Question

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. \(2v\sin\frac{\theta}{2}\)
• B. \(v\sin\theta\)
• C. \(\frac{v\sin2\theta}{2}\)
• D. \(2v\sin\theta\)

Hint:

For two vectors of equal magnitude \(v\) making an angle \(\theta\), the magnitude of their difference is \[ 2v\sin\frac{\theta}{2}. \] This result is frequently used in uniform circular motion to find the change in velocity.

Answer 1

The Correct Option is A

Step 1: Use vector subtraction for speed \( v \) on a circle.
Speed is constant, so \( |\vec{v}_A| = |\vec{v}_B| = v \). The angle between the velocity vectors equals the angle subtended at the center, \( \theta \).

Step 2: Magnitude of change in velocity.
\( |\Delta\vec{v}| = \sqrt{v^2 + v^2 - 2v^2\cos\theta} = 2v\sin\frac{\theta}{2} \)

\[ \boxed{|\Delta v| = 2v\sin\tfrac{\theta}{2}} \]