Of the many ways of looking at this maybe the easiest is to consider Newton’s third law of motion. By that law the car exerts a force on the immovable wall and therefore the wall exerts an equal but oppositely directed force on the car. Similarly, in the two car case, we posit that the second oncoming car exerts an equal and opposite force on the first car, since it has equal mass and equal speed.
You guys are throwing in too many variables for Tom and Ray to squeeze out of this mess. I want to hear them sqirm next week. I like the simple analogy that at the moment of contact, the wall is effectively pushing back with equal and opposite force. Otherwise it would move. It’s the same as another (equally sized) vehicle colliding at the same speed. Bottomline, in either case the car stops in the same amount of time.
Hey guys, forget all the physics jargon [which you don’t seem to understand anyway]. If a car [at any speed] hits a ‘stone wall’ [meaning an immovable object], it comes to a dead stop ‘instantly’, right? Well, if it hits another car of exactly the same kind [meaning the same mass and the same energy absorption characteristics] heading in exactly the opposite direction along the same path [that is a fancy way of saying ‘head on’], it will ALSO stop dead in the same ‘instant’. There is no difference between hitting the stone wall and the other [equal] car at what ever speed you are going [as long as the other car is traveling at the same speed as you are]. ONLY if your car has no mass will it be like 120 mph because in that situation you will go from 60 mph in one direction to 60 mph IN THE OPPOSITE direction in that ‘instant’. THAT will be like hitting a wall at 120 mph because your velocity [OK, there is some physics jargon] has changed by 120 mph. Jac Conaway Olivebridge NY
I like the both-cars-hitting-the-wall-from-opposite-sides thought experiment. Here’s another, in the form I emailed to the show:
You run into a stationary car, identical to yours, at 60mph. Is that the same as running into a (completely immovable) brick wall at 60mph? I trust it’s obvious that the answer is no, because the other car will crumple and bounce away (i.e., it will absorb some of the energy of the crash). Now let’s say the other car is traveling toward you at 1mph when you hit it (while you’re still going 60mph). Is that the same as hitting the brick wall at 61mph? Again, I trust it’s still obvious that the answer is no. How about the car at 5mph and brick wall at 65mph? The car at 20 mph and the wall at 80mph? The car at 60mph and the brick wall at 120mph?
Or how about this. You and your brother are sitting on the living room floor, and you each roll a tennis ball toward the other; the balls hit directly. Do you expect them both to bounce back at twice the speed you rolled them at?
Arghhh, the talk about velocity vectors is all to complicated to me.
Here is how you could think about this:
What will harm the passenger (apart from listening to this show of course) is high DECELERATION. The right question to ask is how does the passenger or the passenger cabin ‘know’ if the car has run into a wall or an approaching car?
Let’s see: How can we intuitively find out the deceleration for head-on crash with an approaching car or for running into a wall?
Think of what happens with the bumper and the passenger cabin: in both cases (approaching car or wall) the bumper stops practically instantly at the point of first contact and will stay more or less put there for the remainder of the crash. Once you can agree with that there is no more debate. The passenger cabin of course continues to move as long as the crumpling crumple-zone allows it to do so. Of course the passenger cabin doesn’t care if the bumper got stopped by a brick wall or by another car and in effect the DECELLERATION of the passenger cabin will be the same in both cases.
In short, it makes practically no difference if you are running into a wall or an approaching car!
After reading most of this I am assuming that most of us agree that our two friends on car talk got it wrong, and Tyler got it right. I posted this in an e-mail right away, but alas they don’t seem to have been corrected during the show.
I have been teaching High School physics for 25 years, hence my login name. The simplest explanation for me is this.
If you hit a car with the same mass going the same speed in the opposite direction you should come to a complete halt, as will the car hitting you head on. 0 to 60 in a certain amount of time and space.
If you hit a wall with the same speed the same thing will happen, 60 to 0 in a certain amount of time and space. I would prefer the head on car as the chances of missing (hence the game of chicken) are greater, and spin can take up energy as well.
I await the professor from Harvard’s appearance on Car Talk. Hopefully he won’t get to complicated.
Tyler, you are right and Click and Clack were wrong. Somewhere in this discussion is a post by myself where I used a rather similar argument as you did in your post.
Keep it simple:
Imagine blindfolded passengers in the two identical cars approaching each other at 60 mph each. When they collide, symmetry and logic dictate that they stop at a fixed point on the road, which is their point of collision. To the passengers this is the same as hitting a wall, essentially 60 to zero in milliseconds.
You guys have neglected the most important experimental finding from this discussion: Tom and Ray have at least 28 listeners. Now that’s really an unintuitive result.
Most of this discussion reminds me of the story about Lincoln. During a debate he challenged his opponent with the following question: “Suppose we call a cow’s tail a leg. Then how many legs does a cow have?” The other fellow answered, “Five, of course!” To which Lincoln replied, “No, its four. Calling it something does not make it so.”
