A particle of mass `m=5 kg` is moving with a uniform speed `v=3 sqrt(2)` in the `XOY` plane along the line `Y=X+4`. The magnitude of the angular momentum of the particle about the origin is

Options

A. `60` unit
B. `40 sqrt(2)` unit
C. zero
D. `7.5` unit

Answer

To find the magnitude of the angular momentum of a particle about the origin, we can follow these steps:

### Step 1: Identify the parameters
Given:
- Mass of the particle, \( m = 5 \, \text{kg} \)
- Speed of the particle, \( v = 3\sqrt{2} \, \text{m/s} \)
- The line along which the particle moves is given by \( y = x + 4 \).

### Step 2: Determine the position vector \( \mathbf{r} \)
The line \( y = x + 4 \) can be rewritten in terms of coordinates:
- When \( x = 0 \), \( y = 4 \) (point (0, 4))
- When \( y = 0 \), \( x = -4 \) (point (-4, 0))

The position vector \( \mathbf{r} \) from the origin to the particle can be represented as:
\[ \mathbf{r} = (x, y) = (x, x + 4) \]

### Step 3: Find the velocity vector \( \mathbf{v} \)
The particle moves along the line with a uniform speed. The slope of the line is 1, which means the velocity components can be determined as follows:
- The angle \( \theta \) of the line with respect to the x-axis is \( 45^\circ \).
- The components of the velocity vector can be calculated using:
\[ v_x = v \cos(45^\circ) = 3\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 3 \, \text{m/s} \]
\[ v_y = v \sin(45^\circ) = 3\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 3 \, \text{m/s} \]
Thus, the velocity vector is:
\[ \mathbf{v} = (3, 3) \]

### Step 4: Calculate the angular momentum \( \mathbf{L} \)
The angular momentum \( \mathbf{L} \) about the origin is given by:
\[ \mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times (m \mathbf{v}) \]
Where \( \mathbf{p} = m \mathbf{v} \) is the linear momentum. The magnitude of angular momentum can be calculated using:
\[ L = m r v \sin(\alpha) \]
Where \( \alpha \) is the angle between \( \mathbf{r} \) and \( \mathbf{v} \).

### Step 5: Finding \( r \) and \( \alpha \)
The distance \( r \) from the origin to the point on the line can be calculated using the y-intercept:
\[ r = 4 \quad (\text{the y-intercept of the line}) \]
The angle \( \alpha \) between \( \mathbf{r} \) and \( \mathbf{v} \) is \( 90^\circ + 45^\circ = 135^\circ \).

### Step 6: Calculate \( L \)
Using the sine of the angle:
\[ \sin(135^\circ) = \sin(180^\circ - 45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
Now substituting the values into the angular momentum formula:
\[ L = m \cdot r \cdot v \cdot \sin(\alpha) = 5 \cdot 4 \cdot (3\sqrt{2}) \cdot \left(\frac{1}{\sqrt{2}}\right) \]
\[ L = 5 \cdot 4 \cdot 3 = 60 \, \text{kg m}^2/\text{s} \]

### Final Answer
The magnitude of the angular momentum of the particle about the origin is \( L = 60 \, \text{kg m}^2/\text{s} \).

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