I was wrong . . .

One of the fun concepts I have been teaching for years has to do with relative velocity. It is easy when one object is stationary and the other is moving. The relative velocity of one to the other is simple the speed of the moving object and its direction. However, it gets more interesting when both objects are moving. If they are both moving in the same direction, you simply subtract the velocity of the faster one from the slower one.

Then here comes my angst. For years I have done a demo where two students walk toward each other and clap hands in the middle as they meet. I then ask what is the relative velocity of the one student moving at 5 m/s to the other moving at 5 m/s. Most everyone answers 0. And then there is that special student who gets it and answers 10 m/s. They get their name on the superstar board for a few days.

And then I say if two cars have a head on collision with the same speed, it is as if one hits a brick wall at the sum of their speeds. Then this year at my new school the student with the top score on the last physics test raises her hand. "Mr. Banks, I remember a mythbusters episode where they did that with two cars and it was not the same as hitting a brick wall at the sum of their speeds." Ouch. Send it to me and I will look at it. Before I got home good she had already emailed it to me.

Here is the link if you would like to see it.

I called in two of my former students (one with a PhD and one almost) for help.

Here are their responses.

First Dr. Foster
(I need to give a brief introduction. He is my first former physics student to get a college major in physics. He invited me to the Air Force Academy for his graduation and I surprised him and went. He surprised me several years later. The doorbell rang one day and there was Michael's mother with a copy of his dissertation. On the fly leaf he thanked me for getting him started in physics and said it was still fun. One of the proudest moments of my teaching career.)

Ok. I've watched the video. Fun stuff.

If I were king for a day, I'd run an additional crash test: 2 moving objects -- 1 car traveling at 50 mph (like before), and one moving wall (50mph also) with extremely high shear strength (titanium?), thickness/volume selected to match the mass of the car while maintaining impact point commonality with prior tests.

What I'd like to test is their premise that "the energy of the crash is transferred to twice the mass, halving it, resulting in a crash that looks like just 1 car, into a wall at 50 mph." By minimizing the kinetic energy transfered to the moving wall (i.e. high shear strength), the impact of doubled KE would be seen in the car (pun intended). My only point is the subtleties of inelastic energy transfer are lost in the quote above.

I agree that the crash test results are initially counterintuitive; but at the end of the day, I think the tenets of relativity are still in tact.

And now for soon to be Dr. Sadjadi.
(In the eleventh grade he was bored with school. He came to my class and never looked back. As a senior, after doing a junior project on a cloud chamber, he told me he wanted to do it on the string theory. At the time I had no idea what he was talking about. I tried to talk him out of a theoretical paper since no one had won with that kind of project. He insisted and I wisely let him go. He won and his class stood and cheered like he had won the super bowl - later I got to hear him present his paper at the AAAS meeting in Boston. Another very proud moment.)

The summary is this: summing the velocities works perfectly. The collision experiment is confusing because both of the cars are destroyed. If you look at the system in terms of energy, there are two ways to frame the experiment: you can be standing on the ground, watching the cars race towards each other, or you can be in one of the cars.

The ground observer sees two objects of mass m moving with velocity v. The total kinetic energy is 2E, where E is the energy of a single car traveling at velocity v. When the cars hit, all of that energy is lost in the heat/sound/damage of the collision. Since the two cars are of similar construction, each car takes half the energy. So the damage is identical to a single car hitting a wall (energy dissipated is E for each car).

From the perspective of one of the drivers, the second car is moving at 2v, while he himself is at rest. So the total energy in this moving reference frame is 4E (energy goes as velocity squared). After the cars hit, the two vehicles come to rest in the ground frame, which is equivalent to the two cars moving backwards in the car's original moving reference frame. So the energy post-collision is 2E (twice the mass, moving at velocity v). This leaves another 2E to be dissipated in the collision itself. This agrees with the ground observer.

I hope that helps.

Thank you for great students past and present and the opportunity to be wrong so I can learn even more.