A particle of mass $m = 5$ is moving with a uniform speed $v = 3\sqrt{2}$ in the XOY plane along the line $Y = X + 4$. The magnitude of the angular momentum of the particle about the origin is: (1991)

$60 \text{ units}$

$40\sqrt{2} \text{ units}$

Zero

$7.5 \text{ units}$

Solution:

Angular momentum $L = mvr_{\perp}$. The line equation is $X - Y + 4 = 0$.
The perpendicular distance $r_{\perp}$ from the origin $(0,0)$ to the line is $\frac{|0 - 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
Thus, $L = 5 \times (3\sqrt{2}) \times (2\sqrt{2}) = 60 \text{ units}$.

Rankers Physics

A particle of mass $m = 5$ is moving with a uniform speed $v = 3\sqrt{2}$ in the XOY plane along the line $Y = X + 4$. The magnitude of the angular momentum of the particle about the origin is: (1991)

Solution: