Question

A particle moves in a circle of radius \(R\) with a constant speed \(v\). The magnitude of the change in velocity after the particle has traveled half of the circular path is:

• A. \(0\)
• B. \(v\)
• C. \(2v\)
• D. \(\sqrt{2}v\)

Hint:

For a particle in uniform circular motion, the change in velocity over an angle \(\theta\) is given directly by the formula \(|\Delta \vec{v}| = 2v \sin(\theta/2)\). For half a circle, \(\theta = 180^\circ\), yielding \(2v \sin(90^\circ) = 2v\).

Answer 1

The Correct Option is C

Step 1: Picture the motion.
A particle runs around a circle of radius $R$ at constant speed $v$. After it covers half the circle we want the magnitude of the change in its velocity.
Step 2: Remember velocity is a vector.
Even though the speed stays $v$, the direction keeps turning. So the velocity vector changes even when its length does not. We need $|\Delta\vec{v}| = |\vec{v}_f - \vec{v}_i|$.
Step 3: Fix a starting point and direction.
Let the particle start at the right edge of the circle moving upward, so \[ \vec{v}_i = v\,\hat{j} \]
Step 4: Find the velocity after half a turn.
Half a revolution lands the particle on the opposite side of the circle, where it is now moving straight down: \[ \vec{v}_f = -v\,\hat{j} \] The direction has completely reversed.
Step 5: Subtract the vectors.
\[ \Delta\vec{v} = \vec{v}_f - \vec{v}_i = -v\,\hat{j} - v\,\hat{j} = -2v\,\hat{j} \]
Step 6: Take the magnitude.
\[ |\Delta\vec{v}| = |-2v\,\hat{j}| = 2v \] So the change in velocity has magnitude $2v$, which is option (C).
\[ \boxed{|\Delta\vec{v}| = 2v} \]