A particle of mass m is travelling with a constant velocity `v=v_(0)hati` along the line `y=b,z=0`. Let dA be the area swept out by the position vector from origin to the particle in time dt and L the magnitude of angular momentum of particle about origin at any time t. Then

Options

A. L=constant
B. `L ne "constant"`
C. `(dA)/(dt)=(2L)/(m)`
D. `(dA)/(dt)=(L)/(2m)`

Answer

To solve the problem step-by-step, we need to analyze the motion of the particle and calculate both the angular momentum \( L \) and the area swept out \( dA \) by the position vector from the origin to the particle.

### Step 1: Understanding the Motion of the Particle
The particle of mass \( m \) is moving with a constant velocity \( \vec{v} = v_0 \hat{i} \) along the line defined by \( y = b \) and \( z = 0 \). This means the particle is moving in the x-direction while maintaining a constant y-coordinate of \( b \).

### Step 2: Position Vector
The position vector \( \vec{r} \) of the particle at any time \( t \) can be expressed as:
\[ \vec{r} = x \hat{i} + b \hat{j} + 0 \hat{k} = v_0 t \hat{i} + b \hat{j} \]

### Step 3: Angular Momentum Calculation
The angular momentum \( \vec{L} \) of a particle about the origin is given by:
\[ \vec{L} = \vec{r} \times m\vec{v} \]
Where \( \vec{v} = v_0 \hat{i} \). The magnitude of angular momentum \( L \) can be calculated as:
\[ L = |\vec{r} \times m\vec{v}| \]
Calculating the cross product:
\[ \vec{r} = (v_0 t) \hat{i} + b \hat{j} \]
\[ m\vec{v} = mv_0 \hat{i} \]
The cross product \( \vec{r} \times m\vec{v} \) is:
\[ \vec{r} \times m\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_0 t & b & 0 \\ mv_0 & 0 & 0 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(b \cdot 0 - 0 \cdot 0) - \hat{j}(v_0 t \cdot 0 - 0 \cdot mv_0) + \hat{k}(v_0 t \cdot 0 - b \cdot mv_0) \]
\[ = -bmv_0 \hat{k} \]
Thus, the magnitude of angular momentum \( L \) is:
\[ L = |bmv_0| \]

### Step 4: Area Swept Out Calculation
The area \( dA \) swept out by the position vector in a small time \( dt \) can be calculated using the formula for the area of a triangle:
\[ dA = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the distance traveled in the x-direction \( v_0 dt \) and the height is the constant distance \( b \):
\[ dA = \frac{1}{2} \times v_0 dt \times b = \frac{1}{2} b v_0 dt \]

### Step 5: Relating Area Swept Out to Angular Momentum
From the previous results, we can relate \( dA \) to \( L \):
\[ \frac{dA}{dt} = \frac{1}{2} b v_0 \]
And since we have \( L = bmv_0 \), we can express \( dA \) in terms of \( L \):
\[ \frac{dA}{dt} = \frac{L}{2m} \]

### Final Results
1. The magnitude of angular momentum \( L \) is given by:
\[ L = bmv_0 \]
2. The area swept out in time \( dt \) is given by:
\[ dA = \frac{1}{2} b v_0 dt \]
3. The relationship between the area swept out and angular momentum is:
\[ \frac{dA}{dt} = \frac{L}{2m} \]