# I Crashed.

Hey’all,

Those who follow my posts in real time will undoubtedly notice that I have not posted anything to this blog since August 26th, 2018. Perhaps I can explain the chain of event which lead to this situation. First, in my last post, I mentioned that I wanted to create a model of the Solar System in which Mars was close enough to Earth on August 27th, 2018 at 12:30 so that it would appear as large as the full moon. I actually did that, and it took about a month (or two?) to really get it right. I shall eventually share this model here, so look forward to that at some point in the future!

Unfortunately, while building this model, I ignored most other things—including the many lab reports and homework assignments that I now had to grade since classes restarted at UC. Needless to say, I spent the rest of the semester (into December) catching up and staying on top of those. Consequently, I didn’t touch this website.

And for some reason, I continued not to touch it throughout UC’s winter break! I frequently thought about it, but frankly, I guess I was having too much fun doing nothing since I, too, was on break.

(That’s not entirely true—I did a lot of work at the gym, which you can follow here.)

Well, UC has started back up again, but my teaching load this semester is a bit lighter. Because of that, I hope to be able to post more regularly, since there is a lot of stuff I want to talk about!

But where to begin? Well, showing off the model solar system I built last September would probably be a good place to start. But, last week, I got an email!

So, for my triumphant return—again—brace yourselves as I spend the next several hours writing about something which is really a “back-of-the-envelope calculation” that I did in a few minutes.

## The Question

If a speeding car T-bones a stationary car, can you determine the velocity of the speeding car?

I have heard that stuff like this is done by investigators trying to reconstruct the events of a traffic accident, and I imagine much knowledge of the practice can be gained from law enforcement, insurance companies, and the NTSB. But that’s a rabbit hole I don’t really have time to explore right now, so I will stick to that with which I am already quite familiar: elementary physics. In physics, (especially in the high-energy realm) collisions are our bread and butter. So, here we go!

As always when dealing with textbook problems like this, let us start by sketching out the scenario in parts, establishing our assumptions, and defining everything carefully.

Figure 1: Car images from openclipart.org

Figure 2: Immediately after the collision, e duobus unum.

Figure 3: fin

In figure 1, we have a striped car of mass $$m_1$$ traveling at an unknown speed $$v$$ directly into the side of a stationary blank car of mass $$m_2$$. In figure 2, we assume that the cars collide perfectly inelastically, meaning instead of bouncing off of each other, the cars remain smashed together once they make contact. We also assume that the car conglomeration continues to move forward with a new speed $$u$$. Finally, in figure 3, after sliding forward some distance $$s$$, everything is brought to a stop by friction between sliding tires and the ground. Note that this hypothetical scenario in which all motion occurs along the same direction is a simplified case. Nonetheless, the analysis which follows can be generalized for arbitrary impact angles and sliding paths.

So, what do we know, and what are we trying to find? We can reasonably determine the masses of both cars by looking up their manufacturer specifications. We can estimate the distance they slide by perhaps measuring the lengths of skid marks on the ground, interviewing witnesses, reviewing camera footage if any exists, etc. We can even look up the friction coefficients between rubber tires and various surfaces in engineering tables. We are asked to find the original speed of the striped car.

Known: $$m_1$$, $$m_2$$, $$s$$, $$\mu$$ (friction coefficient)

Find: $$v$$

As I see it, the most straightforward way of solving this problem is to start at the end and work backward. First, we must determine the weight of both cars on the ground. Since the cars are in mechanical equilibrium with the ground, their weight is equal to the magnitude of the normal force which the ground exerts on the cars. The magnitude of the friction force also exerted on the cars by the ground is then equal to that of the normal force times the friction coefficient. Combining this value with the distance the cars slid tells us how much work was done by friction throughout this process. By the work-energy theorem, that value is then equal to the change in kinetic energy of the cars throughout the slide. From this energy and the cars’ masses, we can derive the initial post-collision speed $$u$$. Finally, from momentum conservation and knowing both car’s masses, we can determine the original speed of the striped car, $$v$$. Piece of cake, right? Let me show you how it’s done.

Find the weight of both cars $$w$$ by multiplying the sum of their masses by the gravitational acceleration magnitude, $$g$$: $$w=\left(m_1+m_2\right)g$$

Equate this weight with the magnitude of the normal force $$N$$ exerted on the cars by the ground: \begin{aligned} N &= w \\ &= \left(m_1+m_2\right)g \end{aligned}

Use the kinetic friction model to convert the magnitude of the normal force $$N$$ into a friction force that acts in the direction opposite the sliding motion and has a magnitude of $$f$$: \begin{aligned} f &= \mu N \\ &= \mu \left(m_1+m_2\right)g \end{aligned}

Calculate the work done by friction during the slide, noting that because friction always acts in the opposite direction of motion, the work it does is negative: \begin{aligned} W &= \mathbf{f}\cdot\mathbf{s} \\ &= -fs \\ &= -\mu \left(m_1+m_2\right)gs \end{aligned}

Calculate the change in kinetic energy from the beginning of the slide, when the car pile’s speed is $$u$$, to the end, when the pile’s speed is $$0$$: \begin{aligned} \Delta \text{KE} &= \text{KE}_f - \text{KE}_i \\ &= \frac{1}{2}\left(m_1+m_2\right)(0^2-u^2) \\ &= -\frac{1}{2}\left(m_1+m_2\right)u^2 \end{aligned}

Use the work-energy theorem to equate the work done by friction to the change in the car pile’s kinetic energy. Simplify the expression and solve for $$u$$: \begin{aligned} &W = \Delta \text{KE} \\ &\Longrightarrow -\mu \left(m_1+m_2\right)gs = -\frac{1}{2}\left(m_1+m_2\right)u^2 \\ &\Longrightarrow u = \sqrt{2\mu gs} \end{aligned}

Use the fact that momentum is conserved throughout the collision to find an expression relating $$u$$ to $$v$$. Then, solve for $$v$$. \begin{aligned} & p_f = p_i \\ &\Longrightarrow \left(m_1+m_2\right) u = m_1 v \\ & \Longrightarrow v = \left(1+\frac{m_2}{m_1}\right)u \\ & \quad\;\;\;\;\;\; = \left(1+\frac{m_2}{m_1}\right)\sqrt{2\mu gs} \end{aligned}

And there is it. If we can determine the masses of both cars, the distance they slide after the collision, and their friction coefficient with the ground, then we can use the last line of the equation above to calculate the impact speed of the moving car. Or at least we can estimate that speed or establish upper and lower bounds depending on how well we know the input parameters. To illustrate this, I have graphed the formula for the case where both cars have the same mass and the friction coefficient between their tires and the ground is 0.7:

Figure 4: An example of the slide distance-impact speed relationship in SI units. Conversion to ‘Mercan units is left as an exercise for the reader.

Thus we see that the answer to the original question, “if a speeding car T-bones a stationary car, can you determine the velocity of the speeding car?“ is yes! At least it’s yes in this simple case. But the same or similar principles apply in more general cases—the calculations might just be more difficult. Again, I have no idea if what I described above is something that accident investigators actually do, but they could.

See you all next time, hopefully soon.

—Aaron