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Q.

Two particles A and B are moving in the XY plane such that their velocity components are $V_{x} = 1 m / s, V_{y} = \frac{1}{\sqrt{3}} m / s$ for A and $V_{x} = 2 m / s V_{y} = 2 m / s$ for B. The angle between their paths is

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a

$45^{0}$

b

$15^{0}$

c

$60^{0}$

d

$30^{0}$

answer is D.

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Detailed Solution

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$\tan α = \frac{1}{\sqrt{3}} \Rightarrow α = 30^{0} t a m β = \frac{2}{2} \Rightarrow β = 45^{0} β - α = 15^{0}$

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