Question

A particle of mass \[ m=0.25\ \text{kg} \] is moving along a straight line parallel to the \(x\)-axis with a constant velocity \[ v=5\ \text{m s}^{-1} \] as shown in the figure. What is the angular momentum of the particle with respect to the origin?

• A. \(0\)
• B. \(1.25\ \text{kg m}^2\text{s}^{-1}\) along \(+z\)-axis
• C. \(1.25\ \text{kg m}^2\text{s}^{-1}\) along \(-z\)-axis
• D. \(1.25\ \text{kg m}^2\text{s}^{-1}\) along \(x\)-axis

Hint:

For a particle moving in a straight line, \[ L = p \times (\text{perpendicular distance from origin to line of motion}). \] Use the right-hand rule to determine the direction.

Answer 1

The Correct Option is C

Step 1: Use the cross product idea for angular momentum.
About the origin, $\vec L = \vec r \times \vec p$. Its size is $L = p\,d$, where $d$ is the perpendicular distance from the origin to the line of motion.
Step 2: Find the linear momentum.
$p = mv = 0.25 \times 5 = 1.25$ kg m per s.
Step 3: Note why only the perpendicular distance matters.
The particle moves on a straight line parallel to the $x$-axis at a fixed height. As it moves, the part of $\vec r$ along the motion contributes nothing to the cross product, so only the steady perpendicular gap $d$ counts.
Step 4: Read the distance from the figure.
The line sits at a perpendicular distance $d = 1.0$ m from the origin.
Step 5: Multiply to get the magnitude.
\[ L = p\,d = 1.25 \times 1.0 = 1.25\ \text{kg m}^2\text{s}^{-1} \]
Step 6: Fix the direction with the right-hand rule.
With $\vec r$ pointing up-left and the velocity along $+x$, the cross product points into the page, that is along the $-z$ axis.
\[ \boxed{L = 1.25\ \text{kg m}^2\text{s}^{-1}\ \text{along } -z} \]