Aug. 24, 2011, 7:37 p.m.

MythBusters and the two car vs. single car/wall myth

I have recently watched the MythBuster episode whereby the "fans" felt Jamie was wrong, when he mentioned that "That's equivalent to a single impact going in to a solid wall at a hundred miles an hour" for the compact-compact episode, whereby he was referring to the equivalence of two trucks slamming into a compact car at 50 mph each at the same time from opposite directions.

In the followup episode they tested slamming a car at 100mph into a solid wall, and then subsequently two cars going at 50mph each, into one another. From their results they deduced that Jamie was wrong, and that two cars crashing into each other at 50mph is equivalent to one car crashing into a solid wall at 50mph based on the amount of crumpling that occurred, as well as the forces at play as measured by a force meter in the back of the car(s) - and not 100mph into a solid wall as Jamie mentioned.

This is not a trivial issue, as it does not depend on standard rigid body physics taught in high school or even entry level university courses. That is where I believe most people (and even the producers) went wrong with this whole thing.

I believe Jamie was right and he was wrong, and the final conclusion also can be seen as both correct and wrong, but let me explain why.

Lets consider the actual experiment. We are basically comparing the equivalence of these three scenario's:

Scenario
Scenario

This begs the question: What is considered "equivalent"? Are we comparing total system momentum? Total energy? Total destructive energy transferred in the collision? That I believe is the root cause of the contradictory results.

If you look at these diagrams, then lets consider the equivalence of the following:

Total system momentum

For scenario A, the total momentum of the system is given by:

Formula 1
Formula 1

hence the total momentum of this system is (with a being car a and b being car b):

Formula 2
Formula 2
Formula 3
Formula 3
(assuming each car has a mass of 1500kg and is traveling at 50mph)
Formula 4
Formula 4
By the same reasoning, for scenario B the total momentum is (with
Formula 5
Formula 5
the mass of the wall, and
Formula 6
Formula 6
the speed of the wall):
Formula 7
Formula 7
Formula 8
Formula 8

And for scenario C,

Formula 9
Formula 9
Formula 10
Formula 10

Therefore, if your definition of equivalence is total momentum, then Jamie was right - scenario A and C are equivalent.

Total Kinetic Energy

For scenario A, the total kinetic energy of the system at the moment of impact (relative to a rest state relative to the observer) is given by:

Formula 11
Formula 11

hence the total kinetic energy of this system is:

Formula 12
Formula 12
Formula 13
Formula 13
Formula 14
Formula 14
By the same reasoning, for scenario B the total kinetic energy is (with
Formula 15
Formula 15
the mass of the wall, and
Formula 16
Formula 16
the speed of the wall):
Formula 17
Formula 17
Formula 18
Formula 18

And for scenario C,

Formula 19
Formula 19
Formula 20
Formula 20

Therefore, if your definition of equivalence is total kinetic energy, then these three systems have all different total kinetic energies. That is because kinetic energy varies with the square of the speed. Since the wall has no kinetic energy, it can be removed from the equation. What this does mean for scenario A is that a total of 726kJ is transferred in the collision, spread evenly between the two cars (since their mass are equivalent they receive 50% of this energy each), therefore each car receives 50% of 726kJ which is 363kJ of energy to be absorbed and converted into heat, sound and elastic deformation.

In scenario B, a total of 363kJ is available to be transferred in the car / wall system. Since the wall is inelastic, most of the energy is transferred to the elastic car (think of two billiard balls colliding - the energy is transferred almost lossless from the moving ball to the stationary ball - same principle applies here). In a perfect world with no losses, the impact would have transferred all the momentum onto the elastic car and none to the wall as it would have been perfectly rigid, matching the outcome of scenario A. In practice, some energy losses would have been apparent as the wall is not truly rigid, and would have been expended by heating the wall, and possibly weakening its structure.

In scenario C, 4 times the energy is transferred in the car/wall system since the speed doubled, the kinetic energy quadrupled. This is why the car crash was so much more severe than the one at 50mph.

This result clearly matches the outcome of that episode, which means they considered "equivalence" to mean "total kinetic energy transferred to the car(s)", which is a fancy way to state how much destruction would occur to a car in each scenario.

