Two particles A and B are moving in the x-y plane such that their velocity components are vx-1 m-s-vy-1-3 m-s for A and vx-2 m-s- vy-2 m-s for B- If both the particles start moving from the same point- what is the angle in degree between their paths-

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Standard XII

Physics

Motion Under Gravity

Two particles...

Question

Two particles $A$ and $B$ are moving in the $x - y$ 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 $B$ ). If both the particles start moving from the same point, what is the angle (in degree) between their paths?

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Solution

$tan α = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \Rightarrow= α = 60^{\circ}$

$tan β = \frac{2}{2} = 1 \Rightarrow β = 45^{\circ}$
The particles are moving in straight lines, hence the angle between their paths
$= α - β = 60^{\circ} - 45^{\circ} = 15^{\circ}$

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Motion Under Gravity

Standard XII Physics

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Two particles A and B are moving in the x−y plane such that their velocity components are vx=1 m/s,vy=1√3 m/s (for A) and vx=2 m/s, vy=2 m/s (for B). If both the particles start moving from the same point, what is the angle (in degree) between their paths?

tanα=11√3=√3⇒=α=60∘ tanβ=22=1⇒β=45∘ The particles are moving in straight lines, hence the angle between their paths =α−β=60∘−45∘=15∘

$tan α = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \Rightarrow= α = 60^{\circ}$

$tan β = \frac{2}{2} = 1 \Rightarrow β = 45^{\circ}$
The particles are moving in straight lines, hence the angle between their paths
$= α - β = 60^{\circ} - 45^{\circ} = 15^{\circ}$