Solution:

The angular momentum $L$ of a particle about the origin is given by:

$$L = m v r \sin\theta$$

where:

• $m = 5$ g (mass of the particle),
• $v = 3\sqrt{2}$ cm/s (speed of the particle),
• $r = 2\sqrt{5}$ cm (perpendicular distance from the origin),
• $\theta = 90^\circ$ (since the velocity is along a straight line parallel to the x-axis, the perpendicular distance is directly used).

Since $\sin 90^\circ = 1$, the equation simplifies to:

$$L = m v r$$

Substituting the given values:

$$L = (5) \times (3\sqrt{2}) \times (2\sqrt{5})$$ $$L = 5 \times 3 \times 2 \times \sqrt{10}$$ $$L = 30 \sqrt{10} \text{ g-cm²/s}$$

Thus, the magnitude of the angular momentum is:

$$\mathbf{30\sqrt{10} \text{ g-cm²/s}}$$