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Two bodies of mass $4 \mathrm{~kg}$ and $5 \mathrm{~kg}$ are moving along east and north directions with velocities $5 \mathrm{~m} / \mathrm{s}$ and $3 \mathrm{~m} / \mathrm{s}$ respectively. Magnitude of the velocity of centre of mass of the system is
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Verified Answer
The correct answer is:
$\frac{25}{9} \mathrm{~m} / \mathrm{s}$
Velocity of centre of mass is
$$
\mathbf{v}_{\mathrm{CM}}=\frac{m_1 \mathbf{v}_1+m_2 \mathbf{v}_2}{m_1+m_2}
$$
Given, $m_1=4 \mathrm{~kg}, m_2=5 \mathrm{~kg}, v_1=5 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$
$$
\begin{aligned}
& v_2=3 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} \\
& \therefore \quad \mathbf{v}_{\mathrm{CM}}=\frac{4 \times 5 \hat{\mathbf{j}}+5 \times 3 \hat{\mathbf{i}}}{5+4} \\
&=\frac{20 \hat{\mathbf{i}}}{9}+\frac{15}{9} \hat{\mathbf{j}}
\end{aligned}
$$
Hence, magnitude $\left|\mathbf{v}_{\mathrm{CM}}\right|=\sqrt{\left(\frac{20}{9}\right)^2+\left(\frac{15}{9}\right)^2}$
$$
=\frac{25}{9} \mathrm{~m} / \mathrm{s}
$$
$$
\mathbf{v}_{\mathrm{CM}}=\frac{m_1 \mathbf{v}_1+m_2 \mathbf{v}_2}{m_1+m_2}
$$
Given, $m_1=4 \mathrm{~kg}, m_2=5 \mathrm{~kg}, v_1=5 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$
$$
\begin{aligned}
& v_2=3 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} \\
& \therefore \quad \mathbf{v}_{\mathrm{CM}}=\frac{4 \times 5 \hat{\mathbf{j}}+5 \times 3 \hat{\mathbf{i}}}{5+4} \\
&=\frac{20 \hat{\mathbf{i}}}{9}+\frac{15}{9} \hat{\mathbf{j}}
\end{aligned}
$$
Hence, magnitude $\left|\mathbf{v}_{\mathrm{CM}}\right|=\sqrt{\left(\frac{20}{9}\right)^2+\left(\frac{15}{9}\right)^2}$
$$
=\frac{25}{9} \mathrm{~m} / \mathrm{s}
$$
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