After This, Veritasium Will Saw a Woman in Half
Some physics sleight-of-hand. I like it.
I have to admit, my first instinct was that the rotating block should not have gone as high, but I quickly realized why this was wrong. I don’t know if the folks in the video who are truly puzzled or acting that way for the sake of the video, because from other science videos I’ve seen, they are smart cookies.
There is not any extra energy that has to be “created” for the rotating case, because this is a completely inelastic collision. Kinetic energy is not a conserved quantity (!), total energy is. KE is only conserved in elastic collisions. There is more kinetic energy in the rotating case, and that’s just fine! And that’s the misdirection — everyone is puzzling over a quantity that’s not constant, implying that it should be. But this is why it’s a great demo — you have to identify and avoid a misconception to get the answer.
So it’s possible they are just playing to the audience to try and get you to think something mystical is going on here, or they “got it” soon after the first reaction was taped.
Now, what is conserved in these interactions is momentum, both linear and angular, since at the moment of impact there are no external forces or torques we have to worry about; we ignore gravity, since the effect during the short time of impact is very, very small. The impulse would be mgt, where t is the time of impact, and that’s small compared to the momentum of a fast-moving projectile. We’re physicists, and we set small effects to be zero. Too messy otherwise.
In the straight-on case, we have a linear momentum p, and momentum and kinetic energy have a simple relationship: KE = p^2/2m. So the kinetic energy is dictated by the momentum of the bullet (nail) just before it strikes the block, and it will go to whatever height where the energy has changes to potential energy, because in that case, mechanical energy is conserved: KE + PE is a constant when no work is done.
Concluding: If the nail has the same momentum in each case, the block must rise to the same height, rotation be damned.
Rotationally, it’s similar. The angular momentum is L = r X p, (or rp, at the point of impact), and angular momentum has a similar relation to rotational KE as the linear case does. Mass becomes moment of inertia, and you have L^2/2I as your rotational energy.
But the rotational energy does not come at the expense of the linear (which is different experience than rolling a ball or disk down an incline, where they must share). Because of the collision, KE is not conserved, and all we are doing in the rotating case is turning little less of the lost KE into sound, deformation of the wood, and a temperature increase, and using it to rotate the block.
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And that is one less topic for the next season of the Mythbusters.
Can we take it a step further: why does the block hit off-center get that extra bit of rotational energy?
Both blocks get an equal impulse (= F t) from the bullet during the impact, but they do not have equal amounts of work (= F d) done on them.
I suspect that during the off-center impact, the side of the block being hit moves upward. Now, the on-center impact also causes that block to move upward, but the impact point for the off-center (rotating) block moves further than the impact point of the on-center (non-rotating) block.
If so, the off-center impact takes place over a longer distance than it does in the head-on impact. That translates into a bit more work being done on the block for the off-center shot, providing some rotational kinetic energy.
I don’t know, CaptainPanic. It would be kind of nice to see this experiment done with a stiff block of ballistics gel so you could see how far the bullet goes in, where it ends up, etc. You could do the same thing with an x-ray machine, I guess, but optical pictures are so much more “intuitive,” “friendly,” etc.
A fun comparison would be to do an elastic collision (or something close to it) with the same momentum. That preserves KE, and then see if the blocks go different heights, as compared to the inelastic, and rotating vs nonrotating (I still have to think about this)