A 15 g bullet moving at 350 m/s goes 6.0 cm into a block of wood and stops.

Step 1. Kinetic energy. $\tfrac12 \times 0.015\ \text{kg} \times (350\ \text{m/s})^2 = 918.75\ \text{J}$.

Step 2. Force. $F = \dfrac{918.75\ \text{J}}{0.060\ \text{m}} = 15312\ \text{N}$, about $1.5 \times 10^{4}\ \text{N}$.

Step 3. The thin plank. It takes $15312\ \text{N} \times 0.025\ \text{m} = 382.8\ \text{J}$, leaving $918.75\ \text{J} - 382.8\ \text{J} = 535.9\ \text{J}$.

Step 4. Speed out. $v = \sqrt{\dfrac{2 \times 535.9\ \text{J}}{0.015\ \text{kg}}} = 267\ \text{m/s}$.