A particle is moving in a circle with uniform speed v . In moving from a point to another diametrically opposite point

Options

A. the momentum change by mv
B. the momentum changes by 2 mv
C. the kinetic energy changes by `((1)/(2))mv^(2)`
D. the kinetic energy changes by `mv^(2)`

Answer

To solve the problem of finding the change in momentum of a particle moving in a circle with uniform speed as it moves from one point to the diametrically opposite point, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial and Final Points**:
- Let the particle start at point A and move to point B, which is diametrically opposite to A on the circle.

2. **Define the Initial Momentum**:
- At point A, the particle has a velocity \( \vec{v} \). The initial momentum \( \vec{P}_{initial} \) can be expressed as:
\[ \vec{P}_{initial} = m \vec{v} \]
- Here, \( m \) is the mass of the particle and \( \vec{v} \) is the velocity vector at point A.

3. **Define the Final Momentum**:
- At point B, the particle still has the same speed \( v \), but the direction of the velocity vector has changed. If we assume the particle moves in a counterclockwise direction, the velocity vector at point B can be represented as:
\[ \vec{v}_{final} = -\vec{v} \]
- Thus, the final momentum \( \vec{P}_{final} \) is given by:
\[ \vec{P}_{final} = m (-\vec{v}) = -m \vec{v} \]

4. **Calculate the Change in Momentum**:
- The change in momentum \( \Delta \vec{P} \) is calculated as:
\[ \Delta \vec{P} = \vec{P}_{final} - \vec{P}_{initial} \]
- Substituting the expressions for initial and final momentum:
\[ \Delta \vec{P} = (-m \vec{v}) - (m \vec{v}) = -2m \vec{v} \]

5. **Magnitude of Change in Momentum**:
- The magnitude of the change in momentum is:
\[ |\Delta \vec{P}| = 2mv \]

6. **Conclusion**:
- The change in momentum of the particle as it moves from point A to point B is \( 2mv \) in the direction opposite to the initial momentum.

### Final Answer:
The change in momentum is \( 2mv \).

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