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I was watching a youtube video the other day where an economist said that he challenged his physics professor on this question back when he was in school. His professor said each scenario is the same, while he said that they are different, and he said he supplied a proof showing otherwise.

He didn't say whether or not the cars are the same mass, but I assumed they were. To state it more clearly, in the first instance each car is traveling at 50mph in the opposite direction and they collide with each other. In the second scenario, a car travels at 100 mph and crashes into a brick wall. Which one is "worse"?

When I first heard it, I thought, "of course they're the same!" But then I took a step back and thought about it again. It seems like in the first scenario the total energy of the system is the KE of the two cars, or $\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$. In the second scenario, it's the KE of the car plus wall, which is $\frac{1}{2}m(2v)^2 + 0 = 2mv^2$. So the car crashing into the wall has to absorb (and dissipate via heat) twice as much energy, so crashing into the wall is in fact worse.

Is this correct?

To clarify, I'm not concerned with the difference between a wall and a car, and I don't think that's what the question is getting at. Imagine instead that in the second scenario, a car is crashing at 100mph into the same car sitting there at 0mph (with it's brakes on of course). First scenario is the same, two of the same cars going 50mph in opposite directions collide. Are those two situations identical?

PS: This scenario is also covered in an episode of mythbusters.

Qmechanic
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smaccoun
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    The "with its brakes on of course" is what makes both situations different, without the brakes they are the exact same, see my answer below. – Jaime Dec 01 '12 at 02:02
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    Possible duplicate of Train crash: are these situations alike? and Classical car collision – voix Dec 01 '12 at 09:18
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    I thought Myth busters tested this, and found that two cars at 50mph equals one car at 50mph hitting a wall. In essence the wall is like a plane of symmetry. – John Alexiou Dec 01 '12 at 14:55
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    As Jamie said, if the second car doesn't have brakes on, the situation are identical : one is viewed from the centre of mass frame, and the second is the same event viewed from a train moving in the same direction and at the same speed of one of the cars – Marco Aita Dec 01 '12 at 19:13
  • ..but a wall is not like a car (i.e. it is like a very massive car), so hitting a wall at 100mph is not like hitting an equally massive car at 100mph (see the chosen answer to Classical car collision for a good explanation). – Marco Aita Dec 01 '12 at 19:35
  • Two cars moving toward each other, each at 50 mph, should be equivalent to one car at 100 mph colliding with a (previously) stationary car (assuming both are in neutral gear, the moving ones are just rolling). But that situation is not equivalent to hitting a wall. – Larry Gritz Dec 05 '12 at 21:40

7 Answers7

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I don't think any of the other answers have made the following point clear enough, so I am going to give it a try. Both scenarios are very similar before the collision, but they differ greatly afterwards...

From a stationary reference, you see the cars driving towards each other at 50mph, but of course if you choose a reference frame moving with the first car, then the second will be headed toward it at 100 mph. How is this different from the wall scenario?

Well, from a stationary reference frame, after the crash both cars remain at rest, so the kinetic energy dissipated is $2\times \frac{1}{2}mv^2$.

From the reference frame moving with the first car, the kinetic energy before the crash is $\frac{1}{2}m(2v)^2=4\times\frac{1}{2}mv^2$, but after the crash the cars do not remain at rest, but keep moving in the direction of the second car at half the speed. So of course the kinetic energy after the crash is $2\times\frac{1}{2}mv^2$, and the total kinetic energy lost in the crash is the same as when considering a stationary reference frame.

In the car against a wall, you do have the full dissipation of a kinetic energy of $4\times\frac{1}{2}mv^2$.

Jaime
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    To put a finer still point on this, and to tie this in with my comment to the OP, the KE before the collision is not the whole story. In the frame in which the car is stationary while the wall is moving, the KE is enormous both before and after. It is the difference that tells the story. – Alfred Centauri Dec 01 '12 at 02:05
  • +1 for showing that the analysis works regardless of reference frame. When I discussed mine, I just assumed it would be easier to keep the reference frame the same in both situations: a 3rd person observation standing at zero mph relative to a wall. – smaccoun Dec 02 '12 at 19:54

  • but after the crash the cars do not remain at rest, but keep moving in the direction of the second car If you consider the reference frame of the first car then isn't the first car always stationary and it is the second car that collides and slows down to 0 velocity?

