# Consider a spherical, frictionless car of mass 1 …

A car of mass m drives at speed v towards a concrete wall. Its kinetic energy is ½ mv2, and all of that energy is dissipated when it hits the wall. Now consider an identical car driving at speed 2v towards the same wall. Its kinetic energy is ½ m(2v)2 = ½ m4v2 = 2 mv— four times as much as the original car, since kinetic energy goes with the square of the velocity. I hope that up to this point, all this is uncontroversial.

Now consider two more identical cars, each driving at the original speed v, but this time directly towards each other, so that their collision speed is 2v. In this case, the total kinetic energy that is dissipated in the collision is (obviously) the sum of the energy the two cars have: that is 2 × ½ mv2 = mv2.

So in one crash at speed 2 v (the head-on collision), there is twice as much energy than in the other (the double-speed car into wall).

How can that be?

### 19 responses to “Consider a spherical, frictionless car of mass 1 …”

1. dafux0r

It’s a trick, brought on by using a multiplier of 2, which just so happens to square at exactly double itself (:

2. Juha Haataja

In two of the cases, there is a concrete wall, in one case there is not.

3. dafux0r

Though I realise that’s a bit of a flippant response; I’d look at how the velocities of both cars are interesting to the power of two, where with the single car, crashing into a stationary wall, there’s only one velocity being squared, making it less significant?

4. Mike Taylor

dafux0r, I don’t see that at all. The issue is not that one of the cases happens to have exactly twice the energy of the other, but that they don’t have identical energy.

Juha Haataja, I don’t understand how the presence of the concrete wall causes the difference. In effect, what you’ve done here is basically to restate the problem. Unless you have some deeper insight that I didn’t pick up.

5. Simon

2 cars, 2 m. 🤓

6. ropata

(a) car of mass M travelling at speed X: KE = 1/2 M X.X
(b) same car going twice as fast: KE = 1/2 M (2X).(2X) = 2 M X.X = four times the energy of (a)

(c) 2 identical cars at speed X: KE = 1/2 (2M).X.X = M.X.X = double the energy of case (a)
(d) 2 identical cars at double speed: KE = 1/2 (2M).(2X).(2X) = 4 M X.X = double the energy of case (b), and four times the energy of case (c)

double the mass, double the energy.

7. Robert

Think about this: how does the mass of one or both of the cars change things? What about the mass of the static, stationary wall?

8. Tim

Why should the two collisions dissipate identical amounts of energy? I don’t see any reason why they should.

There’s also an issue here because the wall doesn’t have infinite mass. If the car ends up at rest then the wall has to start out moving toward the car.

9. Martin McCallion

Velocity is a vector. You have to add the two velocities before you calculate the kinetic energy. So the velocity of the collision is 2V in both cases, giving the same result in both cases.

I think that’s also what ropata is saying above.

10. Vladan

Kinetic energy depends on the reference frame and you are using two different reference frames in the two examples (the 2V velocity in the wall case is not in the same reference frame as the 2V velocity in the two cars case).
The energy dissipated in the collision is always the same, regardless of the reference frame, and is equal to the difference in total kinetic energy before and after the collision, not simply to the total kinetic energy before the collision.
In the second case (two cars), if you want to use 2V as the velocity for the collision, you have to use a reference frame in which one car is stationary and the other is moving with velocity 2V. The total kinetic energy before the collision is 0+(1/2)*m*(2V)^2 = 2*m*V^2.
After the collision the kinetic energy in the same reference frame is not zero, however. In that reference frame, both cars are now moving with velocity V (the first car is moving backwards, while the other decreased its velocity from 2V to V). The total kinetic energy after the collision is 2*(1/2)*m*V^2 = m*V^2.
The difference between the two is m*V^2, which is the correct answer.

11. reenigne

Kinetic energy is dependent on the frame of reference. This is pretty easy to see if you consider a single vehicle – it has zero kinetic energy from your point if view if you’re moving at the same velocity as it, and more if you’re moving at a different velocity. The difference in kinetic energy between the two frames of reference in the two situations accounts for how the two collisions can disperse different amounts of energy despite the colliding objects having the same relative velocity.

12. NickS

FWIW, there was a good mythbusters segment on this question: https://www.youtube.com/watch?v=-W937NM11o8

13. gnustavo

This is tricky. It’s easy to understand the math of it but I had a hard time figuring out what is really going on, because to me the second and third situations are intuitively the same.

