Two bodies of mass 4 ~kg and 5 ~kg are moving | vedclass

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(A) The velocity of the centre of mass is given by the formula:$v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$Given: $m_1 = 4 ~kg$,$m_2 = 5 ~kg$.Veloci

Vedclass · Accepted Answer

$\frac{25}{9} ~m/s$

Vedclass · Answer

(A) The velocity of the centre of mass is given by the formula:$v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$Given: $m_1 = 4 ~kg$,$m_2 = 5 ~kg$.Velocity $v_1$ is along the east direction (let this be $\hat{i}$): $v_1 = 5 \hat{i} ~m/s$.Velocity $v_2$ is along the north direction (let this be $\hat{j}$): $v_2 = 3 \hat{j} ~m/s$.Substituting the values:$v_{CM} = \frac{4(5 \hat{i}) + 5(3 \hat{j})}{4 + 5}$$v_{CM} = \frac{20 \hat{i} + 15 \hat{j}}{9} = \frac{20}{9} \hat{i} + \frac{15}{9} \hat{j}$The magnitude of the velocity of the centre of mass is:$|v_{CM}| = \sqrt{\left(\frac{20}{9}\right)^2 + \left(\frac{15}{9}\right)^2}$$|v_{CM}| = \sqrt{\frac{400 + 225}{81}} = \sqrt{\frac{625}{81}}$$|v_{CM}| = \frac{25}{9} ~m/s$

List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

Two bodies of mass $4 ~kg$ and $5 ~kg$ are moving along east and north directions with velocities $5 ~m/s$ and $3 ~m/s$ respectively. The magnitude of the velocity of the centre of mass of the system is

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