Hi Guys. It’s Tyler. I was the caller on the show. I will not profess to be a physics expert, I got a C in high school and that’s all I ever took of the subject. So reading through what a lot of you have written has confused me to say the least. But I will say this, I was not speaking from what I believe you are referring to as a vector perspective. Clearly, the rate of approach of the two cars is 120. 60 + 60 = 120. I’m talking about the force exerted on each car, the potential energy that gets dissip?ted. Or imagine it like this, if you were in one of the cars, just how much pain would you be in. My argument is that if I was in one of the cars (let’s forget bgmiller8’s original point about the weight of the vehicles, I had established in the call that the two cars in question were the exact same weight) and hit a brick wall at 60 miles an hour, that would be the same as driving 60 miles an hour head on into another car going 60 miles an hour.
Here’s where I get caught up, if the force is the same as driving into an immovable wall 120 (as To and Ray say it is) then in the equation with two cars, each car would have 120 miles an hour worth of potential energy dissipated into it, which adds up to 240 mph.
Or one more way of looking at it. In both scenarios (immovable wall/two cars), there is what I’ll call a “stop point” where the cars cannot go forward any more. In the brick wall scenario the “stop point” is the surface of the wall. With the two cars, the “stop point” is the forward most point of each car when they first come into contact with one another. In both scenarios, the car is going from 60 mph to 0 mph in the same distance.
You are exactly correct, each car only has to dissipate enough energy the decelerate from 60 mph to 0 mph, it does not matter what the other car is doing if the car eventually comes to rest at the point of initial contact.
Think of it this way, you could either build a thick wall at the point of contact or use a layer of tin foil; in theory neither of the cars would penetrate the tin foil because the forces of the impact would be balanced from the other car.
Each car goes from 60 to 0 mph in the same distance (exactly as it does hitting the wall). Each car absorbs equivalent of its weight times the 60 mph in its crumple zone. In case of the brick wall there is 1 crumple zone to absorb weight of 1 vehicle. In head on there are 2 crumple zones to absorb weight of 2 vehicles. ALL weights are multiplied by the speed…all speeds are 60 mph… so its a wash.
Caller is correct. I got an A in physics and have posed this question to our physics teachers for the last several years!!!
I had visited the site to look up the Monty Hall puzzler. Then I remembered my raging fury at the answer Tom and Ray gave to the 60+60 question for the obvious disregard to fundamental physics first brought to light by Galileo (1564-1642).
Imagine two vehicles identical down to the atom each approaching each other at 60 mph head on. The entire setup is so perfect that, as they meet, the crumpling of the front bumpers is identically matched. Each little fleck of paint coming off pops in the same relative direction. The small fragments of shattered glass from the windshields meet one another and bounce off each other in exquisite perfection.
Imagine the plane between the two cars where all the little bits and pieces met and bounced off each other. Now replace the plane with an absolutely immovable object infinitely thin. Fling the two cars at each other once more. Each fragment from each car again meets the immovable infinitely thin plane exactly as before. However, instead of each fragment imparting a force on the other to bounce back, the immovable plane bounces them back with force each had exerted on the other before.
Now, remove one of the cars and again repeat the wonderfully smashing process. The same force is exerted on the one car as it hits the immovable plane.
So two identical cars coming at 60 mph toward each other imparts the same force as a single car hitting an immovable object. We could suffer through the math to reach a rigorous conclusion, but the bigger question is, why does intuition not agree with this conclusion? Because the assumptions are wrong. As others point out, there are no two identical cars. I’ll take the F350 over the Geo, yeah! Even in two Geos, the one in which the driver just visited the McD’s drive-through has an advantage. Also, there is no immovable object. Most building brick walls will “give”. Unless it is a solid block of concrete the size of building, it will give, and this will provide a “softer” collision than hitting an identical car.
I was taken aback when Tom and Ray gave the answer they did, but then again, maybe there is some additional intuition they had that led them to pick the other answer. What would happen if Tom and Ray ran into each other?
Note: elastic/inelastic collision does not alter the conclusion. It only alters the total force imparted to the cars.
OK, so most people got it right - 60 mph into wall equates to 60 mph head-on into identical car also doing 60. Here’s a tougher question. Why is it wrong to state total precollision momentum of one car is MV? Also, why is it wrong to state total kinetic energy precollision is 1/2MVV? This mistake does not change the results, but what is being overlooked when these statements are made?
“My argument is that if I was in one of the cars … and hit a brick wall at 60 miles an hour, that would be the same as driving 60 miles an hour head on into another car going 60 miles an hour.”
Hitting a brick wall at 60 mph is not the same as a head on collision with two cars at 60 mph each. In the latter case the total energy of the system is greater. The relative motion gives “Clearly, the rate of approach of the two cars is 120”. From the perspective of vector mechanics, you are correct, the equivalence you are speaking of was not clear. I can’t recall your actual radio question, but you state it above. The confusing part is that the total energy is not the same, the relative velocity is not the same so from a pure physics analysis, the two situations are not equivalent.
With respect to the question as explained here, then you are correct. The damage sustained by each vehicles in all three cases is the same.
Tom and Ray are still wrong. The momentum equation for a single particle (car) only deals with 60 mph initially. After the crash the velocities are zero but each car has dissipated the amount of momentum equal to its original. And the question you pose here has exactly the same form except that the wall absorbs no energy and it gets transfered back to the car acording to Newton’s first law of Physics.
