1. Given data

• Mass of particle: $m = 5\,\text{kg}$
• Speed: $v = 3\sqrt{2}\,\text{m s}^{-1}$
• Path of motion: $y = x + 4$

We need the angular momentum about the origin.

2. Formula for angular momentum magnitude

For a particle moving along a straight line, the magnitude of angular momentum about the origin is

$L = m v d$

where $d$ is the perpendicular distance of the origin from the line of motion.

3. Find perpendicular distance of origin from the line

Given line:

$y = x + 4$

Rewrite in standard form:

$x - y + 4 = 0$

Distance of point $(0,0)$ from line $ax + by + c = 0$ is

$d = \frac{|a(0) + b(0) + c|}{\sqrt{a^2+b^2}}$

Here, $a=1$, $b=-1$, $c=4$. So,

$d = \frac{|4|}{\sqrt{1^2+(-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\,\text{m}$

4. Compute angular momentum

$L = mvd$

$L = 5 \times 3\sqrt{2} \times 2\sqrt{2}$

$L = 5 \times 3 \times 2 \times 2$

$L = 60\,\text{kg m}^2\text{s}^{-1}$

5. Final answer

$\boxed{60}$

This is the magnitude of angular momentum about the origin.