This brings me back to Jamie's original comment. The scenario they were looking at was NOT one of a head on collision between two cars. It was about the total kinetic energy transferred to the compact car via the two trucks crashing in to it. This is a very different scenario.

Formula 21
Formula 21

In this case, lets consider the system's total kinetic energy again (a = truck 1, b = car, c = truck 2.) - assume the car is at rest:

Formula 22
Formula 22
Formula 23
Formula 23
(assume trucks and car have mass of 1500kg each)
Formula 24
Formula 24

Therefore, the total kinetic energy transferred in the collision is 726kJ, the same as in the two car scenario. But that is where the similarity ends. Since neither of the three vehicles are rigid, the total energy gets absorbed by each one. Depending on complex factors such as the elasticity of the trucks vs. the car, the elasticity of the front of the car being different than the back etc., we can work out how much energy is absorbed by each vehicle. To keep this simple (as my last applied maths class was more than a decade ago), I am going to assume the trucks and the car has exactly the same elasticity, and that the car is symmetrical from an elasticity point of view. I am also going to assume the collision is perfect, in the sense that the left and right trucks impact the car at precisely the same moment - therefore not creating any losses due to dragging of the vehicle or imbalance in impact.

In that case, the total energy transferred to the car is 25% * 2 of 726kJ, which is 363kJ. This can be understood by considering that between the front of the car and truck 1, 50% of the total energy is transferred to both the car and the truck, and the same goes for the impact area between the other side of the car and truck 2. 25% is spent on truck 1, 25% on the front of the car, 25% on the rear of the car and 25% on the rear truck. Based on this result, it is clear that Jamie was incorrect when considering the context of his remark. But the "fans" - though correct in terms of total kinetic energy transferred to the car, were wrong as their remark bears no relevance on the context of the original remark.

Jamie's remark was not about the equivalence of two cars smashing in to each other vs. one into a wall. It was about the equivalence of the energy transfer to the compact car, versus that compact car crashing into a solid wall. At least that is my interpretation. If that is indeed correct, then he and the fans are wrong. Even though the total kinetic energy transferred to the car is the same by the trucks vs. the car against a wall at 50mph, there are some details that changes the effect on the car considerably. Firstly, the car is impacted on both the front and rear, but at 182kJ of energy on each side. That is a very different kind of collision than a head on smash at 363kJ, since the car is semi-elastic. Secondly, the trucks do not have a mass of 1500kg. And this is crucial. I assumed that my two trucks are actually identical in mass to the car. If this is true, then the total energy transferred are identical between the truck compact scenario and the car against a wall at 50mph. But trucks weigh considerably more. In specific, a truck can easily have a mass of 20 000kg. Reconsidering the equation above with real trucks, we get:

Formula 25
Formula 25
Formula 26
Formula 26

Therefore with my real truck scenario, the total kinetic energy transferred to the car at 50mph is 50% of 9680kJ, which is 4840kJ and that is 13.3 times more than the car at 50mph into a brick wall. That is even 3.33 times more than a car at 100mph against a brick wall. Though I do not think one can assume a perfect 50% / 50% distribution of energy between the truck and the car if the truck is that much heavier and has that much momentum versus the 0 momentum of the car. Nevertheless, both the fans and Jamie underestimated the total impact on the car, as is clear from the footage of the total destruction of the car in the truck scene, but only destruction to the front part of the car in the 50mph crash against a wall.

The total momentum of two cars crashing in to each other at 50mph is the same as one car crashing into a wall at 100mph, and double that of a car crashing into a wall at 50mph (which is what Jamie might have been thinking of at the time of his remark). For two cars travelling at 50mph crashing in to each other, the energy transferred to each car is 363kJ. A car travelling at 50mph into a wall receives 363kJ of energy. A car travelling into a wall at 100mph receives 1452kJ of energy. A car that is crushed between two trucks weighing the same as the car, travelling at 50mph each, receives 363kJ of energy. If those were real trucks (as was used in the MythBusters myth), then the car receives closer to 4840kJ (assuming 20 ton trucks).

MythBusters and fans busted!