A particle moves in the x-y plane under the action of a force `vecF` such that the value of its linear momentum `vecP` at any time t is `P_(x)=2` cost and `p_(y)=2sint`. What is the angle `theta` between `vecF` and P at a given time t?

Options

A. `90^(@)`
B. `0^(@)`
C. `180^(@)`
D. `30^(@)`

Answer

To solve the problem, we need to find the angle \( \theta \) between the force vector \( \vec{F} \) and the momentum vector \( \vec{P} \) at any time \( t \).

### Step-by-Step Solution:

1. **Define the momentum vector \( \vec{P} \)**:
- Given:
\[ P_x = 2 \cos t, \quad P_y = 2 \sin t \]
- Therefore, the momentum vector can be expressed as:
\[ \vec{P} = 2 \cos t \, \hat{i} + 2 \sin t \, \hat{j} \]

2. **Calculate the force vector \( \vec{F} \)**:
- The force is defined as the rate of change of momentum:
\[ \vec{F} = \frac{d\vec{P}}{dt} \]
- Differentiate \( \vec{P} \):
\[ \frac{dP_x}{dt} = \frac{d}{dt}(2 \cos t) = -2 \sin t \]
\[ \frac{dP_y}{dt} = \frac{d}{dt}(2 \sin t) = 2 \cos t \]
- Thus, the force vector is:
\[ \vec{F} = -2 \sin t \, \hat{i} + 2 \cos t \, \hat{j} \]

3. **Use the dot product to find the angle \( \theta \)**:
- The dot product \( \vec{F} \cdot \vec{P} \) is given by:
\[ \vec{F} \cdot \vec{P} = (-2 \sin t)(2 \cos t) + (2 \cos t)(2 \sin t) \]
- Simplifying this:
\[ \vec{F} \cdot \vec{P} = -4 \sin t \cos t + 4 \sin t \cos t = 0 \]

4. **Relate the dot product to the angle**:
- The formula for the dot product in terms of the angle is:
\[ \vec{F} \cdot \vec{P} = |\vec{F}| |\vec{P}| \cos \theta \]
- Since \( \vec{F} \cdot \vec{P} = 0 \), we have:
\[ |\vec{F}| |\vec{P}| \cos \theta = 0 \]
- This implies:
\[ \cos \theta = 0 \]

5. **Determine the angle \( \theta \)**:
- The angle \( \theta \) for which \( \cos \theta = 0 \) is:
\[ \theta = \frac{\pi}{2} \text{ radians} = 90^\circ \]

### Conclusion:
The angle \( \theta \) between the force \( \vec{F} \) and the momentum \( \vec{P} \) at any time \( t \) is \( 90^\circ \).