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Particle $A$ moves along $X$ -axis with a uniform velocity of magnitude $10 \mathrm{m} / \mathrm{s}$. Particle $B$ moves with uniform velocity $20 \mathrm{m} / \mathrm{s}$ along a direction making an angle of $60^{\circ}$ with the positive direction of $X$ -axis as shown in the figure. The relative velocity of $B$ with respect to that of $A$ is
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Verified Answer
The correct answer is:
$10 \sqrt{3} \mathrm{m} / \mathrm{s}$ along $Y$ -axis (perpendicular to $X$ -axis)
$$
\begin{aligned}
&\text { The component of velocity of } B \text { along } x \text { -direction }\\
&\begin{aligned}
\mathbf{v}_{B x} &=20 \cos 60^{\circ}=20 \times \frac{1}{2}=10 \mathrm{m} / \mathrm{s} \\
\mathbf{v}_{A} &=10 \hat{\mathbf{i}} \\
\mathbf{v}_{B} &=10 \hat{\mathbf{i}}+10 \sqrt{3} \hat{\mathbf{j}} \\
\mathbf{v}_{B A} &=\mathbf{v}_{B}-\mathbf{v}_{A}=10 \hat{\mathbf{i}}+10 \sqrt{3} \hat{\mathbf{j}}-10 \hat{\mathbf{i}}=10 \sqrt{3} \hat{\mathbf{j}}
\end{aligned}
\end{aligned}
$$
\begin{aligned}
&\text { The component of velocity of } B \text { along } x \text { -direction }\\
&\begin{aligned}
\mathbf{v}_{B x} &=20 \cos 60^{\circ}=20 \times \frac{1}{2}=10 \mathrm{m} / \mathrm{s} \\
\mathbf{v}_{A} &=10 \hat{\mathbf{i}} \\
\mathbf{v}_{B} &=10 \hat{\mathbf{i}}+10 \sqrt{3} \hat{\mathbf{j}} \\
\mathbf{v}_{B A} &=\mathbf{v}_{B}-\mathbf{v}_{A}=10 \hat{\mathbf{i}}+10 \sqrt{3} \hat{\mathbf{j}}-10 \hat{\mathbf{i}}=10 \sqrt{3} \hat{\mathbf{j}}
\end{aligned}
\end{aligned}
$$
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