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Two particles of equal mass have velocities $\overrightarrow{\mathbf{v}}_1=4 \hat{\mathbf{i}}$ and $\overrightarrow{\mathbf{v}}_2=4 \hat{\mathbf{j}} \mathrm{ms}^{-1}$. First particle has an acceleration $\overrightarrow{\mathbf{a}}_1=(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \mathrm{ms}^{-2}$ while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of
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The correct answer is:
straight line
$$
\begin{aligned}
& \mathbf{v}_{\mathrm{CM}}=\frac{m_1 \mathbf{v}_1+m_2 \mathbf{v}_2}{m_1+m_2} \\
& =\frac{\mathbf{v}_1+\mathbf{v}_2}{2}\left(\text { as } m_1=m_2\right) \\
& \text { and } \mathbf{a}_{\mathrm{CM}}=\frac{m_1 \mathbf{a}_1+m_2 \mathbf{a}_2}{m_1+m_2}=\frac{\mathbf{a}_1+0}{2}=\frac{\mathbf{a}_1}{2}
\end{aligned}
$$
The centre of mass of two particles will move with the mean velocity of two particles having common acceleration $\frac{\mathbf{a}}{2}$. Hence, path of CM will be a straight line.
\begin{aligned}
& \mathbf{v}_{\mathrm{CM}}=\frac{m_1 \mathbf{v}_1+m_2 \mathbf{v}_2}{m_1+m_2} \\
& =\frac{\mathbf{v}_1+\mathbf{v}_2}{2}\left(\text { as } m_1=m_2\right) \\
& \text { and } \mathbf{a}_{\mathrm{CM}}=\frac{m_1 \mathbf{a}_1+m_2 \mathbf{a}_2}{m_1+m_2}=\frac{\mathbf{a}_1+0}{2}=\frac{\mathbf{a}_1}{2}
\end{aligned}
$$
The centre of mass of two particles will move with the mean velocity of two particles having common acceleration $\frac{\mathbf{a}}{2}$. Hence, path of CM will be a straight line.
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