A body of mass m_1 = 4 kg moves at 5 hat{i} | vedclass

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(C) The velocity of the centre of mass $(v_{CM})$ is given by the formula: $v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$.Substituting the given value

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$\frac{400}{3} \ J$

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(C) The velocity of the centre of mass $(v_{CM})$ is given by the formula: $v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$.Substituting the given values: $v_{CM} = \frac{4 	imes 5 \hat{i} + 2 	imes 10 \hat{i}}{4 + 2}$.$v_{CM} = \frac{20 \hat{i} + 20 \hat{i}}{6} = \frac{40 \hat{i}}{6} = \frac{20}{3} \hat{i} \ m/s$.The kinetic energy of the centre of mass is given by $K = \frac{1}{2} M v_{CM}^2$,where $M = m_1 + m_2 = 6 \ kg$.$K = \frac{1}{2} 	imes 6 	imes (\frac{20}{3})^2$.$K = 3 	imes \frac{400}{9} = \frac{400}{3} \ J$.

$A$ body of mass $m_1 = 4 \ kg$ moves at $5 \hat{i} \ m/s$ and another body of mass $m_2 = 2 \ kg$ moves at $10 \hat{i} \ m/s$. The kinetic energy of the centre of mass is

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