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?

Explanation

To solve the problem, we need to find the angle \( \theta \) between the force \( \vec{F} \) and the linear momentum \( \vec{P} \) of a particle moving in the x-y plane. The linear momentum components are given as:

\[ P_x = 2 \cos t \]
\[ P_y = 2 \sin t \]

### Step 1: Write the expression for the momentum vector
The momentum vector \( \vec{P} \) can be expressed in terms of its components:

\[ \vec{P} = P_x \hat{i} + P_y \hat{j} = 2 \cos t \hat{i} + 2 \sin t \hat{j} \]

### Step 2: Calculate the force vector
The force \( \vec{F} \) is related to the momentum \( \vec{P} \) by the equation:

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

Now we differentiate the components of \( \vec{P} \):

\[ \frac{dP_x}{dt} = \frac{d}{dt}(2 \cos t) = -2 \sin t \]
\[ \frac{dP_y}{dt} = \frac{d}{dt}(2 \sin t) = 2 \cos t \]

Thus, the force vector \( \vec{F} \) becomes:

\[ \vec{F} = \left(-2 \sin t\right) \hat{i} + \left(2 \cos t\right) \hat{j} \]

### Step 3: Use the dot product to find the angle \( \theta \)
The angle \( \theta \) between two vectors can be found using the dot product formula:

\[ \vec{F} \cdot \vec{P} = |\vec{F}| |\vec{P}| \cos \theta \]

First, we calculate the dot product \( \vec{F} \cdot \vec{P} \):

\[ \vec{F} \cdot \vec{P} = (-2 \sin t)(2 \cos t) + (2 \cos t)(2 \sin t) = -4 \sin t \cos t + 4 \sin t \cos t = 0 \]

Since the dot product is zero, we can conclude:

\[ |\vec{F}| |\vec{P}| \cos \theta = 0 \]

### Step 4: Determine the angle \( \theta \)
From the equation above, since \( \vec{F} \cdot \vec{P} = 0 \), it follows that:

\[ \cos \theta = 0 \]

This implies:

\[ \theta = 90^\circ \]

### Final Answer
The angle \( \theta \) between the force \( \vec{F} \) and the momentum \( \vec{P} \) at any time \( t \) is:

\[ \theta = 90^\circ \]