A particle moves in the `xy` plane under the influence of a force such that its linear momentum is `vecP(t) = A [haticos(kt)-hatjsin(kt)]`, where `A` and `k` are constants. The angle between the force and momentum is

Options

A. `0^(@)`
B. `30^(@)`
C. `45^(@)`
D. `90^(@)`

Explanation

To solve the problem, we need to find the angle between the force and momentum of a particle moving in the xy-plane, given the linear momentum \(\vec{P}(t) = A [\hat{i} \cos(kt) - \hat{j} \sin(kt)]\).

### Step-by-Step Solution:

1. **Identify the Linear Momentum**:
The linear momentum is given by:
\[ \vec{P}(t) = A [\hat{i} \cos(kt) - \hat{j} \sin(kt)] \]

2. **Differentiate Momentum to Find Force**:
The force \(\vec{F}\) acting on the particle can be found using the relation:
\[ \vec{F} = \frac{d\vec{P}}{dt} \]
We differentiate \(\vec{P}(t)\):
\[ \frac{d\vec{P}}{dt} = A \left[-\hat{i} k \sin(kt) - \hat{j} k \cos(kt)\right] \]
Simplifying this, we have:
\[ \vec{F} = -Ak \left[\hat{i} \sin(kt) + \hat{j} \cos(kt)\right] \]

3. **Dot Product of Force and Momentum**:
To find the angle \(\theta\) between the force and momentum, we use the dot product:
\[ \vec{F} \cdot \vec{P} = |\vec{F}| |\vec{P}| \cos(\theta) \]
Calculating the dot product:
\[ \vec{F} \cdot \vec{P} = \left(-Ak \left[\hat{i} \sin(kt) + \hat{j} \cos(kt)\right]\right) \cdot \left(A [\hat{i} \cos(kt) - \hat{j} \sin(kt)]\right) \]
Expanding this:
\[ = -A^2 k \left(\sin(kt) \cos(kt) - \cos(kt) \sin(kt)\right) \]
This simplifies to:
\[ = 0 \]

4. **Conclusion About the Angle**:
Since the dot product \(\vec{F} \cdot \vec{P} = 0\), it implies:
\[ |\vec{F}| |\vec{P}| \cos(\theta) = 0 \]
This means that \(\cos(\theta) = 0\), which indicates:
\[ \theta = 90^\circ \]

### Final Answer:
The angle between the force and momentum is \(90^\circ\).