On MythBusters this week, Adam and Jamie tested the bullet-proofness of various objects. The one that sticks in my mind is the ipod. The said there was a report of a solider being shot by an AK-47, but he was saved because the bullet hit his ipod. To test this, Adam shot an AK-47 at an ipod and it went through. Their conclusion was that he was also wearing body armor. I am not sure I like that conclusion. Why would someone report that the ipod saved him if he was also wearing body armor? Maybe they would, but not sure.

I was thinking, maybe the bullet went through the ipod because they were only 10 feet away. Maybe this doesn't really matter - but how would I know? This is a great example to look at numerical calculations involving air resistance.

For this calculation, I am going to JUST look at horizontal motion and completely ignore vertical motion. This way I won't have to worry about the angle that the gun is shot at and stuff like that. Once the bullet leaves the barrel, I will have the following (modified) free body diagram:

![Screenshot 56](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…)

In this one-dimensional model (which I am sure someone will complain about), there is only one force acting on the bullet - the force of air resistance. Note that this force is in the opposite direction as the velocity (Aristotle would be upset that there is no force in the direction of motion). There is a fairly simple model for the air resistance force that says:

![Screenshot 57](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…)

In this model, ? is the density of air, A is the cross sectional area of the object, C is the drag coefficient for the object (depends on the shape). In case you are familiar with the notation, the v with a "^" over it is called "v-hat" and is a vector in the direction of the velocity, but with magnitude 1 and no units. So, to sum up, the air resistance force is in the opposite direction as the motion and proportional to the square of the magnitude of the velocity. The faster it goes, the greater the air resistance force.

You can get a good feel for this force if you stick your hand out the window of a moving car. The faster the car is driving, the greater the force on your hand.

Now, this should be easy to apply the [momentum principle (or Newton's second law - whichever you prefer)](http://scienceblogs.com/dotphysics/2008/10/basics-forces-and-the-moment…) - actually you could even apply the [work energy theorem](http://scienceblogs.com/dotphysics/2008/10/basics-work-energy/):

![Screenshot 58](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…)

This is actually a problem because the force is not constant. If there is a force on an object, it's velocity will change. Since the force depends on the magnitude of the velocity, it will change also. If you remember the discussion on kinematics, the key was that the average velocity was the final velocity plus the initial divided by 2. This leads to the kinematic equations.

Well, then how can this be examined? There is an analytic way to solve this problem, but I will opt for the numerical approach. Here is the basic idea:

- Break the motion down into tiny little time-steps (really short times)
- Calculate the air resistance force.
- Calculate the acceleration (as though the force were constant)
- Calculate the change in position and velocity during this time as though the acceleration were constant.
- Move to the next "time step"
- Repeat

Ok, let's do it. I could do this in Excel - bleh, or VPython (yeah!), but I have chosen to do this in [Easy Java Simulations](http://fem.um.es/Ejs/). This is a really neato package. Basically, it allows you to build models with little programming fuss. It won't replace [VPython](http://vpython.org), but it can supplement it. The one nice thing is that it can make applets for a webpage (although I am having trouble embedding this in WordPress).

So, here is a picture of the applet and a link to the applet (but not in wordpress - if you know how to do this, let me know).

![Screenshot 59](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…)

[Here is a link to the online calculation](http://scienceblogs.com/dotphysics/bullet2.files/ModelingScience/bullet…)

This is an interactive graph that shows the speed of the bullet as a function of distance. I used data for the [AK-47 from wikipedia](http://en.wikipedia.org/wiki/Ak-47). The applet above allows the user (that would be you) to enter the initial velocity and bullet mass. I really should have also allowed bullet diameter, but this serves it's point. Also, I know there is a huge amount of info on the ballistics of bullets. There are different air resistance models and stuff. The point of this calculation is to get an estimate for how much the bullet will slow down with distance.

So, what is the answer? If you look at the above simulation, at 3 meters (about 10 feet) it is going around 715 m/s (it started at 715 m/s) - so it did really slow down. What about a distance away? How about 50 meters? At 50 meters, it is going around 675 m/s. At 100 meters, it is going 634 m/s. Wikipedia lists the range of the AK-47 at 100-800 (I guess meters), so let me look at 500 meters - the speed would be 390 m/s. So, would this make a difference in the impact? One of the important aspects of the impact (in terms of damage) is the kinetic energy of the bullet. Yes, I know there is a huge amount of stuff on the impact of bullets - but this will give a rough idea. So, let me calculate the kinetic energy of the bullet at these distances:

![Screenshot 60](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/11/screensho…)

If the soldier was shot at close range (less than 100 meters) there is not too much difference in speed or kinetic energy. If it was a long shot (500 meters) then there is a significant difference in both speed and kinetic energy.

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Adam was a poor shot. By hitting the periphery of the iPod he missed the circuit board and the hard drive. Polycarbonate, metal disk, glass fiber/epoxy composite, and metal backing is a good combo for degrading a bullet. The bullet's jacket, if any, makes a big difference at impact.

Calculate deceleration of the bullet with flight. It is impressively large. 715 m/s is Mach 1.43. There can be a significant break in slope as the bullet slows transsonic then proceeds subsonic if it is not streamlined.

I used excel to do a numerical integral in class a while back, to show what I was looking for in a homework problem that no one could do. Turns out the words "numeric" and "numeric integral" and similar don't imply anything no matter how many times you say it, or write out examples, or discuss it with them, if you don't use those magic words "Riemann sum". I'd say this has implications for calculus pedagogy as well as physics but whatever.

Since you are doing this numerically, wouldn't a plot of kinetic energy against distance be appropriate as well?

(The point of using excel was to show them that they indeed had all the pieces of knowledge at hand to solve the problem, and could use a tool they all have at hand courtesy of the community college.)

Bear in mind that there is a difference between shooting something âstand aloneâ, and shooting it against a body. Most body armours will fail if shot standing alone. The reason is that body armour does not merely stop a bullet, but it helps to dissipate the energy, thus the body still absorbs most of the energy but over a larger area. When armour is shot standing alone, there is nothing behind it to absorb the energy and the energy remains concentrated, thus the bullet passes through easier. I am not saying that this is what happened with the soldier case, but it should be considered.