Question

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

• A. Zero
• B. 60 units
• C. 7.5 units
• D. \(40\sqrt{2}\) units

Hint:

For a particle moving in a straight line:
Angular momentum about a point depends only on the perpendicular distance of that point from the line of motion
Formula: \(L = m v r_\perp\)
If the line passes through the origin, angular momentum is zero

Answer 1

The Correct Option is B

Step 1: Recall the formula for angular momentum. The magnitude of angular momentum of a particle about the origin is: \[ L = m v r_\perp \] where \(r_\perp\) is the perpendicular distance of the origin from the line of motion.
Step 2: Find the perpendicular distance from origin to the line of motion. Given line: \[ y = x + 4 \;\;\Rightarrow\;\; x - y + 4 = 0 \] Distance of origin \((0,0)\) from this line: \[ r_\perp = \frac{|0 - 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]
Step 3: Substitute given values. \[ m = 5,\quad v = 3\sqrt{2},\quad r_\perp = 2\sqrt{2} \] \[ L = (5)(3\sqrt{2})(2\sqrt{2}) \] \[ L = 5 \times 3 \times 2 \times 2 = 60 \] Final Answer: \[ \boxed{L = 60\ \text{units}} \]