2. Given that $a$ and $b$ are positive numbers, and $2a + b = 2$. Then the minimum value of $\sqrt{4a^2 + 1} + \sqrt{b^2 + 4}$ is $\qquad$
2. $\sqrt{13}$.

As shown in Figure 6, construct a line segment $AB=2$, with points $C$ and $D$ on opposite sides of $AB$, and $CA \perp AB, DB \perp AB, CA=1, BD=2$, and $E$ is a point on $AB$.

Let $AE=2a, BE=b$. Connect $CE, DE, CD$.
By the Pythagorean theorem, we have
$$
CE=\sqrt{4a^{2}+1}, DE=\sqrt{b^{2}+4}
$$

Since $CE + DE \geqslant CD$, we have
$$
\sqrt{4a^{2}+1} + \sqrt{b^{2}+4} \geqslant CD.
$$

When points $C, E, D$ are collinear, $\sqrt{4a^{2}+1} + \sqrt{b^{2}+4}$ reaches its minimum value, which is $CD$. At this time, $\triangle ACE \sim \triangle BDE$. Therefore, $\frac{AE}{BE} = \frac{AC}{BD}$, which means $\frac{2a}{b} = \frac{1}{2}$.
Thus, $b = 4a$, substituting this into $2a + b = 2$, we get $a = \frac{1}{3}$. Hence, $b = \frac{4}{3}$.
When $a = \frac{1}{3}, b = \frac{4}{3}$, $CD = \sqrt{13}$.
Therefore, the minimum value of $\sqrt{4a^{2}+1} + \sqrt{b^{2}+4}$ is $\sqrt{13}$.