A body of mass m_1 = 4 text{ kg} moves at 5 h | 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 values:

Vedclass · Accepted Answer

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

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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}$$v_{CM} = \frac{20 \hat{i} + 20 \hat{i}}{6} = \frac{40 \hat{i}}{6} = \frac{20}{3} \hat{i} 	ext{ m/s}$The kinetic energy of the centre of mass is calculated as:$K_{CM} = \frac{1}{2} (m_1 + m_2) v_{CM}^2$$K_{CM} = \frac{1}{2} 	imes (4 + 2) 	imes \left( \frac{20}{3} ight)^2$$K_{CM} = \frac{1}{2} 	imes 6 	imes \frac{400}{9}$$K_{CM} = 3 	imes \frac{400}{9} = \frac{400}{3} 	ext{ J}$

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

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