In reality, cars and even brick walls are not totally elastic, nor are they totally inelastic. Recall that the original question concerned two identical cars each moving 60 mph colliding head on, vice one car moving 120 mph colliding with a wall. Presumably, the wall is flat, and the car is moving perpendicular to its surface. (It makes a big difference.) OK since the kinetic energy (KE) is mVV/2, the KE of the two cars is 2mVV/2, while the KE of the single car going twice as fast is m2V2V/2, i.e., twice as much. The magnitude of the total momentum is the same, but when we express the momentum as a vector, the total momentum in the case of the colliding cars is zero, while that of the car hitting the fixed wall is 2mV (In the foregoing, V=60 mph), both in the Galilean reference frame of an observer at rest with respect to the wall.
Now as I said, in the real world the collisions will be neither totally elastic nor totally inelastic. If it were totally elastic, after the collision the cars would be moving at their initial speeds but in the opposite directions, and no energy would have been dissipated. However, since the change in direction would be instantaneous (absent some complicated energy storage then release mechanism), the acceleration would be infinite. Since the human body cannot tolerate more than about 200g acceleration, all the occupants would be dead. In the case of perfectly inelastic collisions, all the kinetic energy would be absorbed, and the question then becomes one of the deceleration rate. The total KE must be dissipated, so the into-the-wall case is potentially more demanding. In the limit (your typical one-star-rated minivan) everybody’s dead. At the other extreme (Audi or Mercedes street cars or real race cars) everybody walks away. I can say this with some authority having put my race car into a wall at 130.
An experiment as suggested by some of the responders would answer the question only for the particular vehicles tested, and not in general.
The question as presented had nothing to do
with forces or vectors. It simply had to do with
energy which is what deforms cars in crashes.
The severity of a crash largely depends on
the amount of kinetic energy that is
converted to vehicle deformation in the
crash. So let?s look at the two scenarios:
Case 1: Two cars ,each weighing m pounds,
traveling at V mph hit head-on.
Energy of car 1: ? mvv (read ?1/2 m v-squared?)
Energy of car 2: ? mvv
Total energy: ? mvv + ? mvv = mvv
This is the total amount of energy coverted to vehicle
deformation in the crash
AND this energy is shared equally by the two
cars at the crash, so each absorbs only half
of it, or ? m vv (not coincidentally,
exactly the energy it brought INTO the
crash!)
Case 2: One car traveling at 2v mph hitting
a wall:
Total energy: ? m (2v)(2V)
Which is ? m (4vv) or 2 mvv
(Note that 2mvv is FOUR time 1/2 mvv)
AND, there is only one car now to absorb all
this energy because the wall is immovable,
so it actually has to absorb
FOUR times as much energy as either car in Case 1,
which means it gets creamed four times as
badly.
The key point is that the severity of the crash
depends on the energy released, and the energy
varies with the square of the speed. Two objects at
60 mph have only half the kinetic energy of one at
120 mph.
So, personally I?d much rather be in the car
in case 1 since only one fourth as much
damage should be done to my car (and to me)!
The energy isn’t 60 mph times something, but the deceleration from 60 to zero. A car that does crush might do it in .02" (a wild guess), one that doesn’t might do it in .01", involving a lot more energy.
As I recall, the original situation was two identical cars, so whether there is crushing or there is a Geo, is immaterial. The issue is whether the “target” moves back or not.
Walls crush and flex letting the vehicle decelerate slower (youtube.com/watch?v=mwk8f_5Yw00). You’d have to know a lot about the wall in order to predict its behavior.
Comparing hitting walls with hitting cars is comparing apples to oranges. Think of comparing only identical cars crashing: going 60 and 60, or 120 and zero. The result will be the same.
With the two 60-mph cars, the “target” doesn’t move during the impact (symmetry) unlike the wall that does move. Ignoring all other factors… like the possibility of glancing off or missing entirely… the wall would be a softer target than the oncoming car.
Whatever you do, don’t hit a wall traveling at 120 mph.
As TACLCT says, it’s the collision energy you need to worry about–
Given: Two vehicles of equal mass; one immovable wall; Kinetic energy is proportional to the velocity squared… (E=1/2*mv^2)
The single car at 120 mph has 4 times the kinetic energy of one car at 60 mph. Meaning, the single car dissipates twice the energy hitting the immovable wall as the two cars combining in a head on collision. It also means a passenger in the 120 mph vehicle has 4 times the energy to give up as any of the passengers in two cars traveling at 60 mph.
Therefore 2 Cars of EQUAL mass hitting head on at a combined speed of 120 mph (60 mph each) IS NOT the same
as one car hitting an immovable wall at 120 mph.
QED
Tom and Ray did flunk mechanics, yes? I side with the caller on this question, and think they got it wrong.
Two cars of equal mass hitting each other head-on produces exactly the same result as one car hitting a stationary wall at 60 mph. In both cases, from the initial point of contact to the rest position, there is no net displacement. That is equivalen to one car hitting an immovable stone wall only if that car is going 60 mph.
If that single car was going 120 mph, it would have 4 times the energy as if it was going 60 mph and would not be equivalent to the two-vehicle 60 mph case.
My physicist son explains it as a matter of using equivalent frames of reference to determine the energy involved. Two vehicles each going 60 mph have twice as much energy as a single vehicle of the same mass going 60 mph, but that energy is dissipated in two cars. A
single vehicle going 120 mph has four times the energy, all of which is dissipated in that one vehicle. Definitely not equivalent.
