Question

A particle of mass $m=5\text{kg}$ is moving with a uniform speed $3\sqrt{2}\text{m/s}$ in $XY$ plane along the line $y=x+4$. The angular momentum about the origin is
[in ${\text{kg m}}^2\text{/s}$]

• A. $40$
• B. $60$
• C. zero
• D. $20$

Answer 1

The correct option is B $60$


Given,
Mass of particle $m=5\text{kg}$
Speed of particle $v=3\sqrt{2}\text{m/s}$
As we know, $L=mv{r}_{\perp }$
where $r_{\perp }$ is the shortest /perpendicular distance, of the particle from the origin.
As we know,
Perpendicular distance between a point and a line is given by,
$r_{\perp }=\left|\frac{A{x}_1+B{x}_2+C}{\sqrt{A^2+B^2}}\right|=\left|\frac{x-y+4}{\sqrt{1^2+(-1{)}^2}}\right|=\left|\frac{0-0+4}{\sqrt{2}}\right|$
(substituting $(x,y)=(0,0)$ for origin)
$r_{\perp }=2\sqrt{2}$

Therefore, from $L=mv{r}_{\perp }$
$L=5\times 3\sqrt{2}\times 2\sqrt{2}$
$L=60$${\text{kg m}}^2\text{/s}$