A particle moves with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin

Options

A. is zero
B. remains constant
C. goes on increasing
D. goes on decreasing

Answer

To solve the problem of determining the behavior of a particle's angular momentum with respect to the origin while it moves with a constant velocity parallel to the X-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Angular Momentum**:
The angular momentum \( L \) of a particle with respect to a point (in this case, the origin) is given by the formula:
\[ L = r \times p \]
where \( r \) is the position vector from the origin to the particle, and \( p \) is the linear momentum of the particle. The linear momentum \( p \) is defined as:
\[ p = mv \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

2. **Set Up the Scenario**:
Since the particle is moving with a constant velocity parallel to the X-axis, we can denote its velocity as \( v = (v_x, 0) \), where \( v_x \) is a constant value. The position vector \( r \) of the particle at any time \( t \) can be expressed as:
\[ r = (x(t), y(t)) = (v_x t, 0) \]

3. **Calculate Angular Momentum**:
The angular momentum \( L \) can be calculated using the cross product of the position vector \( r \) and the momentum \( p \):
\[ L = r \times p = r \times (mv) \]
Since \( p = mv = (mv_x, 0) \), we can express the angular momentum as:
\[ L = (v_x t, 0) \times (mv_x, 0) \]
The cross product of two vectors that are parallel (or one of them is zero) is zero. Therefore, the angular momentum \( L \) will be:
\[ L = 0 \]

4. **Conclusion**:
Since the particle is moving parallel to the X-axis and the angular momentum calculated with respect to the origin is zero, we conclude that the angular momentum remains constant (in this case, it remains zero).

### Final Answer:
The angular momentum of the particle with respect to the origin is **0**.