AP EAMCET · PHYSICS · Motion In Two Dimensions
A particle is moving on a circular path with a constant speed \(v\). Its change of velocity as it moves from \(A\) to \(B\) in the figure is
- A \(2 v \sin \frac{\theta}{2}\)
- B \(v \sin \theta\)
- C \(\frac{v \sin 2 \theta}{2}\)
- D \(2 v \sin \theta\)
Answer & Solution
Correct Answer
(A) \(2 v \sin \frac{\theta}{2}\)
Step-by-step Solution
Detailed explanation
The given situation is shown below From above figure we can observe that velocity vector (tangential to circle) can be resolved into horizontal \(\left(v \cos \frac{\theta}{2}\right)\) and vertical \(\left(v \sin \frac{\theta}{2}\right)\) components. As, particle is executing…
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