Dec 11

Crashing Cars

STEM 9 Comments »
STEM Consider two identical cars, each travelling at a constant velocity v towards each other. Suppose that they crash, and to simplify things, let's not consider any rebound. Then the energy of each is kinetic in nature and so the energy of the crash is given by

\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2

Ok. That's a nice way to show the energy of the collision goes up with the square of velocity. Now, according to the principle of relativity this should be an identical situation if you consider it from the reference frame of one of the cars. In that case, the car you are looking at has a velocity of 0, while the other car has a velocity of 2v. Now the total energy of the crash looks like

\frac{1}{2} m (2v)^2 + 0 = \frac{1}{2} m \times 4v^2 = 2mv^2

So. Why the discrepancy?

Posted by Colin Turner

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Crashing Cars, an answer?
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  1. Proving the Obviously Untrue
    01/06/09 11:26:23

    Crashing Cars, an answer?
    Every year I attempt to visit my old PhD supervisor Brian McMaster (old in the sense that my PhD is now a thing of the past, I am making no reference to the man in question!) at Christmas time to have a quick natter and exchange gifts. I was squeezed for

9 Comments

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  1. JoHo says:
    12/11/08 11:42:55

    Uuuh... can I have a beer please?

  2. Colin Turner says:
    12/11/08 15:05:50

    ButOfCourse (TM). Maybe even a bottle of Black Bush too :-). Will that help you solve the conundrum?

  3. JoHo says:
    12/14/08 22:15:39

    You know it would! :-)

  4. Anthony says:
    08/27/09 17:31:55

    Hi, after (briefly) checking out http://en.wikipedia.org/wiki/Inertial_frames_of_reference I wondered if is the energy input required to "freeze" one car? It may be the case if it is a closed system.

  5. Anthony says:
    08/28/09 15:21:15

    I asked a colleague about this, they explained it to me in terms of momentum and a moving frame of reference. Simple really. I was mostly wrong, I was on track that it is a closed system alright so energy is conserved. Now I know a little more about moving frames of reference, thanks to Eddie. But now I'm confused again http://faraday.physics.utoronto.ca/PVB/Harrison/SpecRel/MassEnergyEquivalence.pdf I only looked at it briefly though

  6. Colin Turner says:
    08/28/09 15:43:04

    Hi Anthony,

    Thanks for the comments, you might not have seen, browsing through various posts, that there's a trackback to a possible answer here:

    http://www.piglets.org/serendipity/archives/72-Crashing-Cars,-an-answer.html

    See if this agrees with your ideas. I'm not sure about momentum, that caused even more oddities when I played with it.

    CT.

  7. Colin Turner says:
    08/28/09 15:44:07

    But I'd be quite interested in your colleague's explanation if you have it.

  8. Amir says:
    07/08/11 11:45:22

    Nothing is wrong in here.
    The KE is not invariant under the the change of frame of reference.
    Consider a stone falling down a building, if you travel with the stone it has zero KE.

    The thing is number you measure for energy is not invariant under frame transformation.

  9. Colin Turner says:
    07/13/11 06:24:31

    Hi Amir,

    You'll see you weren't the first to spot this in the comments, and there's an updated post somewhere.

    Thanks!

    CT.

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