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  • 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

Answers (1)

best_answer

 

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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