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Two particles of masses $1 \mathrm{~g}$ and $2 \mathrm{~g}$ move towards each other with velocities $10 \mathrm{~ms}^{-1}$ and $20 \mathrm{~ms}^{-1}$ respectively. The velocity of the centre of mass of the system of the two particles is
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The correct answer is:
$10 \mathrm{~ms}^{-1}$
Masses, $\mathrm{m}_1=1 \mathrm{~g} ; \mathrm{m}_2=2 \mathrm{~g}$
Velocity, $\mathrm{v}_1=10 \mathrm{~m} / \mathrm{s} ; \mathrm{v}_2 20 \mathrm{~m} / \mathrm{s}$
The velocity of the centre of mass
$\begin{aligned}
& V_{c m}=\frac{m_1 v_1+m_2 v_2}{m_1+m_2} \\
& =\frac{-1 \times 10+2 \times 20}{1+2}=\frac{30}{3}=10 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Velocity, $\mathrm{v}_1=10 \mathrm{~m} / \mathrm{s} ; \mathrm{v}_2 20 \mathrm{~m} / \mathrm{s}$
The velocity of the centre of mass
$\begin{aligned}
& V_{c m}=\frac{m_1 v_1+m_2 v_2}{m_1+m_2} \\
& =\frac{-1 \times 10+2 \times 20}{1+2}=\frac{30}{3}=10 \mathrm{~m} / \mathrm{s}
\end{aligned}$
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