Basic. You draw an invisible box around the system of particles. Then apply the conservation laws to the system. One need not invoke a MIT degree, a professorship, physics grades, the spirit of our Physics ancestors or their mother to know that.
However, got it right refers solely to the question as re-stated in the callers posting. It all depends on the question that is being asked. The total energy in both of these cases is clearly different when you do the thermodynamical thing and consider the system which you can define arbitrarily such that the conservation laws do apply when all forces are considered. Shall we invoke papa Einstein or assign chapter 1 of the book Relativity to everyone?
The point about elastic v. inelastic is to show that when the cars stop, where the energy goes. It does not matter, but people need an intuitive way to verify their hypothesis. If the cars stop, then the final condition needs to reflect the fact that the energy and momentum disappeared. Basic high school and college physics ignores the dissipation terms.
“So two identical cars coming at 60 mph toward each other imparts the same force as a single car hitting an immovable object. We could suffer through the math to reach a rigorous conclusion, but the bigger question is, why does intuition not agree with this conclusion?” Because the two situations are not equivalent. The velocities and energies of the system are not the same for both cases. That is why it does not make intuitive sense. Hit anything at the same speed and come to a stop and you remove the same quantity of energy. The problem is not clearly understanding the question or the question to be asked.
You cannot talk about “total” momentum or kinetic energy without defining the system you are looking at. Are you referring to a single car, or does your system include both cars?
Do you guys think I wasn’t clear when i asked the question on the show? I downloaded the Podcast and listened to it a few times. While I probably could have done a better job explaining it, it seemed to me that all the essential elements were presented.
SOLVE BY SYMMETRY: You can assume the two cars are identical, and hit exactly in the center. In fact, if one driver is smoking a cigar, the other driver is also smoking a cigar. When they hit, both cigars fly out the window and collide exactly. It’s like there is a mirror. (One of the cars is from England) So, as the cars crush sideways, even the rear view mirrors hit exactly. Your tailpipe flies up in the air, oops it hits the other tailpipe! A lot of energy is dispersed sideways but both cars stay on their own side. Guys - that is also what happens when you hit a wall at 60.
The reason your gut says you would rather hit a car than a wall, is because you know there is a good chance you will hit off center and spin around or fly over the other guy.
Sorry I was not clear, but I did explicitly state “total precollision momentum of one car” in my question. That’s what I am referring to. One car, traveling 60 mph, precollision. This is not a trick question.
Physics equations give me a headache. Make that a double headache when trying to apply them to a real situation.
Here’s a quote from the April 1991 edition of Consumer Reports:
‘NHTSA crashes cars into a fixed barrier at 35 mph. That’s equivalent to two cars of equal weight hitting each other head-on while each is traveling at 35 mph.’ (page 219 paragraph 2)
This seems to suggest the guys may have been wrong when they told the caller the damage to each car would be representative of twice the speed of each car.
I believe the only way to prove this out is to replicate this head-on crash test of two identical vehicles (MG TD’s may be perfect for this) and then crash a third similar vehicle into a wall at 35 mph. Better yet, crash a fourth similar car at 70 mph into a wall for further comparison.
From what I think I know about Newtonian mechanics, Consumer Reports was right about this. Think about it: If the force was doubled, we’d have the solution to all of our energy problems, and we know that ain’t going to happen.
Any analysis of this problem starts by thinking about simple two body collisions. Consider 3 impact examples involving pool balls and a concrete wall.
Case 1: Consider first, one stationary ball struck by a ball moving at 60mph, analogous to stationary car struck by a moving car. The result of a perfectly elastic collision between identical balls is that the moving ball comes to rest and the stationary ball leaves the collision at 60mph. The result is that both the moving and stationary balls changed velocity by 60mph.
Case 2: Now think about two balls hurtling towards each other at 60mph (120mph relative to each other). After impact, both balls leave the scene at 60mph in the opposite direction. The moving balls changed velocity by 120mph.
Case 3: Let’s add the concrete wall. Fling a ball at a wall 60mph; the ball bounces back at 60mph. The moving ball changes velocity by 120mph as a result of the impact.
Simple, elestic, two-body collisions show that case 2 and 3 result in the same change in momentum for the moving balls. Two objects colliding with equal but opposite momentum is the same as one of those objects hitting a wall.
There are a few conclusions to be drawn:
The son who posed this problem was correct
The father was wrong
Click and Clack were embarassingly WRONG
MIT educations are highly overrated
While I understand the science, I would still prefer to hit a wall than on-coming traffic
Your Case 1 (pool balls) is only true if the shooter put enough bottom english on the struck ball so that it had zero angular momentum when it hit the stationary ball. If it rolled into the ball (as is the case in most pool shots), then the angular momentum will cause the struck ball to continue rolling forward after the collision, but at a reduced speed. This is nitpicking - your basic premise is sound.
I agree with Jamckenna. By varying the the speed of each car…the point of concact is different from the point of rest. So the rate of deacceleration has changed.
A better demonstration of this principle would use that executive toy from the 1970’s with those steel ball bearings hanging from a wooden frame. With those, you could see how the momentum of one ball was cleanly transferred to the next one that it hit. If two balls were swung towards each other, they would bounce back like they hit a wall.