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: Understanding the Question:
The question asks for the magnitude of the change in the velocity vector of a particle moving in uniform circular motion after it completes exactly half of a revolution.

Step 2: Key Formula or Approach:
Velocity is a vector quantity, meaning it has both magnitude and direction.
The change in velocity (\(\Delta \vec{v}\)) is the vector difference between the final velocity and the initial velocity:
\[ \Delta \vec{v} = \vec{v}_f - \vec{v}_i \] We need to find its magnitude \(|\Delta \vec{v}|\).

Step 3: Detailed Explanation:
Consider a particle moving counterclockwise in a circle in the \(xy\)-plane.
Let the particle start at the rightmost point of the circle on the x-axis, at \((R, 0)\).
At this starting point, the velocity vector is directed tangentially upwards along the y-axis:
\[ \vec{v}_i = v\hat{j} \] After traveling half of the circular path, the particle reaches the diametrically opposite point on the negative x-axis, at \((-R, 0)\).
At this diametrically opposite point, the velocity vector is directed tangentially downwards along the negative y-axis:
\[ \vec{v}_f = -v\hat{j} \] Now, calculate the change in velocity:
\[ \Delta \vec{v} = \vec{v}_f - \vec{v}_i = -v\hat{j} - (v\hat{j}) = -2v\hat{j} \] The magnitude of this change in velocity is:
\[ |\Delta \vec{v}| = |-2v\hat{j}| = 2v \]

Step 4: Final Answer:
The correct choice is (C).