Want a flower with that touchie feelie bad answer?
Assuming that both hitting the wall and hitting the other vehicle will decelerate your car from 60 mph to 0 mph, the forces (F=ma) in each situation should be same, right? I ran into a tree once, and it wasn’t pleasant. I wish there had been a GEO in front of that tree.
Yes!
Momentum is a vector quantity just like velocity. Along the one dimension of the road along which both cars are traveling one of the cars (clay balls, billiard balls) have momentum mv and the other m(-v), so the total momentum is mv-mv=0. Clearly when the clay balls are both at rest after the collision the momentum is also 0. The billiard balls which recoil at the same but opposite directions also sum up to zero. So as turbonium states momentum is preserved in either case, whereas kinetic velocity is only preserved for the elastic case.
If you are in a car moving 60 MPH and collide head-on with an identical car moving 60MPH, the closing speed is 120 MPH. However, that does not mean the situation is like hitting a solid wall at 120MPH. That would be the same closing speed, but a different obstacle. The equivalent situation would be going 120 MPH and hitting an identical car that is stopped.
If you want to talk about walls, it’s the same as hitting a solid wall at 60MPH. The 2-car collision is fully symmetric. If you put up a giant sheet of rubber across the road where the cars collide, each car moves towards the sheet at 60MPH, collides, and both may bounce back with a smaller speed Vb. The sheet won’t move. You can take away the 2nd car, put a solid wall behind the sheet (which didn’t move, remember?) and the remaining car would experience the same forces, hitting the sheet at 60MPH and bouncing back with speed Vb.
Probably no coincidence. Car Talk is available as a podcast too.
You seem to have struck a nerve with Crash. Perhaps the Monty Hall memory was too much. This question was off to a bad start right away. You can see below that Wilber and Ortho both reached the same conclusion that you, many others, and dozens of search results all did. Some die hards who think these answers are bad have even suggested that there is a typo on the website of a well-known testing organization. The crash test people say 35 mph into a wall is equivalent to a 35 mph head on collision with an identical car going 35 mph. The real “bad answer” guys suggest the first 35 is a typo and should read 70. Maybe some people are confused about bumper damage and passenger safety. It is, indeed, the impulse (FdT = MdV).
Viewing the two-car collision from a frame of reference with one car at rest is a nice way of making the point.
Also, discussions about energy are somewhat interesting, but the significant effect on occupants of each car will be determined by the acceleration experienced by each part of each person’s body during the impact. A rough indication of the relative effect of different impact speeds on the resulting acceleration can be calculated from the loss of momentum of the car and its initial mass. Thus, in an impact with an immovable wall, the occupants of a car at 120 mph will undergo roughly twice the acceleration as they would in an impact with an identical car, each going 60 mph. The acceleration will be the determining factor for physical damage to the occupants.
Tom and Ray got it wrong.
There are a number of factors in this:
- velocity
- mass (essentially weight) but all the cars are the same, so this won’t matter
- momentum = mass times velocity; this is useful in actual car crashes where mass, velocities, and crash angles are different, but not in this case because we only have one type of car and it is head on.
- energy; this is the key factor. A 60 mph car hitting a wall or identical car headon will have to dissapate (1/2) times mass times (velocity squared) of kinetic energy into deformation, heat, and sound.
A vehicle hitting a wall at 120 mph will have four times the energy to disapate: speed kills. - Force = mass times deceleration; not useful in this question.
- deceleration: this is critical for what happens to the driver, but with identical cars at 60 mph, the deceleration should be about the same because the crumple zones are the same. At 120 mph, the energy is four times that at 60 mph and the time-to-stop-the-car will be less: You don’t want to be in this car.
Assuming that both hitting the wall and hitting the other vehicle will decelerate your car from 60 mph to 0 mph, the forces (F=ma) in each situation should be same, right?
Only if that other vehicle is the same (mirror image) as yours traveling at the same speed directly at yours. In real life that usually does not happen, but a head on is close enough to give the same basic result as hitting an immovable wall.
I like you would prefer there be a stationary GEO without driver or passenger in front of that tree.
If it had some kind of impact meter, it would register twice the impact from the two cars as it would from the car vs wall because both cars would impart their force to it."
True and each car would absorb half the force. In the case of an immovable wall the wall would not absorb any of the force (only resist it) so the force on the car would be the same. Sort of like hitting your fist against a concrete block and hitting your fist against against each other.
I sent the following e-mail message to Tom & Ray on May 25:
"Apprezzati Fratelli Magliozzi:
I write to criticize your surprisingly dismal performance in your discussion of automotive collisions with Tyler of Crested Butte (aired Saturday, May 24). It sounded as though you both had forgotten what you learned in Physics 101.
It became clear that the discussion was going off the rails when Tyler began speaking of the “force” of the collision and you did not correct him. The physical parameter that is the key to analyzing each of the proposed collisions is the ENERGY (not “force”) released therein. No doubt, automotive engineers use a number of criteria in designing the crash survivability of their products but the amount of kinetic energy the structure can absorb while still protecting the occupants within surely is high on their list. For two cars of the same mass, moving in opposite directions with the same velocity, the kinetic energy of each is 1/2 mv2. (Your e-mail page does not provide for superscripts.) For a head-on collision between the two, the total energy released is the sum of these, or mv2, that is, TWICE AS MUCH as that for one of these cars crashing into an unyielding object (e.g. wall, tree). Tyler’s point about the “force” (i.e. crash energy) being shared by the two cars has some merit – in principle. But suppose one car is built more robustly than the other or has some components that will become penetrating “warheads” in a crash. Then the occupants of the other car will be at a severe disadvantage. Would any rational person rely on the theoretically equal distribution of crash energy and component reliability between the two vehicles? On this score, I well remember the words of our very excellent instructor at an AAA-sponsored Driver’s Safety Course: “You must do ANYTHING to avoid a head- on collision!” He elaborated that “anything” included going off the road and into a tree, if necessary.