     – 1110101001 Aug 15 '15 at 21:05
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    Nope. The reference frame is not tied to the car, it is an inertial reference frame that happens to be one that the first car is stationary in at the starting time. A stationary car hit by another car will not stand still, but start moving backwards at half the speed of the incoming car. – Jaime Aug 15 '15 at 21:57
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    Oh I see! Also, I assume the reason you can't use the reference frame of the car itself is that upon collision the car decelerates, hence the reference frame of the car is not strictly inertial. – 1110101001 Aug 15 '15 at 22:55
  • Yes, that's the way I see it. – Jaime Aug 16 '15 at 01:44
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    This is silly but I am a mathematician who has been confused how this works for YEARS, and this comment finally made me understand. so thanks. – pancini Dec 08 '23 at 04:08
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Actually, assuming that the oncoming car is the same mass as yours, colliding with an oncoming car at 50 MPH is equal to colliding with an ideal immovable wall at 50 MPH. Consider this:

I'm going to set up one of two experiments. I'm either going to ram car A into car B, both of them moving 50 MPH in opposite directions, or I'm going to ram car A into a solid wall at 50 MPH. However, I'm going to put up a shroud so that you can only see car A, you will be unable to see either car B or the wall, whichever one my coin-flip tells me to use.

Because you can now only see at car A and its contents, how would you tell which experiment I'd decided to do?

Aric TenEyck
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    I would ask Schrödinger's Cat... – Mik Cox Dec 04 '12 at 00:23
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    I agree that the situations are identical, but fail to see how the shroud experiment proves that. It seems like you assumed they're the same at the beginning, and then concluded they're the same at the end based on your beginning assumption. Can you explain more how the shroud illustrates that? – smaccoun Dec 04 '12 at 00:25
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    Some people believe that head-on-at-50 is equivalent to brick-wall-at-100, while in fact it's equivalent to brick-wall-at-50. The shroud experiment encourages them to consider the different scenarios and figure out why head-on-at-50 is equivalent to brick-wall-at-50. – Aric TenEyck Dec 04 '12 at 02:19
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    You might say something about the energy. One car crashing into a boulder at 50mph absorbs so-much kinetic energy by twisting metal and breaking bones. Two cars crashing head on have twice as much total KE (in the reference frame of the resulting wreck), but there is also twice as much car to absorb the energy. OTOH, one car crashing into a boulder at 100mph must absorb four times as much energy as one car crashing into a boulder at 50mph. – Solomon Slow May 25 '17 at 15:54
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I think that it makes sense to move away from the specific walls and cars and consider simply an inelastic collision on of two masses, $m_1$ and $m_2$. Otherwise we get stuck in the details.

When two bodies collide, the devastating effect in collision depends only on their relative velocity $v_1-v_2$. Kinetic energy, which has the destructive effect is equal to

$$\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(v_1-v_2)^2$$

The rest of the kinetic energy is associated with the movement of the center of mass of the system. This energy in the collision does not change, and has no effect of destruction.

In given case, if faced two identical cars moving toward each other with one and the same velocity $v$ the energy of destruction is

$$\frac{1}{2}\frac{mm}{m+m}(v+v)^2=mv^2$$

Now, consider the case where a car collides with a massive barrier at speed $2v$.In this case $m_1=m$, $v_1=2v$, $m_2=\infty$, $v_2=0$

The energy of destruction:

$$\frac{1}{2}m(2v)^2=2mv^2$$

I.e. the latter case is much more dangerous.

Martin Gales
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Certainly they are not exactly the same - a wall is not the same thing as a car, and a crash is a very complicated physical event. Even if simple calculations involving momentum and energy or descriptions involving reference frames suggest that aspects of a car-car and car-wall collision are the same, the real collisions will be fairly different.

In this case, though, simple considerations do reveal that the car-car crash at 50 mph is almost certainly safer than crashing 100 mph into a wall. Your energy calculation is a fine way to see this.

Another is to consider the car-car collision from a frame co-moving with the second car. In this frame, you're going 100 mph and crash into a stationary car. So the question is like asking whether it is worse to crash into a stationary wall or a stationary car when going 100 mph (apart from the fact that the movement relative to the road is a little different). Of course crashing into the car is less dangerous than crashing into the wall, confirming your earlier result.

I have often heard the same problem rephrased so that you consider crashing into a wall at 100 mph or crashing into a car when you're both going 100 mph. It may be that this was the original problem the physics professor mentioned, and it got distorted somewhere in the game of telephone it played since then.

