A particle of mass `80` units is moving with a uniform speed `v=4 sqrt 2` units in `XY` plane, along a line `y=x+5`. The magnitude of the angular momentum of the particle about the origin is

Options

A. `1600` units
B. `160sqrt(2)` units
C. `152sqrt(2)` units
D. `16sqrt(2)` units

Explanation

To find the magnitude of the angular momentum of a particle about the origin, we can use the formula:

\[ L = m \cdot v \cdot r_p \]

where:
- \( L \) is the angular momentum,
- \( m \) is the mass of the particle,
- \( v \) is the speed of the particle,
- \( r_p \) is the perpendicular distance from the line of motion to the origin.

### Step 1: Identify the given values
- Mass of the particle, \( m = 80 \) units
- Speed of the particle, \( v = 4\sqrt{2} \) units
- The line of motion is given by \( y = x + 5 \).

### Step 2: Determine the perpendicular distance \( r_p \)
The equation of the line is \( y = x + 5 \). To find the perpendicular distance from the origin (0,0) to this line, we can use the formula for the distance from a point to a line given by \( Ax + By + C = 0 \):

\[ r_p = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For the line \( y = x + 5 \), we can rewrite it in the form \( Ax + By + C = 0 \):

\[ x - y + 5 = 0 \quad \Rightarrow \quad A = 1, B = -1, C = 5 \]

Substituting \( (x_0, y_0) = (0, 0) \):

\[ r_p = \frac{|1(0) - 1(0) + 5|}{\sqrt{1^2 + (-1)^2}} = \frac{|5|}{\sqrt{2}} = \frac{5}{\sqrt{2}} \]

### Step 3: Calculate the angular momentum \( L \)
Now substituting the values into the angular momentum formula:

\[ L = m \cdot v \cdot r_p = 80 \cdot (4\sqrt{2}) \cdot \left(\frac{5}{\sqrt{2}}\right) \]

### Step 4: Simplify the expression
\[ L = 80 \cdot 4 \cdot 5 = 1600 \text{ units} \]

### Conclusion
The magnitude of the angular momentum of the particle about the origin is:

\[ \boxed{1600} \text{ units} \]

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