A particle of mass 0.75 kg is moving in the XY plane parallel to Y-axis, with a uniform speed of 4 m/s. It crosses the X-axis at 3 m from the origin. What is the angular momentum of the particle about the origin?

Options

A. `"3 kg m"^(2)//s`
B. `"9 kg m"^(2)//s`
C. `"6 kg m"^(2)//s`
D. `"1.5 kg m"^(2)//s`

Explanation

To find the angular momentum of a particle moving in the XY plane about the origin, we can use the formula for angular momentum (L):

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

where:
- \( L \) is the angular momentum,
- \( m \) is the mass of the particle,
- \( v \) is the linear velocity of the particle,
- \( r \) is the perpendicular distance from the line of motion to the point about which we are calculating the angular momentum.

### Step-by-Step Solution:

1. **Identify the given values**:
- Mass of the particle, \( m = 0.75 \) kg
- Speed of the particle, \( v = 4 \) m/s
- The distance from the origin to the point where the particle crosses the X-axis is \( r = 3 \) m.

2. **Determine the perpendicular distance**:
Since the particle is moving parallel to the Y-axis and crosses the X-axis at 3 m from the origin, the perpendicular distance \( r \) from the origin to the line of motion is simply the x-coordinate of the crossing point, which is 3 m.

3. **Calculate the angular momentum**:
Using the formula for angular momentum:
\[ L = m \cdot v \cdot r \]
Substitute the values:
\[ L = 0.75 \, \text{kg} \cdot 4 \, \text{m/s} \cdot 3 \, \text{m} \]

4. **Perform the multiplication**:
\[ L = 0.75 \cdot 4 \cdot 3 = 9 \, \text{kg m}^2/\text{s} \]

5. **Conclusion**:
The angular momentum of the particle about the origin is \( 9 \, \text{kg m}^2/\text{s} \).