A particle moves in the xy-plane under the action of a force F such that the components of its linear momentum p at any time t and `p_(x)=2cos` t, `p_(y)=2sin` t. the angle between F and p at time t is

Options

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

Explanation

To find the angle between the force \( F \) and the linear momentum \( p \) of the particle at any time \( t \), we can follow these steps:

### Step 1: Write the expressions for momentum components
The components of the linear momentum \( p \) are given as:
\[ p_x = 2 \cos t \]
\[ p_y = 2 \sin t \]
Thus, we can express the momentum vector \( \mathbf{p} \) as:
\[ \mathbf{p} = p_x \hat{i} + p_y \hat{j} = 2 \cos t \hat{i} + 2 \sin t \hat{j} \]

### Step 2: Differentiate momentum to find force
The force \( \mathbf{F} \) acting on the particle is the rate of change of momentum:
\[ \mathbf{F} = \frac{d\mathbf{p}}{dt} \]
Now, we differentiate \( \mathbf{p} \):
\[ \frac{d\mathbf{p}}{dt} = \frac{d}{dt}(2 \cos t \hat{i} + 2 \sin t \hat{j}) \]
Using the derivatives of sine and cosine:
\[ \frac{d}{dt}(2 \cos t) = -2 \sin t \quad \text{and} \quad \frac{d}{dt}(2 \sin t) = 2 \cos t \]
Thus, the force vector \( \mathbf{F} \) becomes:
\[ \mathbf{F} = -2 \sin t \hat{i} + 2 \cos t \hat{j} \]

### Step 3: Calculate the dot product of \( \mathbf{F} \) and \( \mathbf{p} \)
Next, we calculate the dot product \( \mathbf{F} \cdot \mathbf{p} \):
\[ \mathbf{F} \cdot \mathbf{p} = (-2 \sin t \hat{i} + 2 \cos t \hat{j}) \cdot (2 \cos t \hat{i} + 2 \sin t \hat{j}) \]
Calculating the dot product:
\[ = (-2 \sin t)(2 \cos t) + (2 \cos t)(2 \sin t) \]
\[ = -4 \sin t \cos t + 4 \sin t \cos t = 0 \]

### Step 4: Determine the angle between \( \mathbf{F} \) and \( \mathbf{p} \)
The dot product of two vectors is given by:
\[ \mathbf{F} \cdot \mathbf{p} = |\mathbf{F}| |\mathbf{p}| \cos \theta \]
Since we found that \( \mathbf{F} \cdot \mathbf{p} = 0 \), it implies:
\[ |\mathbf{F}| |\mathbf{p}| \cos \theta = 0 \]
This means that \( \cos \theta = 0 \), which indicates that:
\[ \theta = 90^\circ \]

### Final Answer
The angle between the force \( F \) and the momentum \( p \) at time \( t \) is:
\[ \theta = 90^\circ \]

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