Mathematically, it’s obvious that since the kinetic energy is proportional to the mass and to the square of the velocity, doubling the mass doubles the energy but doubling the velocity quadruples the energy. So, in the first case, when the car hits the wall with velocity v, the initial kinetic energy of the system is mvv/2. In the second case, when the car hits the wall with velocity 2v, the initial kinetic energy is m(2v)(2v)/2, which is 2mvv. And in the third case, when both cars hit each other with velocity v, the initial kinetic energy is (2m)vv/2, which is mvv. In all cases, the total kinetic energy after the collision is zero. So, since what matters is the kinetic energy lost, it’s clear that the second case (2mvv) is harder than the third case (mvv). But how?

One trick that allowed me to see this more clearly is to change the point of reference. Given the descriptions above it’s clear that we’re considering the ground at rest in our reference system, which is the most obvious thing to do. But Relativity being what it is, the physics must not change if we change our reference system to another one, as long as it is inertial. So, let’s change it for the third case and consider as the origin of our reference system the front of one of the cars, so that it is “moving” along with the car. No… it’s not moving… in this new reference system this car is at rest relative to the origin but the other car is moving towards the origin with velocity 2v.

Note that in this new system the total kinetic energy is bigger. In fact, it’s twice as it was considering the ground as our reference. Let’s see… the car at rest in the origin has zero kinetic energy, but the other car’s kinetic energy is m(2v)(2v)/2, or 2mvv.

But isn’t energy supposed to be conserved? Not exactly… the “total” kinetic energy can change when we change the reference, since the velocities are always relative to the origin. What is conserved is the “difference” between the initial and the final kinetic energy of the system. So, let’s see what we have so far:

* In the case of the car hitting the wall with velocity 2v and the reference on the ground, the initial kinetic energy is 2mvv and the final kinetic energy is 0. So, the difference is (2mvv – 0), or 2mvv.

* In the case of the cars hitting each other with veolcity v and the reference on the ground, the initial kinetic energy is mvv and the final kinetic energy is 0. So, the difference is (mvv – 0), or mvv.

* In the case of the cars hitting each other with reference on one of them, the initial kinetic energy is 2mvv and the final kinetic energy is… mvv! Why? Because after the collision both cars will stick to each other and both of them will continue to move in the direction of the moving car but with velocity v. (Supposing there is no friction between the cars and the ground, of course.) So, the final kinetic energy is (2m)vv/2, or mvv. And the difference is, again, (2mvv – mvv), or mvv.

This was a revelation to me. Only because of my prejudice I was thinking about the experiments from the point of view of a person standing on the ground. When I imagined myself riding inside one of the cars I was able to see that the ending result of the experiment was very different. When the car hits the wall is stops completely, because the wall is for all intents and purposes a body with infinite mass. But when the cars hit each other they don’t stop, but continue moving together with half the initial velocity of the moving car.

I remember to have been thinking about this very problem more than a few times in the last two decades. Since a friend of mine told me that he saw two cars at 40Km/h hitting each other and thought that at such a low speed they shouldn’t have been so damaged. At that time I told him that the collision would be the same as if one car hit a wall at 80Km/h. But I never was completely sure about that. Now I know that I was wrong… In fact, to the people inside the cars, two cars hitting each other at 40Km/h has the same intensity of one car hitting a wall at 40Km/h. Damn it!

Thanks so much for the thought provoking post!

14. Daniel

A further interesting perspective, which I’ve found illuminating in a variety of circumstances ever since reading one or another of the articles on this website—http://www.projectrho.com/public_html/rocket/—is to consider the wall as it truly is: not *quite* immovable, but very *nearly* so (being rigidly attached to the mass of the entire Earth). Then in the hitting-a-wall-at-2v case, using the Earth’s initial velocity as our reference frame, we have:

The earth has mass M >> m.
Before the collision:
E[car] = 1/2m(2v)^2 = 2mv^2
E[earth] = 1/2M(0)^2 = 0

After the collision:
v[car and earth] = mv/(m + M) (I think—have not done physics in a loooong time. It’s something very close to this.) In the version where we consider the wall fixed, we round this all the way to zero—which is not an unreasonable thing to do! Instead, let’s consider it some very small value ϵ (epsilon).

E[car] = 1/2mϵ^2
E[earth] = 1/2Mϵ^2

The energy of the car is obviously very small. Looking back at what ϵ actually is, it has an M in the denominator—so the Earth’s energy is very small also, specifically on the order of 1/M. Thus the total energy of the system is very small but not zero, and the total collision energy is *very nearly* 2mv^2, but not quite.

