Question

A particle is moving along a line $y=x+a$ with a constant velocity $v$. Find the angular momentum of the particle about the origin.

• A. $mva$
• B. $mva\sqrt{2}$
• C. $\frac{mva}{\sqrt{2}}$
• D. $\frac{mvy}{x\sqrt{2}}$

Answer 1

The correct option is C $\frac{mva}{\sqrt{2}}$

Step-1: Draw a rough diagram of path of particle.
Given: path equation $y=x+a$
Comparing with equation $y=mx+c$
We get $m=1,c=a$
So, angle with $x$ axis is ${45}^{\circ }$ and it intersects $y$ axis at $y=a$
Step-2 : Find angular momentum of particle about origin
Formula used: $L=mvr$
$r$ is perpendicular distance between line of motion and origin
From figure $r=a\cos {45}^{\circ }=\frac{a}{\sqrt{2}}$
$L=\frac{mva}{\sqrt{2}}$
Final answer (c)