In that scenario, some people say they are equally bad because the energy dissipated per car is the same. Personally, I would probably go for the wall because at least some of the car's energy should go into the wall, but here the details become important (e.g. what if I fly through the window and then hit the wall?), and the energy alone is not a strong enough difference to say what which is worse. I imagine that either crash is very likely to be fatal at that speed.

Addressing your new question, two cars crashing head-on each at 50 mph is essentially the same as one car going 100 mph and crashing into a stationary car, by the relativity principle. However, relativity is broken by the existence of the road, so to the extent that the cars interact with the road during the collision there may be some differences.

Mark Eichenlaub
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    I think the idea here is to get at the fundamental physics concepts, not whether or not hitting a car or a wall is worse. For that reason, I don't think your second point is at all convincing, and it is the very reason why I originally thought the answer would be the same (and why I think most people would). For the second proposition, and that may very well have been the original question, I think that one is most certainly te cars. Using the same energy calculations, there is twice as much energy that needs to be dissipated for the cars colliding, so I think the cars are clearly worse there. – smaccoun Dec 01 '12 at 01:15
  • I don't know what you're talking about because phrases like "second such-and-such" are ambiguous; I don't know how you're counting. Please rephrase to say what specific concepts you are trying to refer to. – Mark Eichenlaub Dec 01 '12 at 01:28
  • By second proposition, I meant your rephrased version of the problem (your 3rd paragraph, which is a different problem) where the 2 cars colliding are each going a 100, and the second scenario the car is going 100 and crashes into the wall. For an analysis of that, I would say the two cars colliding is worse, because the total energy is twice that of a single car going 100 and crashing into the car. – smaccoun Dec 01 '12 at 01:37
  • I thin your third paragraph, where you use reference frames, is possibly the misconception that this problem is getting at. I originally thought that too, and it is true that relative to the other car the first car is going 100mph. This is why I had to revert to an energy argument to see a difference. You could just as easily make the problem be only walls crashing at each other, or only cars crashing at each other. Again, I'm not sure about this, but my suspicion is that the reference frame argument isn't sufficient in its own right. But, I posted it on here to see some other perspectives:) – smaccoun Dec 01 '12 at 01:40
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    Sorry man, in my 2nd comment I meant your 4th paragraph, LOL. Whoo, I need to learn how to count – smaccoun Dec 01 '12 at 01:46
  • If two cars collide head-on at 100mph, there is twice the kinetic energy involved as if one car collides with a wall. However, that energy is used to damage both cars. Presumably, when one car collides with a wall, most of the energy goes into destroying the car and little goes into destroying the wall. Thus, in each case roughly the same amount of energy goes towards destroying your car. This is what I meant by "the energy dissipated per car is the same." – Mark Eichenlaub Dec 01 '12 at 02:22
  • The argument in which I say that two cars colliding head on at 50 mph each is equivalent to one car hitting a stationary one at 100 mph is perfectly sufficient. You say it isn't (or that you suspect it isn't), but since you don't give any reasons for your opinion, I have nothing to clarify. – Mark Eichenlaub Dec 01 '12 at 02:25
  • Updated answer to address the new question. – Mark Eichenlaub Dec 01 '12 at 02:32
  • Rereading the comments, I think I see what you were concerned about. Although hitting a car and a wall are different in terms of the details of how the collision goes, they are different in terms of fundamental physics as well. So when you talk about getting at "fundamental physical concepts", my original answer actually did that. The reason is that a wall is very massive, and thus essentially immobile. When you hit a wall, it doesn't move or absorb much energy. That is quite different from hitting a car, and that is why my original answer was germane as it stood. – Mark Eichenlaub Dec 01 '12 at 02:36
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The most straightforward way to see how different the two scenarios are is to:

(1) consider two cars crashing into each other from a reference frame in which one of the cars is stationary and the other has a speed of 100mph

(2) consider the one car crashing into the wall with a speed of 100mph.

Assuming the wall is substantial enough that its mass and physical strength far exceeds that of the stationary car in (1), it's clear that the two scenarios significantly differ.

In (2), the car crashes into a stationary and effectively immovable, indestructible object at 100mph while in (1), the moving car crashes, at 100mph, into a stationary but otherwise identical object that importantly, can both move and deform.