In any event, if I had the choice as to which situation I’d rather be in, the answer is easy. I’d rather hit the stone wall because fewer people would die!
Speed bump got it right. If the wall is capable of absorbing energy, then the single car could be going faster than the cars in the head on collsion, but since the argument assumes an immovable object, we assume that all kinetic energy is absorbed by the car. The statement that a car traveling twice as fast actually carries 4 times the energy is also correct, which means that a single car going 120 mph must dissipate twice the sum of the energy in a 60 mph head on collision. But in the head-on collision the cars split the sum of their kinetic energy, while a single vehicle must dissipate the entire energy of the 120 mph single car collision, which is 4 times what either of the head on cars must dissipate (assuming equal mass of all cars). Only in the example of a single car collision with an immovable rock wall where the vehicle is traveling 60 mph does the vehicle dissipate the same energy of either of the two vehicle in the head-on collision.
On an interesting note: the energy in an impact of a single car crashing into an immovable wall at roughly 85 mph is the same as the energy of a two car head on collision where each car is traveling at 60 mph (85 = 60 times the square root of 2 - the result of the kinetic equivalence equation). In this case, the single car collision still has to dissipate twice the energy of what either car in a head on collision must dissipate.
When you get stopped after hitting the GEO,
which would probably “go mostly under”
the 350 Ford / with little loss of speed,
you would want to back up and see if they are ok…
You would be correct if you assume that ONE single team has as many horses as TWO teams together.
Most of the answers are far more complex than necessary, and some even ignore the second part of the scenario. It’s two equal cars at 60 mph each vs one car approaching a wall at 60 mph (not 120 mph).
Let’s picture the first part as two identical cars traveling toward each other at 60 mph on a collision course, one from the right and one from the left. Let’s take the perspective of the one coming from the right. (I’m the driver, so I choose the one on the right.) At collision, this car will go from 60 mph to 0 mph in an instant, converting all that momentum to heat and light. (Exactly the same thing happens to the car coming from the left, but that’s for some other poor slob to drive.) No need to calculate momentum – just remember 60 mph to 0 mph. Now assume that the car on the left is replaced by an immovable wall. My car coming from the right will still go from 60 mph to 0 mph. Same change in momentum, same crunching of the car. What apparently confuses some people with problems like this is that if you replace the immovable wall with another car or movable object just sitting there, the car moving from the right will dissipate some of its momentum by moving the other car backward (with both moving to the left at some reduced speed). This is not the same as hitting an immovable wall (although I would not like to be in that collision, either). Momentum is still conserved, just not in the same way since we no longer have two moving cars or a car and a fixed wall. So from a physics perspective, hitting a wall is the same as hitting an identical car coming
at you at the same speed (at least if we ignore the bodily damage if we are hit by flying debris).
The intuitive feeling that hitting a “brick wall” at 60 mph would be somehow less severe than hitting another car coming in the opposite direction is only because immovable walls do not exist in the real word. An immovable wall absorbs no energy. This just does not happen in real life. So, hitting any real object at any speed would be less (if only infinitesimally so) severe than hitting an oncoming car with the same speed. However, the point of the original question is: is a 120 mph collision equivalent to a head-on 60 mph collision? The answer is clearly no. The 120 mph collision represents twice the energy of the head-on 60 mph collision of two cars, or 4-times the energy of a single 60 mph car. As for the Geo, who would drive one of those in the first place? It’s a nonsense question.
This is a great way to understand the amount of energy in either of the crashes. However, it isn’t clear that in the case of the head on crash, two cars will dissipate the crash energy while the energy of a single car accident with an immovable rock wall will have to absorbed by a single car. And this comes down to the consideration of the original question. The energy of a two car crash where each is going 60 mph is equal to a single car crash into an immovable rock wall at 85 mph. However, the damage to the single car will be twice the damage of either of the head-on crash cars.
In order for all cars to have equal damage, they must be going at the same speeds. As KiwiPhil mentioned, any 2 ton car traveling at 60 mph must dissipate 7194 kJ. In the case of a head on, each car brings 7194 kJ to the accident and each car absorbs 7194 kJ. To come out of the single car crash with the same amount of damage, the single car must bring and absorb 7194 kJ, meaning that it too must be going 60 mph.
If I may be so boring, allow me to present the equivalent crash energy math in another fashion without assigning values (since I went to the trouble to do it 
Vehicle 1 & 2, each at 60 mph, impact head on
Vehicle 3 is the single vehicle hitting a wall
M1 is mass of vehicle 1
V1 is the speed of vehicle 1
…
M1 = M2 = M3 (all cars equal)
V1 = V2 (60 mph)
Solve for V3
? * M1V1^2 + ? * M2V2^2 = ? * M3*V3^2
substitute M1 for M2 and M3 and divide both sides by ? * M1
V1^2 + V2^2 = V3^2
substitute V1 for V2
2*(V1^2) = V3^2
V3 = V1* (square root of 2)
If V1 = 60 mph, then V3 = 85 mph