WBJEE · Physics · Motion In Two Dimensions
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
- A \(10 \mathrm{m} / \mathrm{s}\) along \(X\) -axis
- 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
Answer & Solution
Correct Answer
(B) \(10 \sqrt{3} \mathrm{m} / \mathrm{s}\) along \(Y\) -axis (perpendicular to \(X\) -axis)
Step-by-step Solution
Detailed explanation
\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…
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