Question

A particle of mass $10$ kg is moving with a uniform speed of $6$ m/sec. In x-y plane along the line $3y=4x+10$, what is the magnitude of its angular momentum about the origin in $kg-m^2/s$?

Answer 1

$3y=4x+10$
$\Rightarrow$ slope $m=\frac{4}{3}=\tan \theta$
$\Rightarrow \cos \theta =\frac{3}{5}$;
$\Rightarrow \sin \theta =\frac{4}{5}$;
$|\overrightarrow{L}|=mv\cos \theta \times \perp distance=10\times 6\times \frac{3}{5}\times \frac{10}{3}=120kg{m}^2/s$