Alfred Centauri
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  • Please see my edit. Do you think it would be the same if it was just walls crashing into walls, or just cars crashing into cars? Is my energy argument incorrect? Good points otherwise, but I don't think that's what the question is getting at. – smaccoun Dec 01 '12 at 01:50
  • @smaccoun, the total KE of the system is a minimum in the COM reference frame. Consider, for example, the frame in which the car is stationary and the wall is moving at 100mph. The KE in that frame is enormously larger than in the frame in which the wall is stationary. Yet, the damage to the car (and perhaps the wall) does not depend on the reference frame. Your argument cannot depend on just the KE before the crash. – Alfred Centauri Dec 01 '12 at 01:59
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Damage should be the same if two cars colliding at $50$mph and if a car travelling $(50*\sqrt{2})$ mph crashes into a wall.

The energy of destruction is an internal energy so:

First case

Equation of energy conservation

$\frac{mv^2}{2}+\frac{mv^2}{2}= T$

where $T$ - internal energy, $m$ - car mass

so energy of destruction in first case:

$T = mv^2$

Second case

Equation of energy conservation

$\frac{m(v\sqrt{2})^2}{2}=\frac{(m+M)u^2}{2}+ T$

where $T$ - internal energy, $m$ - car mass, $M$ - wall mass

Equation of momentum conservation

$mv\sqrt{2}=(m+M)u$

so internal energy $T= \frac{m(v\sqrt{2})^2}{2}\Big(1-\frac{m}{m+M}\Big)$

$M>>m$ so energy of destruction in second case:

$T \approx {mv^2}$

voix
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  • The calculations are correct but they solve the problem for the total energy of the system. If you are only interested in the damage caused on one car (because in the second case there is no "other car"), the same calculations show that to obtain the same damage (=change in internal energy=deformation+heat) you need to have the same speed $v$ (again considering the wall very massive, $M>>m$ so that $u ≈ 0$). – Marco Aita Dec 01 '12 at 19:08
  • @Marco Aita, you are not right. Imagine a car standing before the wall, so damage will be equal for two cars in the second case. – voix Dec 02 '12 at 14:38
  • In the fist case each car dissipates $\frac{1}{2}mv^2$. In the case of an impact of a car with a wall of a very large (infinite) mass, the final speed is zero and therefore the car dissipates again an energy of $\frac{1}{2}mv^2$. If all this energy goes into deformation of the car you will get the same damage to the car as in the first case. The problem is that we can't really tell how much of the dissipated energy goes into deforming the car, so my reasoning is partial..

    ..but I don't understand your point with the car in front of the wall, can you elaborate?

     – Marco Aita Dec 03 '12 at 00:10
  • @Marco Aita, if in the second case we put another car in front of the wall we will get the same damage to the car as in the first case. – voix Dec 03 '12 at 04:54
  • ..I think we are talking about different things.. Anyway, the part with the wall is practically unsolvable unless we specify more about the wall characteristics.. Imagine the difference in hitting a massive solid steel wall and an equally massive marshmallow wall.. :-). It all depends on how much energy the wall absorbs.. – Marco Aita Dec 03 '12 at 13:09
  • @Marco Aita, maybe better to speak about two lead cannonballs in the first case and lead cannonball with lead wall in the second :) – voix Dec 03 '12 at 16:46
  • Agreed! Assume the lead wall does not absorb energy in the impact, and for clarity let's use the CM reference frame. First case: total energy dissipated: $mv^2$ between the two cannonballs. Second case (the ball travels at velocity v before impact against wall): total energy dissipated: $\frac{1}{2}mv^2$ which goes all into deformation of the cannonball. Case with two balls, one at rest in front of the wall: total energy dissipated: $\frac{1}{2}mv^2$ which goes into deformation of two cannonballs (although in this case I would argue that the ball at rest gets a worse "whack"! :-).. – Marco Aita Dec 03 '12 at 17:30
  • @Marco Aita, "Assume the lead wall does not absorb energy in the impact"- it's not reasonable assumption – voix Dec 04 '12 at 05:29
  • Well, if you don't make an assumption on how the wall absorbs energy, you cannot tell what the result of the impact is on the incoming car (see lead vs marshmallow). As simple as that.. – Marco Aita Dec 04 '12 at 10:19
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I will assume that:

  1. There is car-car collision in both the cases.
  2. The car is a wooden lifeless cart with no engine.

Now, the Classical difference between the two collisions is-

The wheels of the car moving at 100mph is twice as energetic as the wheels of both the cars combined in the 50mph case. So, in the 100mph case (2nd case), the accident will be worse.

Now, there will also be special relativistic effect, although very tiny. The combined energy of the cars in the second case (100mph) is larger than the first one.

So in both the cases, the second case will be worse, for the reasons stated above.

Prem
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