A particle of mass 1 kg is moving along the line `y = x + 2` (here x and y are in metres) with speed `2 m//s`. The magnitude of angular momentum of paticle about origin is -

Options

A. `4 kg - m^(2)//s`
B. `2sqrt(2)kg - m^(2)//s`
C. `4sqrt(2) kg-m^(2)//s`
D. `2kg - m^(2)//s`

Explanation

To find the magnitude of the angular momentum of a particle of mass 1 kg moving along the line \( y = x + 2 \) with a speed of \( 2 \, \text{m/s} \) about the origin, we can follow these steps:

### Step 1: Determine the position of the particle
The equation of the line is given as \( y = x + 2 \). We can express the position of the particle in terms of \( x \):
- Let \( x = t \) (where \( t \) is time).
- Then, \( y = t + 2 \).

Thus, the position vector \( \vec{r} \) of the particle at any time \( t \) can be written as:
\[ \vec{r} = (t, t + 2) \]

### Step 2: Calculate the velocity of the particle
The velocity \( \vec{v} \) of the particle is given as \( 2 \, \text{m/s} \). Since the particle is moving along the line, we can find the components of the velocity. The slope of the line \( y = x + 2 \) is 1, which means the particle moves equally in both \( x \) and \( y \) directions. Therefore, we can express the velocity as:
\[ \vec{v} = (v_x, v_y) = \left( \frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}} \right) = (1, 1) \]

### Step 3: Find the perpendicular distance from the origin to the line
To find the perpendicular distance \( R \) from the origin to the line \( y = x + 2 \), we can use the formula for the distance from a point to a line given by \( Ax + By + C = 0 \):
\[ \text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( y = x + 2 \), we can rewrite it as:
\[ x - y + 2 = 0 \]
Here, \( A = 1, B = -1, C = 2 \), and the point is the origin \( (0, 0) \):
\[ \text{Distance} = \frac{|1(0) - 1(0) + 2|}{\sqrt{1^2 + (-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 4: Calculate the angular momentum
The angular momentum \( L \) about the origin is given by the formula:
\[ L = m \cdot v \cdot R \]
Where:
- \( m = 1 \, \text{kg} \)
- \( v = 2 \, \text{m/s} \)
- \( R = \sqrt{2} \)

Substituting the values:
\[ L = 1 \cdot 2 \cdot \sqrt{2} = 2\sqrt{2} \, \text{kg m}^2/\text{s} \]

### Final Answer
The magnitude of the angular momentum of the particle about the origin is:
\[ L = 2\sqrt{2} \, \text{kg m}^2/\text{s} \]