Center of mass of a system of three particles | vedclass

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(D) Let the first system have total mass $M_1 = 1 + 2 + 3 = 6 \, kg$ and center of mass $\vec{R}_1 = (1, 2, 3) \, m$.Let the second system have total

Vedclass · Accepted Answer

$(3 \, m, 1 \, m, 8 \, m)$

Vedclass · Answer

(D) Let the first system have total mass $M_1 = 1 + 2 + 3 = 6 \, kg$ and center of mass $\vec{R}_1 = (1, 2, 3) \, m$.Let the second system have total mass $M_2 = 2 + 3 = 5 \, kg$ and center of mass $\vec{R}_2 = (-1, 3, -2) \, m$.Let the third particle have mass $m_3 = 5 \, kg$ at position $\vec{r} = (x, y, z)$.The total mass of the system is $M = 6 + 5 + 5 = 16 \, kg$.The center of mass of the combined system is $\vec{R}_{cm} = \frac{M_1\vec{R}_1 + M_2\vec{R}_2 + m_3\vec{r}}{M}$.We want $\vec{R}_{cm} = \vec{R}_1 = (1, 2, 3)$.Substituting the values: $\frac{6(1, 2, 3) + 5(-1, 3, -2) + 5(x, y, z)}{16} = (1, 2, 3)$.$6(1, 2, 3) + 5(-1, 3, -2) + 5(x, y, z) = 16(1, 2, 3)$.$(6, 12, 18) + (-5, 15, -10) + 5(x, y, z) = (16, 32, 48)$.$(1, 27, 8) + 5(x, y, z) = (16, 32, 48)$.$5(x, y, z) = (16-1, 32-27, 48-8) = (15, 5, 40)$.$(x, y, z) = (3, 1, 8) \, m$.

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