Finally, I turn to your outrageous statement that the two- car collision was the equivalent of one car crashing into a wall at twice the velocity. Whatever happened to the v- squared law you learned in freshman or even high school physics? To which dark recesses of your brains did it disappear? Twice the velocity yields FOUR TIMES the crash energy! Please don’t try this in practice. I enjoy your show too much to lose you to such idiocy.
Va bene!
With affection,
Barry Freed, Ph.D.
Bronx, NY
(Your writer holds bachelor’s and master’s degrees in physics and a doctorate in bioengineering.)"
I finally listened to that podcast, the brothers were definitely off base with that answer (not just a misunderstanding of the question, they really missed it).
In my view, there is a plane of symmetry at the point of collision between the two cars. If the cars are absolutely identical with same mass, same velocity and same energy dissipation capability, then no part of either car will cross this plane of symmetry. In effect, the plane of symmetry is the proverbial immovable object. Therefore, I conclude that the caller was correct and the effect of two IDENTICAL cars crashing at 60 mph each is the same as one car crashing into a rigid wall at 60 mph. Of course, I’m not sure I would want to personally test this hypothesis. Aren’t there some automotive test engineers out there who know the answer to this question.
When I heard the podcast, I did not believe the caller was concerned with pieces of cars turning into projectiles, filing insurance claims, the possible difference in the number of deaths, the extent of damage to cars, walls, and trees, or whether you should brake, swerve, or speed up. It was posed as a physics question and a decade-old dispute with his dad. The son believes a head-on collision between two cars each going the same speed is “the same as” one of those cars crashing into an immovable wall. The dad holds that the head-on would be equivalent to one car hitting the wall at twice the speed, presumably because the second car is moving at that same speed. The son is right, of course, even though he did not do a good job of making his case.
There is nothing I can add to the many fine explanations given here in support of the son; however, I would like to offer my kudos to what I found to be two intuitively pleasing solutions. First: to those many who included in their explanations an immovable wall, sheet, or elastic/rubber barrier between the two colliding vehicles. You very clearly make the case that the two car crash is equivalent to the one car crash. Second: to those who point out that the destructive impulse of the sudden stop is equal in both crashes as the car’s occupants lose their equal velocities in equal times in either crash. Plainly this is not the case for a car with twice the speed hitting an immovable wall. The physics equations work, but you must have the correct setup.
Finally, for the person who intimated that physics doesn’t always work by commenting that physics says wide tires don’t matter; this is not so. Normally we think of the surface area not mattering when we calculate frictional forces. This is because, for hard, smooth (at the macroscopic level) surfaces the larger contact area results in less pressure per unit of area, resulting in less friction per unit of area. Soft, sticky surface areas do not share this quality. Conventional problems usually do not deal with “sticky” friction. Physicists are well-aware that cars use wide, slick tires–and heat them up before the drag race to soften them further and increase the friction.
I waited a week before going to the Car Talk website hoping that the mistake would be fixed on the Air. Then I came here and found the mistake had been properly dealt with by many before me. After reading over one-hundred entries, flower child’s comment here stood out as negative nonsense. Why not contribute to the discussion and respect others who do? I don’t want to ignore my own advice, but your response cannot be dealt with on a relevant basis, since you address nothing on topic. You would not be right, but you could say, “You’re answer is wrong because the Kinetic Energy of the wall is . . .” Maybe then you’ll feel better!
Tyler, you explained it well enough. The writers who referred to ONE vehicle going 120 Mph were not paying enough attention, and those who were exploring the flexibility of the wall were nitpicking the question, as the phrase ?solid wall? is meant to refer to a completely immoveable object. Your question certainly brought out the nitpickers, including me, for nitpicking the nitpickers. Thanks.
I also emailed Car Talk, and I have been told that: ?In a week or two, (Car Talk Radio) hope(s) to feature Tom and Ray’s friend, and Harvard Physics Professor, Wolfgang, chiming in on this very topic.?
I had posted my comments on this topic to a general car-talk e-mal address, and Dougie told me about this thread. Reading through the answers, I see at least one person who agrees with me(!) – the two cases are equivalent. I did not see mention of Newton’s 3rd law, which is relevant when talking about the wall case. Also, being trained in high-energy theory, I like to siimplify the problem, so cars and walls are infinitely rigid and incompressible and energy is conserved. I think the simplest way to esplain it is with an analogy: two pendula attached at the same point with identical balls at the end of wires of equal length. Start with both balls initially at rest displaced by the same angle on opposite sides of the center line, and then release them. The balls bounce off each other and come to rest at the initial starting positions (assuming no friction, air resistance, infinite rigidity and incompressibility, etc.). If one ball is replaced by a wall, the first ball would bounce back again coming to rest at its original starting position – just like a billiard ball bouncing off the cushion (or whatever it is called).
I agree on the mirror symmetry argument that a car hitting an immovable wall at 60 MPH is indistinguishable from two cars hitting head on, each of which is going 60 MPH. I was puzzled for a while by an alternate way of looking at the problem. If viewed from the occupant of one car, assuming that this car is stationary, in one case a wall is coming toward him/her at 60 MPH, and in the second case a car is coming at him/her at 120 MPH. At first glance, being hit by a wall at 60 MPH would seem less damaging than being hit by a car at 120 MPH. However, after the collision with the wall, both the car and the wall are going 60 MPH. After the collision with the second car, both cars are going 60 MPH, the same as the collision with the wall. The two accidents are identical!