The momentum of a body is `vec P =2 cos t hati +2 sin t hatj.` What is the angle between the force `vec F` acting on the body and the momentum `vec P` ?

Options

A. `90^(@)`
B. `45^(@)`
C. `135^(@)`
D. `120^(@)`

Explanation

To find the angle between the force \(\vec{F}\) acting on a body and its momentum \(\vec{P}\), we will follow these steps:

### Step 1: Identify the momentum vector
The momentum of the body is given as:
\[ \vec{P} = 2 \cos t \, \hat{i} + 2 \sin t \, \hat{j} \]

### Step 2: Differentiate the momentum vector to find the force
The force \(\vec{F}\) is defined as the rate of change of momentum:
\[ \vec{F} = \frac{d\vec{P}}{dt} \]
Differentiating \(\vec{P}\) with respect to time \(t\):
\[ \vec{F} = \frac{d}{dt}(2 \cos t \, \hat{i} + 2 \sin t \, \hat{j}) = -2 \sin t \, \hat{i} + 2 \cos t \, \hat{j} \]

### Step 3: Calculate the dot product of \(\vec{P}\) and \(\vec{F}\)
The dot product \(\vec{P} \cdot \vec{F}\) is given by:
\[ \vec{P} \cdot \vec{F} = (2 \cos t)(-2 \sin t) + (2 \sin t)(2 \cos t) \]
Calculating this:
\[ \vec{P} \cdot \vec{F} = -4 \cos t \sin t + 4 \sin t \cos t = 0 \]

### Step 4: Find the magnitudes of \(\vec{P}\) and \(\vec{F}\)
The magnitude of \(\vec{P}\) is:
\[ |\vec{P}| = \sqrt{(2 \cos t)^2 + (2 \sin t)^2} = \sqrt{4 \cos^2 t + 4 \sin^2 t} = \sqrt{4(\cos^2 t + \sin^2 t)} = \sqrt{4} = 2 \]

The magnitude of \(\vec{F}\) is:
\[ |\vec{F}| = \sqrt{(-2 \sin t)^2 + (2 \cos t)^2} = \sqrt{4 \sin^2 t + 4 \cos^2 t} = \sqrt{4(\sin^2 t + \cos^2 t)} = \sqrt{4} = 2 \]

### Step 5: Use the dot product to find the angle \(\theta\)
The cosine of the angle \(\theta\) between the two vectors is given by:
\[ \cos \theta = \frac{\vec{P} \cdot \vec{F}}{|\vec{P}| |\vec{F}|} \]
Substituting the values we found:
\[ \cos \theta = \frac{0}{2 \cdot 2} = 0 \]

### Step 6: Determine the angle \(\theta\)
Since \(\cos \theta = 0\), we have:
\[ \theta = 90^\circ \]
Thus, the angle between the force \(\vec{F}\) and the momentum \(\vec{P}\) is \(90^\circ\).

### Final Answer
The angle between the force \(\vec{F}\) and the momentum \(\vec{P}\) is \(90^\circ\).

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