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: Understand the velocity vectors.
The particle moves with constant speed \[ v. \] At points \(A\) and \(B\), the velocity vectors are tangential to the circular path.
Since the radius vectors at \(A\) and \(B\) subtend an angle \[ \theta \] at the centre, the corresponding velocity vectors also subtend the same angle \[ \theta. \] Thus, the initial and final velocity vectors have: \[ |\vec v_A|=|\vec v_B|=v \] and the angle between them is \[ \theta. \]

Step 2: Use vector subtraction.
The change in velocity is \[ \Delta \vec v=\vec v_B-\vec v_A. \] The magnitude of the difference of two vectors of equal magnitude \(v\) making an angle \(\theta\) is obtained using the cosine rule. \[ |\Delta \vec v| = \sqrt{v^2+v^2-2v^2\cos\theta}. \] \[ = \sqrt{2v^2(1-\cos\theta)}. \]

Step 3: Apply the trigonometric identity.
Using \[ 1-\cos\theta = 2\sin^2\frac{\theta}{2}, \] we get \[ |\Delta \vec v| = \sqrt{2v^2\cdot 2\sin^2\frac{\theta}{2}}. \] \[ = \sqrt{4v^2\sin^2\frac{\theta}{2}}. \] \[ = 2v\sin\frac{\theta}{2}. \]

Step 4: Final conclusion.
Hence, the magnitude of the change in velocity is \[ \boxed{2v\sin\frac{\theta}{2}} \] Therefore, the correct option is \[ \boxed{(1)} \]