A particle moves in a circle with a uniform speed. When it goes from a point A to a diametrically opposite point B, the momentum of the particle changes by `vecP_(A)-vecP_(B)=2` kg m/s `(hatj)` and the centripetal force acting on it changes by `vecF_(A)-vecF_(B)=8N(hati)` where `hati,hatj` are unit vectors along X and Y axes respectively. The angular velocity of the particle is

Options

A. dependent on its mass
B. 4 rad/sec
C. `(2)/(pi)` rad/sec
D. `16pi` rad/sec

Explanation

To solve the problem, we will analyze the motion of a particle moving in a circle with uniform speed and determine its angular velocity based on the given changes in momentum and centripetal force.

### Step-by-Step Solution:

1. **Understanding the Problem**:
The particle moves from point A to point B, which are diametrically opposite points on a circular path. The change in momentum and centripetal force are provided.

2. **Change in Momentum**:
The change in momentum is given as:
\[ \vec{P}_A - \vec{P}_B = 2 \, \text{kg m/s} \, \hat{j} \]
Since momentum is defined as \( \vec{P} = m \vec{v} \), we can express the change in momentum as:
\[ m \vec{v}_A - m \vec{v}_B = 2 \hat{j} \]

3. **Velocity at Points A and B**:
Given that the particle moves with uniform speed, the magnitudes of velocities at points A and B are the same, but their directions differ. If we assume the particle moves clockwise:
- At point A, the velocity vector \( \vec{v}_A \) can be represented as \( V \hat{j} \) (upward).
- At point B, the velocity vector \( \vec{v}_B \) will be \( -V \hat{j} \) (downward).

4. **Substituting Velocities**:
Substituting the velocity vectors into the momentum equation:
\[ m(V \hat{j}) - m(-V \hat{j}) = 2 \hat{j} \]
This simplifies to:
\[ 2mV \hat{j} = 2 \hat{j} \]
Dividing both sides by \( \hat{j} \):
\[ 2mV = 2 \implies mV = 1 \implies mV = 1 \, \text{kg m/s} \]

5. **Change in Centripetal Force**:
The change in centripetal force is given as:
\[ \vec{F}_A - \vec{F}_B = 8 \, \text{N} \, \hat{i} \]
The centripetal force at points A and B can be expressed as:
\[ F_A = \frac{mV^2}{R} \quad \text{and} \quad F_B = -\frac{mV^2}{R} \]
Therefore, we have:
\[ \frac{mV^2}{R} - \left(-\frac{mV^2}{R}\right) = 8 \hat{i} \]
This simplifies to:
\[ 2\frac{mV^2}{R} = 8 \implies \frac{mV^2}{R} = 4 \]

6. **Finding Angular Velocity**:
We know that \( V = \omega R \). Substituting this into the centripetal force equation:
\[ m \frac{(\omega R)^2}{R} = 4 \implies m \omega^2 R = 4 \]
From the previous step, we found \( m = 1 \):
\[ 1 \cdot \omega^2 R = 4 \implies \omega^2 R = 4 \]
Thus, we can express \( \omega \):
\[ \omega = \frac{4}{R} \]

7. **Final Result**:
Since \( R \) is a constant radius, we conclude that the angular velocity \( \omega \) is:
\[ \omega = 4 \, \text{radians/second} \]