Q

Need explanation for: A particle is moving along a circular path with constant speed of 10ms-1 . What is the magnitude of change in velocity of the particle, when it moves through an angle of 600 around the centre of the circle?

A particle is moving along a circular path with constant speed of 10ms-1 . What is the magnitude of change in velocity of the particle, when it moves through an angle of 600 around the centre of the circle?

• Option 1)

$10\sqrt{2}m/s$

• Option 2)

10m/s

• Option 3)

zero

• Option 4)

$10\sqrt{3}m/s$

Views

Relative Velocity -

Relative velocity of a body, A with respected body B when the to bodies moving at an angle $\Theta$.

$V_{AB}= \sqrt{V_{A}^{2}+V_{B}^{2}+2V_{A}V_{B}\cos \left ( 180-\theta \right )}$

$= \sqrt{V_{A}^{2}+V_{B}^{2}-2V_{A}V_{B}\cos \left ( \theta \right )}$

- wherein

$\\*V_{A}= velocity\: of\: A\\* V_{B}= velocity\: of\: B\\* \Theta = angle \: between \: A \: and \: B$

$\left | \Delta \vec{V} \right |=$$\sqrt{\vec{V}_{1}^{2}+\vec{V}_{2}^{2}+2\vec{V}_{1}\vec{V}_{2}\cos \left ( \pi -\theta \right )}$

Since $\left | \vec{V_{1}} \right |=\left | \vec{V_{2}} \right |=\left | V \right |$

$\left |\Delta \vec{V} \right |=2 V \: \sin \frac{\Theta }{2}$

$=20\: \sin \: 30$

$= 10 m/s$

Option 1)

$10\sqrt{2}m/s$

Option 2)

10m/s

Option 3)

zero

Option 4)

$10\sqrt{3}m/s$

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