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NEET PHYSICSEasy

The figure shows a body of mass mm moving with a uniform speed vv along a circle of radius rr. The change in velocity in going from AA to BB is:

A

v2v\sqrt{2}

B

v/2v/\sqrt{2}

C

vv

D

zero

Step-by-Step Solution

  1. Vector Difference: The change in velocity Δv\Delta \mathbf{v} is the vector difference between the final velocity vB\mathbf{v}_B and the initial velocity vA\mathbf{v}_A: Δv=vB−vA\Delta \mathbf{v} = \mathbf{v}_B - \mathbf{v}_A .
  2. Uniform Circular Motion: In uniform circular motion, the speed remains constant (∣vA∣=∣vB∣=v|\mathbf{v}_A| = |\mathbf{v}_B| = v), but the direction changes .
  3. Calculation: Assuming points A and B subtend an angle of 90∘90^\circ at the center (implied by the standard result v2v\sqrt{2} for a quarter circle path), the angle between the velocity vectors vA\mathbf{v}_A and vB\mathbf{v}_B is also 90∘90^\circ. Using the law of vector subtraction: ∣Δv∣=v2+v2−2v2cos⁡(90∘)|\Delta \mathbf{v}| = \sqrt{v^2 + v^2 - 2v^2 \cos(90^\circ)} ∣Δv∣=v2+v2−0=2v2=v2|\Delta \mathbf{v}| = \sqrt{v^2 + v^2 - 0} = \sqrt{2v^2} = v\sqrt{2}.
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