An axe of mass $m$ is swung with kinetic energy $E$ directly into a wooden block, also of mass $m$. In a first trial the block is rigidly clamped to the ground, and the axe buries a depth $L$ into the wood. In a second trial the block is identical but rests on a frictionless surface, free to slide, and the axe sticks where it comes to rest relative to the wood. To what depth is the axe buried in the second trial? Assume the axe is wedge-shaped with a constant angle and that the force per unit contact area between the axe and the wood during the impact is proportional to the depth of penetration.