The position vectors of two particles of mass | vedclass

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(A) The formula for the position vector of the center of mass is $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$.Given $m_1 = 1 \ kg$,$

Vedclass · Accepted Answer

$-2\hat{i} - \hat{j} + \hat{k}$

Vedclass · Answer

(A) The formula for the position vector of the center of mass is $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$.Given $m_1 = 1 \ kg$,$\vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k}$ and $m_2 = 3 \ kg$,$\vec{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k}$.Substituting these values:$\vec{r}_{cm} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3}$$\vec{r}_{cm} = \frac{\hat{i} + 2\hat{j} + \hat{k} - 9\hat{i} - 6\hat{j} + 3\hat{k}}{4}$$\vec{r}_{cm} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4}$$\vec{r}_{cm} = -2\hat{i} - \hat{j} + \hat{k}$.

The position vectors of two particles of masses $1 \ kg$ and $3 \ kg$ are $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} - 2\hat{j} + \hat{k}$ respectively. The position vector of the center of mass of this system is:

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