Two particles of equal mass have velocities 2 | vedclass

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(D) Given,masses $m_1 = m_2 = m$.Initial velocities are $\vec{v}_1 = 2\hat{i}\,ms^{-1}$ and $\vec{v}_2 = 2\hat{j}\,ms^{-1}$.Accelerations are $\vec{a}

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straight line

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(D) Given,masses $m_1 = m_2 = m$.Initial velocities are $\vec{v}_1 = 2\hat{i}\,ms^{-1}$ and $\vec{v}_2 = 2\hat{j}\,ms^{-1}$.Accelerations are $\vec{a}_1 = (3\hat{i} + 3\hat{j})\,ms^{-2}$ and $\vec{a}_2 = 0$.The velocity of the centre of mass is $\vec{v}_{CM} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2} = \frac{m(2\hat{i} + 2\hat{j})}{2m} = (\hat{i} + \hat{j})\,ms^{-1}$.The acceleration of the centre of mass is $\vec{a}_{CM} = \frac{m_1\vec{a}_1 + m_2\vec{a}_2}{m_1 + m_2} = \frac{m(3\hat{i} + 3\hat{j} + 0)}{2m} = \frac{3}{2}(\hat{i} + \hat{j})\,ms^{-2}$.Since the initial velocity vector $\vec{v}_{CM} = (\hat{i} + \hat{j})$ and the acceleration vector $\vec{a}_{CM} = \frac{3}{2}(\hat{i} + \hat{j})$ are parallel (i.e.,$\vec{a}_{CM} = k\vec{v}_{CM}$ where $k = 1.5$),the centre of mass moves in a straight line.

Two particles of equal mass have velocities $2\,\hat{i}\,ms^{-1}$ and $2\,\hat{j}\,ms^{-1}$. The first particle has an acceleration $(3\,\hat{i} + 3\,\hat{j})\,ms^{-2}$ while the acceleration of the second particle is zero. The centre of mass of the two particles moves in

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