How Deep Does Veritasium’s Bullet Go?

After a couple of very productive days where I closed my Twitter tab because it was too freakin’ annoying to read, I checked in briefly Wednesday morning, and found Rhett Allain and Frank Noschese discussing this Veritasium bullet-in-block experiment:

Tom at Swans On Tea offers some analysis, and Rhett offers a video response doing out some of the math:

I basically agree with their explanations, and that should have been that, except that in his discussion Rhett mentioned that the bullet probably doesn’t go as far into the spinning block, which prompted Frank to ask whether you could measure and/or calculate the difference. Which sucked me in, and rather than working on the book chapter I’d planned to do this morning, I found myself scribbling equations. Which just goes to show that Twitter is the work of the devil.

Having gone through the analysis, though, I might as well get a blog post out of the deal. So, how much of a difference would you see in the depth of the bullet for the spinning and non-spinning cases? The full calculation would be very messy, but we can at least make a crude estimate.

First and foremost, let’s work out the bits that we can do exactly: as Rhett explains in his video, we know that momentum needs to be conserved, here, so we can relate the initial velocity of the bullet to the final velocity of the bullet-and-block:

$latex mv_i = (m+M)v_f $

I’m using m for the mass of the bullet and M for the mass of the bullet. Since the thing we can measure from the video is the final velocity of the bullet-and-block, v_f let’s solve this for the thing we don’t measure, namely the initial speed of the bullet v_i:

$latex v_i = \frac{(m+M)}{m}v_f $

(As a check: this is clearly going to be larger than the final speed, which makes sense, since the bullet is going too fast to see.)

So far, so good. But what we really care about is the energy, and specifically the change in the energy, which is different in the rotating-block case than the straight-up case. In the straight-up case, you only have two kinds of energy to worry about, kinetic and “internal,” which is a catch-all for the energy that goes into sound, deformation of the wood, heating of the block, etc. The final energy is just the kinetic energy of bullet-and-block plus this internal energy:

$latex E_{final} = \frac{1}{2}(m+M)v_f^2 + E_{int}$

This has to be equal to the initial energy, which is just the kinetic energy of the bullet:

$latex E_{initial} = \frac{1}{2}mv_i^2 = \frac{1}{2}(m+M)v_f^2 \frac{m+M}{m} $

where I’ve used the value for v_i from up above to put this in terms of measurable quantities. If you set these two equal, and do a bit of algebra, you can get a simple expression for the internal energy after the bullet buries itself into the wood:

$latex E_{int} = \frac{1}{2}(m+M) v_f^2 \frac{M}{m} $

(Do the algebra for homework, and send it to Rhett to be graded…). As a sanity check, this suggests that the vast majority of the initial kinetic energy is converted to internal energy– M is presumably considerably larger than m. Which makes sense– this is doing quite a bit of violence to the block.

In the rotating case, the initial energy is the same, but you pick up some additional terms in the final energy, due to the rotation. In very general terms, this looks like:

$latex E_{final, rot}=\frac{1}{2}(m+M)v_f^2 + E_{int} + K_{rot,block} + K_{rot,bull} $

Those two K terms represent the rotational kinetic energy of the block spinning about its center of mass, and the bullet embedded in the block. I have these separated because Frank asked on Twitter about the effect of the “impact parameter,” namely the distance off-center at which the bullet is fired, and this let me calculate it. These kinetic energies will depend on the “moment of inertia” for the block and bullet, which is going to be equal to the mass of the rotating object multiplied by some distance squared– in the case of the bullet, it’s just the “impact parameter” which traditionally gets the symbol b. In the case of the block, it depends on the size and shape of the block, and I’m just going to use a single number for the size, which I’ll call R, and stick in a fudge factor β to cover the effect of the shape, because I’m too lazy to look it up. Doing that, we get:

$latex E_{final, rot}=\frac{1}{2}(m+M)v_f^2 + E_{int} + \frac{1}{2}\beta M R^2 \omega^2 + \frac{1}{2}mb^2 \omega^2 $

where we’ve been forced to introduce another new variable, the “angular velocity” ω. This is something we can measure from the video, though, so it’s just a new parameter like v_f.

The internal energy in the rotating-block case, then, is:

$latex E_{int,rot} = \frac{1}{2}(m+M) v_f^2 \frac{M}{m} – \frac{1}{2}\beta M R^2 \omega^2 – \frac{1}{2}mb^2 \omega^2 $

As you would expect, this is lower than the energy in the non-rotating case, because some of the initial kinetic energy got turned into rotational energy of the block and bullet.

Up to this point, everything is exact, except for my lazy β thing. This doesn’t let us do anything about the penetration depth, though. To get that information, we need a crude approximation. This is not a thing that will make a lot of physics education research types happy, but we’re going to say that this internal energy is associated with some work done by the frictional-type forces that resist the bullet’s entry into the block. “Work done by friction” is a Bad Phrase to a lot of folks these days, but it’s the only way to get the penetration depth, so we’ll assume that the internal energy change comes from this work, which is equal to some force times some distance:

$latex E_{int} = F_{avg} \Delta x $

This isn’t really a constant force, of course, but you can finesse that by calling it some “average” force, which isn’t too awful an approximation. What’s less justifiable, but absolutely necessary to make this work, is the assumption that the “average” force is the same for both rotating and non-rotating cases. That’s probably not a very good approximation, but there’s no other simple way to attack it, so we’ll boldly forge ahead.

