Two particles of equal masses have velocities | vedclass

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(C) The velocity of the center of mass is given by $\overrightarrow{v}_{com} = \frac{m_1 \overrightarrow{v_1} + m_2 \overrightarrow{v_2}}{m_1 + m_2}$.

Vedclass · Accepted Answer

straight line

Vedclass · Answer

(C) The velocity of the center of mass is given by $\overrightarrow{v}_{com} = \frac{m_1 \overrightarrow{v_1} + m_2 \overrightarrow{v_2}}{m_1 + m_2}$.Since $m_1 = m_2 = m$,we have $\overrightarrow{v}_{com} = \frac{m(\overrightarrow{v_1} + \overrightarrow{v_2})}{2m} = \frac{\overrightarrow{v_1} + \overrightarrow{v_2}}{2} = \frac{2\hat{i} + 2\hat{j}}{2} = (\hat{i} + \hat{j}) \ m/s$.Similarly,the acceleration of the center of mass is $\overrightarrow{a}_{com} = \frac{m_1 \overrightarrow{a_1} + m_2 \overrightarrow{a_2}}{m_1 + m_2} = \frac{\overrightarrow{a_1} + \overrightarrow{a_2}}{2}$.Given $\overrightarrow{a_1} = (3\hat{i} + 3\hat{j}) \ m/s^2$ and $\overrightarrow{a_2} = 0$,we get $\overrightarrow{a}_{com} = \frac{3\hat{i} + 3\hat{j}}{2} = 1.5(\hat{i} + \hat{j}) \ m/s^2$.Since the initial velocity vector $\overrightarrow{v}_{com} = (\hat{i} + \hat{j})$ and the acceleration vector $\overrightarrow{a}_{com} = 1.5(\hat{i} + \hat{j})$ are parallel to each other,the center of mass moves in a straight line.

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