A particle of mass `m=5` units is moving with a uniform speed `v = 3 sqrt(2)` units in the XY-plane along the `y=x+4`. The magnitude of the angular momentum about origin is

Options

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

Explanation

To find the magnitude of the angular momentum of a particle moving in the XY-plane, we can follow these steps:

### Step 1: Identify the parameters
We are given:
- Mass of the particle, \( m = 5 \) units
- Speed of the particle, \( v = 3\sqrt{2} \) units
- The line along which the particle is moving is given by the equation \( y = x + 4 \).

### Step 2: Determine the slope and angle of the line
The equation of the line can be rewritten in the slope-intercept form \( y = mx + c \), where:
- \( m = 1 \) (slope)
- \( c = 4 \) (y-intercept)

The angle \( \theta \) that the line makes with the x-axis can be found using:
\[ \tan(\theta) = m = 1 \implies \theta = 45^\circ \]

### Step 3: Find the position vector of the particle
To find the position vector \( \vec{r} \) of the particle, we can choose a point on the line. For instance, when \( x = 0 \):
\[ y = 0 + 4 = 4 \implies \text{Point} = (0, 4) \]
Thus, the position vector \( \vec{r} \) from the origin to the point is:
\[ \vec{r} = (0, 4) \]

### Step 4: Determine the momentum of the particle
The momentum \( \vec{p} \) of the particle is given by:
\[ \vec{p} = m \vec{v} \]
The velocity vector \( \vec{v} \) can be determined from the speed and direction. Since the particle moves at an angle of \( 45^\circ \):
\[ \vec{v} = v(\cos(45^\circ), \sin(45^\circ)) = 3\sqrt{2}\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = (3, 3) \]
Thus, the momentum is:
\[ \vec{p} = m \vec{v} = 5(3, 3) = (15, 15) \]

### Step 5: Calculate the angular momentum
The angular momentum \( \vec{L} \) about the origin is given by the cross product:
\[ \vec{L} = \vec{r} \times \vec{p} \]
In 2D, the magnitude of the angular momentum can be calculated as:
\[ L = r p \sin(\phi) \]
where \( \phi \) is the angle between \( \vec{r} \) and \( \vec{p} \). The angle \( \phi \) can be found as follows:
- The angle between the position vector \( \vec{r} = (0, 4) \) and the momentum vector \( \vec{p} = (15, 15) \) is \( 45^\circ \).

Thus,
\[ L = |\vec{r}| |\vec{p}| \sin(45^\circ) \]

### Step 6: Calculate the magnitudes
The magnitude of \( \vec{r} \):
\[ |\vec{r}| = \sqrt{0^2 + 4^2} = 4 \]
The magnitude of \( \vec{p} \):
\[ |\vec{p}| = \sqrt{15^2 + 15^2} = \sqrt{450} = 15\sqrt{2} \]
Now substituting these values:
\[ L = 4 \cdot 15\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 4 \cdot 15 = 60 \text{ units} \]

### Final Answer
The magnitude of the angular momentum about the origin is \( 60 \) units.

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