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(B) Let the masses be $m_1 = 1 \, g, m_2 = 2 \, g, m_3 = 3 \, g$ and $m_4 = 4 \, g$.Since the center of mass of the first three particles is at the or

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$5/2$

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(B) Let the masses be $m_1 = 1 \, g, m_2 = 2 \, g, m_3 = 3 \, g$ and $m_4 = 4 \, g$.Since the center of mass of the first three particles is at the origin $(0, 0, 0)$,their combined position vector is $\vec{R}_{123} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3} = (0, 0, 0)$,which implies $m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3 = 0$.The position vector of the fourth particle is $\vec{r}_4 = \alpha(\hat{i} + 2\hat{j} + 3\hat{k})$.The center of mass of the new system is $\vec{R}_{cm} = (1, 2, 3)$.The formula for the center of mass is $\vec{R}_{cm} = \frac{(m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3) + m_4\vec{r}_4}{m_1 + m_2 + m_3 + m_4}$.Substituting the known values: $(1, 2, 3) = \frac{0 + 4\alpha(\hat{i} + 2\hat{j} + 3\hat{k})}{1 + 2 + 3 + 4} = \frac{4\alpha(\hat{i} + 2\hat{j} + 3\hat{k})}{10}$.Comparing the $x$-components: $1 = \frac{4\alpha}{10} \implies 10 = 4\alpha \implies \alpha = \frac{10}{4} = 2.5 = 5/2$.Similarly,for $y$-components: $2 = \frac{8\alpha}{10} \implies 20 = 8\alpha \implies \alpha = 2.5 = 5/2$.Thus,the value of $\alpha$ is $5/2$.

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