Crashing Cars

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?






6 responses to “Crashing Cars”

  1. Proving the Obviously Untrue avatar
    Proving the Obviously Untrue

    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

  2. Anthony avatar

    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 I only looked at it briefly though

  3. Colin Turner avatar
    Colin Turner

    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:,-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.


  4. Colin Turner avatar
    Colin Turner

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

  5. Anthony avatar

    Hi again Colin, I have just realised you printed an answer, my fault for not reading your post thoroughly.

    As I said a colleague (thanks Eddie) explained it to me after I got it wrong.
    The way it was explained to me was in terms of the frame of reference. If the frame of reference is travelling at velocity v with one of the cars then after the collision the two cars with mass 2m will be travelling at -v relative to the fixed frame of reference (think of the observer floating through the mangled car in their fixed frame of reference). The fixed frame of reference always travels at v. So the energy will be 1/2(2m)(-v)^2 = mv^2. QED.

  6. Amir avatar

    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.

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