The velocity `vecv` of a particle of mass 'm' acted upon by a constant force is given by `vecv (t) = A[cos (kt) bari - sin (kt) barj]` . Then the angle between the force and the momentum of the particle is (Here A and k are constants)

Options

A. `90^(@)`
B. `0^(@)`
C. `180^(@)`
D. `45^(@)`

Answer

To solve the problem, we need to find the angle between the force acting on the particle and the momentum of the particle. Let's break this down step by step.

### Step 1: Write down the expression for velocity
The velocity of the particle is given by:
\[ \vec{v}(t) = A \left[ \cos(kt) \hat{i} - \sin(kt) \hat{j} \right] \]

### Step 2: Find the momentum of the particle
The momentum \(\vec{p}\) of the particle is given by:
\[ \vec{p}(t) = m \vec{v}(t) = m A \left[ \cos(kt) \hat{i} - \sin(kt) \hat{j} \right] \]

### Step 3: Differentiate momentum to find force
The force \(\vec{F}\) acting on the particle is the time derivative of momentum:
\[ \vec{F} = \frac{d\vec{p}}{dt} \]
Calculating the derivative:
\[ \vec{F} = \frac{d}{dt} \left( m A \left[ \cos(kt) \hat{i} - \sin(kt) \hat{j} \right] \right) \]
Using the chain rule:
\[ \vec{F} = m A \left[ -k \sin(kt) \hat{i} - k \cos(kt) \hat{j} \right] \]
Thus, we can write:
\[ \vec{F} = -m A k \left[ \sin(kt) \hat{i} + \cos(kt) \hat{j} \right] \]

### Step 4: Calculate the dot product of force and momentum
Next, we need to find the dot product \(\vec{F} \cdot \vec{p}\):
\[ \vec{F} \cdot \vec{p} = \left(-m A k \left[ \sin(kt) \hat{i} + \cos(kt) \hat{j} \right] \right) \cdot \left(m A \left[ \cos(kt) \hat{i} - \sin(kt) \hat{j} \right] \right) \]
Calculating the dot product:
\[ \vec{F} \cdot \vec{p} = -m^2 A^2 k \left[ \sin(kt) \cos(kt) - \cos(kt) \sin(kt) \right] = 0 \]

### Step 5: Find the magnitudes of force and momentum
The magnitudes of \(\vec{F}\) and \(\vec{p}\) are:
\[ |\vec{F}| = m A k \sqrt{\sin^2(kt) + \cos^2(kt)} = m A k \]
\[ |\vec{p}| = m A \sqrt{\cos^2(kt) + \sin^2(kt)} = m A \]

### Step 6: Calculate the angle between force and momentum
The angle \(\theta\) between the two vectors can be found using the formula:
\[ \cos(\theta) = \frac{\vec{F} \cdot \vec{p}}{|\vec{F}| |\vec{p}|} \]
Since \(\vec{F} \cdot \vec{p} = 0\):
\[ \cos(\theta) = \frac{0}{(m A k)(m A)} = 0 \]
Thus, \(\theta = \cos^{-1}(0) = 90^\circ\).

### Final Answer
The angle between the force and the momentum of the particle is \(90^\circ\).

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