Question

A particle is moving in a circular path with a constant speed \(v\). If \(\theta\) is the angular displacement. Then starting from \(\theta = 0\), the maximum and minimum changes in the momentum will occur, when value of \(\theta\) is respectively:

• A. \(45^\circ \text{ and } 90^\circ\)
• B. \(90^\circ \text{ and } 180^\circ\)
• C. \(180^\circ \text{ and } 360^\circ\)
• D. \(90^\circ \text{ and } 270^\circ\)

Hint:

Minimum change in momentum occurs when initial and final momenta are same → full revolution.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Momentum is a vector quantity (\( \vec{p} = m\vec{v} \)). Even if speed is constant, the change in direction in circular motion causes a change in momentum.
: Key Formula or Approach:
The magnitude of the change in momentum for a particle moving in a circle with constant speed \( v \) through an angle \( \theta \) is:
\[ |\Delta \vec{p}| = 2mv \sin\left(\frac{\theta}{2}\right) \]
Step 2: Detailed Explanation:
- Maximum change occurs when \( \sin(\theta/2) \) is maximum (i.e., equal to 1).
\[ \frac{\theta}{2} = 90^\circ \implies \theta = 180^\circ \]
At \( \theta = 180^\circ \), \( |\Delta \vec{p}| = 2mv \sin(90^\circ) = 2mv \). This happens when the particle is at the diametrically opposite point.
- Minimum change occurs when \( \sin(\theta/2) \) is minimum (i.e., equal to 0).
\[ \frac{\theta}{2} = 180^\circ \implies \theta = 360^\circ \]
At \( \theta = 360^\circ \), the particle returns to its starting point, its velocity vector is identical to the initial one, so the change is zero.
Step 3: Final Answer:
Maximum change occurs at \( 180^\circ \) and minimum at \( 360^\circ \).