The momentum of a moving particle is vectorially given a, `vecp=p_(0)(costhati+sinthatj)`where `t` stands for time. Choose the correct option:

Options

A. The applied force is constant.
B. The momentum is constant.
C. The applied force always remains perpendicular to the momentum.
D. The applied force is always parallel to the momentum.

Explanation

To solve the problem, we need to analyze the momentum vector given by:

\[ \vec{p} = p_0 (\cos t \hat{i} + \sin t \hat{j}) \]

where \( p_0 \) is a constant and \( t \) represents time.

### Step 1: Understand the momentum vector
The momentum vector \(\vec{p}\) is expressed in terms of time \(t\). The components of the momentum in the \(x\) and \(y\) directions are \(p_0 \cos t\) and \(p_0 \sin t\), respectively.

### Step 2: Differentiate momentum to find force
According to Newton's second law, the force \(\vec{F}\) is the time derivative of momentum:

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

### Step 3: Differentiate each component of momentum
We will differentiate \(\vec{p}\):

\[ \vec{F} = \frac{d}{dt}(p_0 \cos t \hat{i} + p_0 \sin t \hat{j}) \]

Using the chain rule, we differentiate each component:

\[ \vec{F} = p_0 \left(-\sin t \hat{i} + \cos t \hat{j}\right) \]

### Step 4: Analyze the force
The force vector is given by:

\[ \vec{F} = -p_0 \sin t \hat{i} + p_0 \cos t \hat{j} \]

Since the force depends on \(t\) (due to the sine and cosine functions), we conclude that the force is **not constant**.

### Step 5: Check if momentum is constant
Since \(\vec{p}\) depends on \(t\) (as it contains \(\cos t\) and \(\sin t\)), we can conclude that the momentum is also **not constant**.

### Step 6: Check if force is perpendicular to momentum
To check if the force is always perpendicular to the momentum, we need to verify if the dot product of \(\vec{F}\) and \(\vec{p}\) is zero:

\[ \vec{F} \cdot \vec{p} = (-p_0 \sin t \hat{i} + p_0 \cos t \hat{j}) \cdot (p_0 \cos t \hat{i} + p_0 \sin t \hat{j}) \]

Calculating the dot product:

\[ = (-p_0 \sin t)(p_0 \cos t) + (p_0 \cos t)(p_0 \sin t) \]
\[ = -p_0^2 \sin t \cos t + p_0^2 \cos t \sin t = 0 \]

Since the dot product is zero, the force is indeed **perpendicular** to the momentum.

### Conclusion
Based on the analysis:
1. The applied force is not constant.
2. The momentum is not constant.
3. The applied force is always perpendicular to the momentum.

Thus, the correct option is that the applied force always remains perpendicular to the momentum.