I think you mean that they are "denser" than I figure (i.e., not just "more mass" but "more mass per cubic inch"). But even if I assumed they had the density of solid lead, I'm still not picturing them getting far into the soil. I did a little googling and found that a bullet will penetrate about 16 inches into dry sand at maximum. I don't know if you can scale that up for the FDR, but I'd be interested in the maths that shows a radically different result.
If I had a suspicion about all this it would be that the 25' figure was the total depth to which they were required to dig away the site to make sure that they got everything, and that this figure got attached to the FDRs by the press because it's a dramatic number ("Wow, they buried that far!").
No, I really did mean "more mass." A lead bullet won't penetrate more than 16 inches, perhaps, but a lead cannonball fired with the exact same muzzle velocity will go quite a bit further than that.
When the projectile (in this case, a Flight Data Recorder) plows through a yielding solid (in this case, dirt), it's more of a drag effect, with the
surface area rather than the volume determining how quickly kinetic energy is dissipated. As objects scale up, the volume (scales as the third power of diameter) grows faster than the surface area (scales as the second power), so a scaled-up object will experience less resistance per unit mass, and thus penetrate farther.
To put hard numbers on it, take a real FDR, assume it's a cube 20 cm on a side, and weighs 8 kilograms total. Its volume is 20cm x 20cm x 20cm = 8,000 cubic centimeters, thus it has a density of 1 gram per cubic centimeter.
Now scale that FDR down to dollhouse proportions, 1cm on a side, but with the same density. You have an FDR with volume 1cm x 1cm x 1cm = 1 cubic centimeter, and thus weighs 1 gram.
Now accelerate both to the impact speed, roughly 260 meters per second. Which one will penetrate farther? Well, the mini-FDR has only 1cm x 1cm of cross-sectional area, or 1 square centimeter, through which you're pushing 1 gram x (260 meters / second)
2 of kinetic energy, or 67.6 Joules. That's 67.6 Joules of energy per square centimeter of crater to dissipate in the dirt.
The big FDR has an area of 20cm x 20cm = 400 square centimeters, so it makes a hole 400 times bigger. But its energy is 8,000 grams x (260 meters / second)
2 of kinetic energy, or 540,800 Joules. That's 1,352 Joules per square centimeter of crater. Consequently, it tunnels a lot farther.
Hope that made sense. Yes, it is more complex than that, since we don't punch a square hole in the ground, and the FDR will deform as well, but the basic concept is correct.
Also, it isn't clear just how they measured recovery depth. It could have been 25 feet below the level plain, but only a couple of feet below the bottom of the crater. We really need a more detailed report to investigate much further.
ETA: Grammar c.f. further/farther. If it's a distance, it's "farther." Otherwise, it's "further." I knew that before I posted, honest.