Now, how is this connected with rockets? Well, what if the car hits, not an immovable wall, but a much larger vehicle, still at a combined speed of 2v? We will get a result somewhere in between—as M approaches m, the collision energy approaches the two-cars case of mv^2. Now, play the situation in reverse—the large car is the rocket, and the small one is its propellant, which it accelerates out the back. The more propellant mass, the less energy it takes to accelerate the rocket to a given velocity—but this would be better stated as, the faster the propellant moves, the less *propellant* it takes to accelerate to a given velocity (at the cost of vastly higher energy use), as propellant mass, not energy consumption, is generally the limiting factor in hard-sci-fi rocket design.

15. kaleberg

You have to measure the relevant velocity in terms of the center of mass. You want a reference frame so that the combined kinetic energy after the crash is zero.

The wall is much more massive than the car, so the wall is effectively at the center of mass. The kinetic energy to be dissipated in the collision depends on the square of the car’s velocity relative to the wall.

The two cars have the same mass and each the same speed, so their center of mass is the point of the collision. That means the kinetic energy depends on the sum of the squares of the velocity of each car.

This also explains another problem that puzzles many physics students. Consider those two cars heading towards each other. If you change the frame of reference, one car is sitting still while the other car is moving towards it at twice the stated speed. That would imply that there is twice as much kinetic energy in the system, just as if one car had been heading twice as fast towards a brick wall. The difference is that after the collision, the car that appeared to be sitting still would now be moving while the other car would have halved its speed but not stopped. This can be fixed by subtracting the energy after from the energy before, but it makes sense to choose a better reference frame.

16. Neo

your working it out as 2 separate incidents. where it is 1 impact

17. Neo

better explination the impact is 2x the speed and 2x the mass. not 2x 1/2 mv squared but 2x 2x 1/2 mv squared

18. Neo

went away had a cup of tea turn on my pc as a phone is no good. Your theory is inccorect as it does not allow for the wall mass . a wall of no mass would have no impact . also hardness / density wolud be another factor. a brick wall of mass one would have more impact than a soft -custard- wall of any weight . and cross sectional area must come into it as well.

19. araybold

Firstly, someone should note that you have stated the paradox the wrong way around – you calculate twice the dissipation in the car vs. wall crash as in the head-on collision.

There is a problem with your analysis of the brick wall case in that it does not respect the conservation of momentum: Initially, you have a momentum of m2v, but afterwards? If you model the brick wall as an infinite mass initially at rest, you have to accelerate it infinitesimally on impact, which means that some as-yet undetermined energy may not have been dissipated.

To avoid the infinite infinitesimals, let’s consider a car at 2v (so kinetic energy 2mv^2) hitting a stationary target of mass T. The collision is inelastic, so the objects recoil together with speed r in the direction the car was initially moving.

From the conservation of momentum,
m2v = r ( m + T )

The final energy is
(m + T) r^2 / 2
= 2 (mv)^2 / (m + T)

Let’s do a couple of checks:
T = 0: final energy = 2mv^2
i.e. the initial energy, as expected when the car collides with nothing.

T = m: final energy = 2.m^2.v^2 / 2m = mv^2
i.e. the dissipated energy is 2mv^2 – mv^2 = mv^2
This is the result of your head-on collision, as expected, as this is just that case analyzed in a frame of reference that is travelling with the second car’s initial velocity.

With T >> m, m + T ~= T, and the final kinetic energy approaches being inversely proportional to T, and so as T approaches infinity, it approaches zero. Thus, we get your result, but hopefully in a way that throws more light on what is happening.

The first paradox is thinking that your two cases are equivalent, but I think it is clear that they are not when you look at the head-on collision from the frame of reference in which one of the vehicles is initially stationary (like the wall) and the other is travelling at 2v (like the car that hit the wall). The difference is that the vehicles in the head-on collision do not end up stationary.

I think there is a deeper paradox here, which I think is that we are intuitively thinking of kinetic energy as a blob of stuff that’s attached to the car, but that simple analogy falls apart when we think of multiple observers of the same event travelling at different speed (until Einstein abolished it, we could imagine an extant but unknown universal and absolute frame of reference in which the ‘reference blob’ of k.e. could be calculated, though one always worked with changes because we did not know our velocity in that frame.)

I guess schools don’t typically explore these issues because of the risk of just confusing most pupils.

This site uses Akismet to reduce spam. Learn how your comment data is processed.