[jluetjen]

I'd say the rubber bullet since I would expect the lead bullet more likely to travel through the wood. I think (I'm a salesman, not a physicist) the the critical parameter is the density of the bullet, since this will determine the surface pressue at any given time during the impact.

1) Both bullets will have the same energy when leaving the barrel (in a simplified world which ignores barrel friction and bullet deformation within the barrel)

2) Since both bullets have the same mass, the lead bullet would be smaller since lead is denser then rubber, as a result the friction generated as the bullet travels through the air (assuming the same shape) will largely be a function of the frontal area and surface area, both of which would be larger with the rubber bullet. So by the time the projectiles reach the wood, the rubber bullet would have expended some of it's kinetic energy being heated by the air (as well as heating the air).

3) If the energy level is high enough, both bullets could just punch through the wood without knocking it over, just like the old Marx Brothers' trick where they pull the table cloth out from underneath a fully set table. (Heck! A car could even fly 200 feet and into a 2nd story appartment if the energy level was high enough -- but I digress..) But just thinking intuitively, I would expect that in most cases the rubber bullet would be more likely to knock the wood over.

4) Not to mention the surface drag as the bullet travels through the wood. Rubber is generally quite grippy when compared to a smooth metal, so once again, this would tend to increase the friction (compared to the lead bullet) as the rubber bullet travels through the wood, increasing the likelyhood that the rubber bullet would pull the wood over.