## Soda Cup Killer

The Mythbusters are back! In their first episode of the new season, Adam and Jamie tested the myth that a cup of soda casually thrown out a car on the highway can smash through a windshield and kill the person sitting behind it. (Technically busted, but still really dangerous!)

One of the major factors that determines whether or not a projectile (like the cup of soda) will be able to smash a windshield is the force exerted by the cup on the glass. On the show, the Mythbusters used a piezoelectric force sensor to directly measure the force exerted when a cup of soda/ice/slush impacted on a steel plate, but we can also try to calculate the force using Newton’s second law,

$$\vec{F} = \ud{\vec{p}}{t}$$

First some simplifying assumptions are in order. I’m going to start by assuming that (1) each part of the cup and its contents continues to move at constant velocity up to the moment it impacts the plate, and that (2) after impacting the plate, the soda/ice/slush flies out in a direction perpendicular to its initial trajectory, along the plane of the plate. Of course, I wouldn’t expect that either of these assumptions would be correct in reality, but let’s see where they take us.

In order to calculate the force using Newton’s second law, we’ll need to know the change in momentum in the direction in which the force is exerted, as well the time it takes that momentum to change. The change in momentum is easy enough: we’re given the mass of the cup and its velocity on its way into the collision, and under assumption (2), the final momentum in the direction of the force is zero, so the change in momentum — or impulse — is just

$$I = \Delta p = -mv$$

Assumption (1) allows us to calculate the amount of time the collision takes as $$\Delta t = \Delta x/v$$, where $$\Delta x$$ is the length of the cup. Basically we’re assuming that the collision time is the same as the time it would take for the length of the cup to pass by the target plate. Putting these together, we can find the average exerted force as

$$F = \frac{\Delta p}{\Delta t} = -\frac{mv^2}{\Delta x}$$

Notice that just as Jamie predicted on the show, the force (in this approximation) is proportional to mass. Higher mass means higher force.

The Mythbusters were kind enough to post their data for $$m$$ and $$v$$, and together with my estimate of $$\Delta x \approx \unit{15}{\centi\meter}$$, I can calculate the average force that this model predicts for each cup. Here’s the math:

$$\abs{F_\text{soda}} = \frac{(\unit{632}{\gram})(\unit{132}{\mileperhour})^2}{\unit{15}{\centi\meter}} = \unit{14.7}{\kilo\newton} = \unit{3300}{\pound}$$
$$\abs{F_\text{soda/ice}} = \frac{(\unit{649}{\gram})(\unit{134}{\mileperhour})^2}{\unit{15}{\centi\meter}} = \unit{15.5}{\kilo\newton} = \unit{3490}{\pound}$$
$$\abs{F_\text{slush}} = \frac{(\unit{683}{\gram})(\unit{134}{\mileperhour})^2}{\unit{15}{\centi\meter}} = \unit{16.3}{\kilo\newton} = \unit{3670}{\pound}$$

The experimental results from the show were, respectively, $$\unit{3368}{\pound}$$, $$\unit{4331}{\pound}$$, and $$\unit{4386}{\pound}$$. Not too bad of an estimate, given our rough assumptions! I do notice that the result for soda is a lot closer than for ice or slush, which I suspect is because ice and slush are made of harder pieces which would probably bounce back more than the liquid soda.

If you were paying attention, you might have noticed that the tests Adam and Jamie ran with the steel plate were not quite the same as the real situation — a car windshield is slanted at a pretty significant angle. That means that when a soda cup impacts the windshield, instead of flying off in the perpendicular direction, the soda (or slush, or ice) retains some of its forward momentum, meaning that the resulting force on the windshield is less than what was measured using the force sensor.

We can calculate just how much less using the same procedure as above, with one modification: the assumption that the contents of the cup flies off perpendicular to its initial trajectory has to be dropped, because that’s obviously not what happens when a windshield is involved. Instead, I’ll assume that the stuff in the cup slides along the windshield and off the top. That means that, as seen by the windshield (or by the hapless victim), the final momentum in the horizontal direction will be $$(mv\cos\theta)\unitx + (mv\sin\theta)\unity$$ instead of zero, where $$\theta$$ is the angle of inclination of the windshield above the horizontal. That makes the impulse

$$\vec{I} = -mv(\cos\theta - 1)\unitx + mv\sin\theta \unity$$

and the force becomes

$$\vec{F} = -\frac{mv^2}{\Delta x}(\cos\theta - 1)\unitx + \frac{mv^2}{\Delta x}\sin\theta\unity$$

Notice that I’ve left the force in vector notation this time. That’s because the breaking strength of a windshield would most likely be measured in terms of force exerted perpendicular to the glass, which is not the same as the horizontal force. (The parallel component of force doesn’t really contribute to breakage). In order to compare the force we calculate to the reported breaking strength of a windshield, we’re going to have to find that perpendicular component of force. Fortunately, it’s not hard to do; all we need to do is take the dot product with the normal vector to the glass, which is $$\sin\theta\unitx - \cos\theta\unity$$. The result is

$$\abs{F_{\perp}} = \frac{mv^2}{\Delta x}[(\cos\theta - 1)\sin\theta - \sin\theta\cos\theta] = \frac{mv^2}{\Delta x}\sin\theta$$

For a windshield inclined at, say, about $$\unit{40}{\degree}$$ (I just guessed that), the $$\sin\theta$$ factor works out to about $$0.6$$, so the force exerted on a real windshield will be about that much times the equivalent force as measured by a force plate in the Mythbusters’ setup.

Unfortunately, after going through all that, I’ve still been unable to find a numeric value for the breaking strength of a typical windshield to compare the calculation with. That’s probably to be expected because whether a windshield will break depends on a lot more than just how much force is applied to it; things like the area covered by the impact and the time taken for the force to build up to full strength would probably also have an effect. And this is just a rough calculation anyway. In this case, the math doesn’t have quite the same impact (no pun intended?) as the Mythbusters’ demonstration!