac{10(\hat{i} + \hat{j} + \hat{k}) + 30(-\hat{i} - \hat{j} - \hat{k})}{10 + 30}$.$\vec{r}_{cm} = \frac{10(\hat{i} + \hat{j} + \hat{k}) - 30(\hat{i} + \hat{j} + \hat{k})}{40}$.$\vec{r}_{cm} = \frac{-20(\hat{i} + \hat{j} + \hat{k})}{40}$.$\vec{r}_{cm} = -\frac{(\hat{i} + \hat{j} + \hat{k})}{2}$.Therefore,the correct option is $A$.

Two particles whose masses are $10\,kg$ and $30\,kg$ and their position vectors are $\hat{i} + \hat{j} + \hat{k}$ and $-\hat{i} - \hat{j} - \hat{k}$ respectively would have the centre of mass at:

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