**Answer:**

The momentum of a particle is given by the vector **P** = **Px** **i** + **Py** **j**, where **Px** and **Py** are the x and y components of the momentum vector and **i** and **j** are the unit vectors along the x and y axes respectively.

Given:
**Px** = 2cos(t)
**Py** = 2sin(t)

To find the angle between the force vector **F** and the momentum vector **P** at time t, we need to calculate the dot product of **F** and **P** and then use the dot product formula to find the angle.

**Finding the Dot Product:**
The dot product of two vectors **A** and **B** is given by **A · B = |A||B|cos(θ)**, where |A| and |B| are the magnitudes of vectors **A** and **B**, and θ is the angle between them.

In this case, **A** = **F** and **B** = **P**.
So, **F · P = |F||P|cos(θ)**

**Calculating the Magnitude of Force:**
The magnitude of the force vector **F** can be calculated using the Pythagorean theorem:
|F| = √(Fx^2 + Fy^2)

**Calculating the Magnitude of Momentum:**
The magnitude of the momentum vector **P** can be calculated using the Pythagorean theorem:
|P| = √(Px^2 + Py^2)

**Calculating the Dot Product:**
Substituting the given values of **Fx**, **Fy**, **Px**, and **Py** into the dot product formula:
(FxPx + FyPy) = |F||P|cos(θ)

**Calculating the Angle:**
We are given that the angle between **F** and **P** is 90 degrees, which means cos(θ) = 0.
Therefore, the dot product (FxPx + FyPy) must also be zero.

**Calculating Fx and Fy:**
To find the x and y components of the force vector, we need to differentiate the given momentum components with respect to time.
Fx = d(Px)/dt = -2sin(t)
Fy = d(Py)/dt = 2cos(t)

**Substituting the Values:**
Substituting the values of Fx and Fy into the dot product formula:
(-2sin(t) * 2cos(t)) + (2cos(t) * 2sin(t)) = 0

Simplifying the equation:
-4sin(t)cos(t) + 4sin(t)cos(t) = 0

The equation simplifies to:
0 = 0

Since the equation is true for all values of t, the angle between the force vector **F** and the momentum vector **P** is indeed 90 degrees.