Now, we’ve got two different values for the internal energy, each involving an unknown average force and an unknown penetration depth, so this might seem hopeless. But all we really care about is the change in the penetration depth– that is, if we take the depth of the bullet in the straight-up case as another empirical parameter, we can find the depth of the bullet in the rotating case in terms of that, and see how much they differ.

To do this, we first solve for the average force in terms of the internal energy in the straight-up case:

$latex F_{avg} = \frac{E_{int}}{\Delta x_s} = \frac{1}{2}\frac{M}{m}\frac{(m+M)v_f^2}{\Delta x_s} $

Then we solve for the penetration depth in the rotating case using that internal energy:

$latex \Delta x_{rot} = \frac{E_{int,rot}}{F_{avg}} = \frac{\frac{1}{2}(m+M) v_f^2 \frac{M}{m} – \frac{1}{2}\beta M R^2 \omega^2 – \frac{1}{2}mb^2 \omega^2}{F_avg} $

That looks pretty awful, but we can plug in the value we found for F_avg in the straight-up case, and do a bunch of algebra to get:

$latex \Delta x_{rot} = \Delta x_s(1-\beta\frac{R^2 \omega^2}{v_f^2}\frac{m}{m+M} – \frac{b^2 \omega^2}{v_f^2} \frac{m^2}{M(m+M)})$

(Again, work this out on your own for homework, and send it to Rhett to be graded.)

Take a deep breath, the worst of the math is over. Let’s look at what we’ve got, here: we have an expression for the penetration depth in the rotating case that looks like the penetration depth in the straight-up case minus two factors, one related to the rotation of the block, the other related to the bullet within the block. These are kind of complicated, but the basic form makes sense: the faster the block spins after the collision, the larger these correction terms get, and the smaller the penetration depth. that’s exactly what we expect.

Now, you’d have a major problem if those corrections get too big, but there’s good reason to suspect that they’ll be small– β is going to be less than 1, and both terms contain a ratio of masses that is guaranteed to be less than 1. As long as the rotation rate isn’t absurdly large, we’re probably fine.

So, does this give a measurable difference between the two cases? (If you remember back before all the math, that’s the question that got me sucked into this whole mess…) Well, to estimate that, we need to know the speeds. If I were Rhett, I would download the YouTube video, crank it into Tracker Video, and measure those exactly, but that would require installing software, and anyway, I’m tired from all that algebra. So I’m going to do a really crude back-of-the-envelope approximation, and see what that suggests.

To get the final speed of the block, we can look at how long it spends in the air. Since we know the acceleration due to gravity, the time in the air depends simply on the velocity– looking at the video, a bit less than two seconds pass between the gun going off and the block falling back to its original height; if we call it two seconds, that’s an initial speed of 10 m/s, a nice round number.

To get the angular speed, we need to know how fast the block spins. In the high-speed video, it looks like maybe 10 revolutions on the way up, which, again, is a nice round number. That would be an angular speed of 20π radians per second.

But really, the velocities only come in as a ratio of some length times the angular speed divided by the linear speed. Which means we need an estimate of the size of the block– in the spirit of picking nice round numbers, I’m going to say that both R and b are 10cm, or 0.1m. Which makes that ratio equal to 2π/10, and it’s squared, which my calculator tells me is about 0.4.

That leaves only the masses to worry about, which is hard to figure. Because I’m just doing order-of-magnitude type things here, I’ll say that the block is about a kilogram and the bullet about ten grams. Those are wild guesses, but probably not off by more than an order of magnitude, which is good enough for my purposes. These really come in as a ratio, which is about 100 to 1 for the numbers I picked.

So, then, plugging these crude numbers in, we have:

$latex \Delta x_{rot} = \Delta x_s (1 – \beta (0.4) \frac{1}{100} – (0.4)\frac{1}{100^2}) $

This requires some kind of estimate for the mystery fudge factor β— if I call it 1/4, that cancels the 4 in the 0.4, and leaves us with a difference between the two penetration depths of about one part in a thousand. If the bullet made it ten centimeters into the block (which would probably be just about all the way through), that’d be a difference of a tenth of a millimeter, which would be extremely difficult to measure.

Now, I could be wrong about the masses– if it’s a 50g bullet and a half-kilogram block, you’d be looking at a difference of order one percent of the total depth. And you could maybe pick up another bit by more accurately estimating the speeds. But this is going to be a small number, any way you look at it.

Does that make sense? Well, yes, or I wouldn’t be posting this. See, in the crude work-by-friction approximation, a 1% change in the penetration depth means that 1% of the internal energy gets turned into rotational kinetic energy. And the internal energy was something like 100 times the kinetic energy of the bullet-and-block after the collision (for the 100 to 1 mass ratio I used above), which would require a really fast rotation of the block to achieve– it’s hard to pack much rotational kinetic energy into a small object.

So, that’s my overthought analysis of the bullet-and-block video, and where the energy goes. Which is way more than was required, but that’s why I’m a physicist, I guess…