The velocity of the centre of mass of the sys | vedclass

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(B) Given:Mass $m_1 = 1 	ext{ kg}$,velocity $\vec{v}_1 = 2\hat{i} 	ext{ m/s}$.Mass $m_2 = 2 	ext{ kg}$,velocity $\vec{v}_2 = 2 \cos(30^{\circ})\hat

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$\left( \frac{2 + 2\sqrt{3}}{3} \right)\hat{i} - \frac{2}{3}\hat{j}$

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(B) Given:Mass $m_1 = 1 	ext{ kg}$,velocity $\vec{v}_1 = 2\hat{i} 	ext{ m/s}$.Mass $m_2 = 2 	ext{ kg}$,velocity $\vec{v}_2 = 2 \cos(30^{\circ})\hat{i} - 2 \sin(30^{\circ})\hat{j} 	ext{ m/s}$.The velocity of the centre of mass $\vec{v}_{CM}$ is given by:$\vec{v}_{CM} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$Substituting the values:$\vec{v}_{CM} = \frac{1(2\hat{i}) + 2(2 \cos 30^{\circ}\hat{i} - 2 \sin 30^{\circ}\hat{j})}{1 + 2}$Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\sin 30^{\circ} = \frac{1}{2}$:$\vec{v}_{CM} = \frac{2\hat{i} + 2(2 \cdot \frac{\sqrt{3}}{2}\hat{i} - 2 \cdot \frac{1}{2}\hat{j})}{3}$$\vec{v}_{CM} = \frac{2\hat{i} + 2(\sqrt{3}\hat{i} - 1\hat{j})}{3}$$\vec{v}_{CM} = \frac{2\hat{i} + 2\sqrt{3}\hat{i} - 2\hat{j}}{3}$$\vec{v}_{CM} = \left( \frac{2 + 2\sqrt{3}}{3} ight)\hat{i} - \frac{2}{3}\hat{j} 	ext{ m/s}$.

The velocity of the centre of mass of the system as shown in the figure is

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