A particle of mass m = 5 units is moving with a uniform speed $\mathrm{v}=3\sqrt{2}$, units in the XOY plane along the line y = x + 4. The magnitude of the angular momentum of the Particle about the origin is

Momentum of the particle,
$\mathrm{mV}=5\times \left(3\sqrt{2}\right)=15\sqrt{2}$
The direction of momentum in XOY plane is given by
y=x+4
The slope of the line is
$1=\tan \mathrm{\theta } \mathrm{or} \mathrm{\theta }={45}^{\circ }$
Intercept of the line =4
Length of the perpendicular z from the origin on straight line = 4sin 45°
$4/\sqrt{2}=2\sqrt{2}$
$\therefore$ Angular momentum = (m v) x z
$=\left(15\sqrt{2}\right)\times \left(2\sqrt{2}\right)=60 \mathrm{units}$