A bullet with a mass of 20 g travelling horizontally at a speed of...

Given mass of the bullet $m = 20 \text{ g} = 0.02 \text{ kg}$ and its initial velocity $u = 300 \text{ m/s}$.

Mass of the wooden block $M = 500 \text{ g} = 0.5 \text{ kg}$.

Let $v$ be the velocity of the bullet as it emerges and $V$ be the velocity of the block just after the bullet passes through it.

By conservation of mechanical energy for the block after the collision:

$\dfrac{1}{2} M V^2 = M g h$

$V = \sqrt{2 g h}$

Substituting $g = 10 \text{ m/s}^2$ and $h = 0.2 \text{ m}$:

$V = \sqrt{2 \times 10 \times 0.2} = 2 \text{ m/s}$

By conservation of linear momentum during the collision:

$m u = m v + M V$

$0.02 \times 300 = 0.02 \times v + 0.5 \times 2$

$6 = 0.02 v + 1$

$0.02 v = 5$

$v = \dfrac{5}{0.02} = 250 \text{ m/s}$