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

Options

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

Explanation

To find the angle between the force and the momentum of a particle moving in the X-Y plane, we start with the given linear momentum vector:

\[ \vec{p}(t) = A \left( \hat{i} \cos(kt) - \hat{j} \sin(kt) \right) \]

### Step 1: Differentiate the momentum to find the force
The force \(\vec{F}\) acting on the particle is given by the rate of change of momentum:

\[ \vec{F} = \frac{d\vec{p}}{dt} \]

Now, we differentiate \(\vec{p}(t)\):

\[ \frac{d\vec{p}}{dt} = A \left( \frac{d}{dt} \left( \hat{i} \cos(kt) \right) - \frac{d}{dt} \left( \hat{j} \sin(kt) \right) \right) \]

Using the chain rule, we find:

\[ \frac{d}{dt} \left( \cos(kt) \right) = -k \sin(kt) \quad \text{and} \quad \frac{d}{dt} \left( \sin(kt) \right) = k \cos(kt) \]

Thus, we have:

\[ \vec{F} = A \left( -k \hat{i} \sin(kt) - k \hat{j} \cos(kt) \right) \]

This simplifies to:

\[ \vec{F} = -Ak \left( \hat{i} \sin(kt) + \hat{j} \cos(kt) \right) \]

### Step 2: Find the angle between force and momentum
To find the angle \(\theta\) between the force \(\vec{F}\) and momentum \(\vec{p}\), we use the dot product formula:

\[ \vec{F} \cdot \vec{p} = |\vec{F}| |\vec{p}| \cos(\theta) \]

If the dot product is zero, then \(\cos(\theta) = 0\), which implies that:

\[ \theta = 90^\circ \]

### Conclusion
Thus, the angle between the force and the momentum is:

\[ \theta = 90^\circ \]

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

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