Three particles A, B and C of masses m, 2m an | vedclass

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(B) Given masses are $m_1 = m$, $m_2 = 2m$, and $m_3 = 3m$.The velocities are directed as follows:Particle $A$ moves north: $\vec{v}_1 = 6 \hat{j} \,m

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$5$

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(B) Given masses are $m_1 = m$, $m_2 = 2m$, and $m_3 = 3m$.The velocities are directed as follows:Particle $A$ moves north: $\vec{v}_1 = 6 \hat{j} \,ms^{-1}$Particle $B$ moves south: $\vec{v}_2 = -12 \hat{j} \,ms^{-1}$Particle $C$ moves east: $\vec{v}_3 = 8 \hat{i} \,ms^{-1}$The velocity of the centre of mass $\vec{v}_{cm}$ is given by:$\vec{v}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3}{m_1 + m_2 + m_3}$Substituting the values:$\vec{v}_{cm} = \frac{m(6 \hat{j}) + 2m(-12 \hat{j}) + 3m(8 \hat{i})}{m + 2m + 3m}$$\vec{v}_{cm} = \frac{6m \hat{j} - 24m \hat{j} + 24m \hat{i}}{6m}$$\vec{v}_{cm} = \frac{24m \hat{i} - 18m \hat{j}}{6m} = 4 \hat{i} - 3 \hat{j} \,ms^{-1}$The magnitude of the velocity of the centre of mass is:$v_{cm} = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,ms^{-1}$

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