Three masses of 2,kg,4,kg,and 4,kg are placed | vedclass

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Question

(A) Given masses are $m_1 = 2\,kg$,$m_2 = 4\,kg$,and $m_3 = 4\,kg$.Their position vectors are $\vec{r}_1 = (1, 0, 0) = \hat{i}$,$\vec{r}_2 = (1, 1, 0)

Vedclass · Accepted Answer

$\frac{3}{5}\,\hat{i} + \frac{4}{5}\,\hat{j}$

Vedclass · Answer

(A) Given masses are $m_1 = 2\,kg$,$m_2 = 4\,kg$,and $m_3 = 4\,kg$.Their position vectors are $\vec{r}_1 = (1, 0, 0) = \hat{i}$,$\vec{r}_2 = (1, 1, 0) = \hat{i} + \hat{j}$,and $\vec{r}_3 = (0, 1, 0) = \hat{j}$.The center of mass position vector $\vec{r}_{cm}$ is given by $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$.Substituting the values: $\vec{r}_{cm} = \frac{2(\hat{i}) + 4(\hat{i} + \hat{j}) + 4(\hat{j})}{2 + 4 + 4}$.$\vec{r}_{cm} = \frac{2\hat{i} + 4\hat{i} + 4\hat{j} + 4\hat{j}}{10} = \frac{6\hat{i} + 8\hat{j}}{10}$.$\vec{r}_{cm} = \frac{6}{10}\hat{i} + \frac{8}{10}\hat{j} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}$.

Column-$I$	Column-$II$
$(1)$ $\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$	$(a)$ Reduced mass of a two-particle system
$(2)$ $\frac{{{r_1} + {r_2}}}{2}$	$(b)$ Position vector of the center of mass for a system of two equal masses

Three masses of $2\,kg$,$4\,kg$,and $4\,kg$ are placed at the three points $(1, 0, 0)$,$(1, 1, 0)$,and $(0, 1, 0)$ respectively. The position vector of its center of mass is:

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Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$
$(1)$ $\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$ $(a)$ Reduced mass of a two-particle system
$(2)$ $\frac{{{r_1} + {r_2}}}{2}$ $(b)$ Position vector of the center of mass for a system of two equal masses

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