Q141 / 160 · WBJEE 2015 · Physics

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 Figure

  • A
    $10 \mathrm{m} / \mathrm{s}$ along $X$ -axis
  • Correct answer
    B
    $10 \sqrt{3} \mathrm{m} / \mathrm{s}$ along $Y$ -axis (perpendicular to $X$ -axis)
  • C
    $10 \sqrt{5} \mathrm{m} / \mathrm{s}$ along the bisection of the velocities of A and $B$
  • D
    $30 \mathrm{m} / \mathrm{s}$ along negative $X$ -axis

Solution

$$ \begin{aligned} &\text { The component of velocity of } B \text { along } x \text { -direction }\\ &\begin{aligned} \mathbf{v}_{B x} &=20 \cos 60^{\circ}=20 \times \frac{1}{2}=10 \mathrm{m} / \mathrm{s} \\ \mathbf{v}_{A} &=10 \hat{\mathbf{i}} \\ \mathbf{v}_{B} &=10 \hat{\mathbf{i}}+10 \sqrt{3} \hat{\mathbf{j}} \\ \mathbf{v}_{B A} &=\mathbf{v}_{B}-\mathbf{v}_{A}=10 \hat{\mathbf{i}}+10 \sqrt{3} \hat{\mathbf{j}}-10 \hat{\mathbf{i}}=10 \sqrt{3} \hat{\mathbf{j}} \end{aligned} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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