Tyler, you are absolutely right. By analyzing kinetic energy we could say that a single car accident at 85 mph is equivalent to a head on collision where each car is doing 60 mph, but in this case, the single car must absorb all the energy. The only way to have similar damage on all cars (all being equal) is if all of the cars are traveling the same speed. Each car brings the same amount of energy into the equation and each car absorbs the energy it brings.
In the case of a 120 mph into a rock wall, the energy is nowhere near equivalent to a head on collision where each car is going 60 mph. The energy involved in the single car accident is twice the energy of the 60 mph head-on collision, although the energy that the single car absorbs is actually 4 times what either of the head-on cars absorbs.
Here’s a simple way to think about the problem, without involving any math. Imagine one car traveling at 60m/h. An external force is applied to the front end of the car to cause it to stop suddenly. Does it matter what it was that delivered that force? It could have been an immovable wall, a car of equal mass traveling at the same speed in the opposite direction, or a car of twice the mass traveling at half the speed in the opposite direction.
Of course, I’m assuming a perfectly inelastic collision, but the logic is the same for a perfectly elastic collision, or anything in between.
AMEN!
Had the lamebrains thought a second about what the supposed father’s theory implied–i.e., that to be equal,
the 60mph vehical should hit a 0mph veehickle (with that specious reasoning of adding velocities), it would be
obvious that that was boneheaded reasoning.
As part of my job I do shock testing for a living at a high-tech company. Lots of discussion of physics going on here, but only a few people have touched upon what seems most interesting to me…Change in velocity over time as experienced by the all-too-squishy passengers.
The proverbial brick wall has very little give: whatever hits it stops moving RIGHT NOW, and the only thing limiting the deceleration felt by the passengers is the crumpling of their own vehicle. Hitting another vehicle introduces 2 crumple zones, BUT, off-angle hits, going over or under, or different weights of the 2 vehicles make the head-on scenario a real ‘situational’ thing. For sure, head-on into a much heavier vehicle means you experience a much worse situation as the heavier vehicle may not crumple much, AND, it most likely will drive your vehicle backwards after the impact, causing lucky you to experience much greater accelerations, as you rapidly go from 60MPH in one direction to, say 30MPH in the other. Now you have a 90MPH velocity change…
So, head-on impact is better than the brick wall IF you can choose your target to be the weenie in the lightweight gas sipper versus your massive SUV. The gas sipper driver probably would be better off choosing the wall instead of the SUV as his kamikaze target.
The initial momentum is mV + m(-V) = 0. The same as after the collision. Energy can be dissipated meaning it transfers to other forms such as heat and work down on crumpling cars. Momentum just transfers motion from one object to motion in another object. Momentum doesn’t dissipate. In the case where the car hits a massive rock wall, the momentum lost by the car is tranfered to the rock and planet earth. The latter having such a large mass that its change in velocity is hard to detect.
You may be confused because Tom and Ray gave you the wrong answer. You are right. I just listened again to your question. It is quite clear and unambiguous, and Tom and Ray instantly gave you the wrong answer. Most of the answerers here agree that your car would stop almost instaneously regardless whether you hit the wall or hit the other car. The effect on you would be the same.
Tyler,
I can’t resist weighing in on this subject, probably because I teach this stuff for a living (did you ever know a professor who could keep his mouth shut?). Lots of posters have correctly analyzed this question and concluded (properly) that it doesn’t matter whether you hit a wall or an identical car. The energy conservation principle seems the most useful to me. Both cars start with identical kinetic energy, after the crash they have none, the kinetic energy has been converted to heat and material deformation. One car has half as much initial kinetic energy, but also only half the metal deformation…its damage is the same in either case.
Many posters talk about the 120 MPH closing speed, then make some kind of energy argument. The fallacy is that the initial kinetic energy is different if you change the frame of reference (that is, if you change the speed of the observer). The observer on the ground sees two masses approaching at 60 MPH, with each kinetic energy proportional to 60 MPH squared. An observer in one of the two cars sees twice as much initial kinetic energy in the one car approaching at 120 MPH than the two cars combined at 60 (120^2=twice 2x60^2). Hey, an observer on an airplane flying by would say that both cars have a whole lot more kinetic energy than that. So stick with the non-moving (ground) frame of reference, where each car is moving at 60 before impact, and not moving afterward.
After reading 86 answers, I can see that about 70 of you need to go back to Physics 101, along with Tom and Ray. You can’t throw around scientific terms such as (energy, momentum, velocity) loosely, and you can’t apply conservation principles with abandon. So, listen to your basic physics teacher!
Here is a sample “bad physic” answer: one person suggests that, in a collision, two cars will apply equal and opposite forces ONLY when they are identical. When the Mack truck hits the Geo, the forces at any instant are ALWAYS equal and opposite! I believe Newton called that Law #3. And, if you don’t believe that, you might as well not be commenting here.
I disagree with your physics conclusions.> The cars will have kinetic (motion) energy> which is proportional to their speed> squared. A car will continue to move until> all its kinetic energy is absorbed by> deformation and heating of the car. When the> car hits an immoveable wall, the car will> absorb all of the kinetic energy.
Two cars> colliding head on, if they each have the same> mass,will each absorb the same energy as> the wall hitting car. Thus the damage should> be the same.>
A car going twice as fast (120 mph) would> have 4 times the kinetic energy and would> have 4 times the damage as the 60 mph car>> --------------------------------------------
Absolutely, but the analogy is to a brick wall where everything comes to a dead stop. In that case the cars have to have equal masses and equal, but oppositely directed, velocities. Otherwise the equal forces change the velocities unequally.
Your answer to the 60+60 collision question is the final proof you have very strong grounds to demand a refund of the tuition paid to MIT.
Some tutoring in high school physics from Wolfgang will do you a world of good if you persist in answering this type of questions.
I will continue to enjoy the show enormously anyway!
He may have gotten an A in Vector Mechanics, but he’s forgotten how he did it. Velocity ain’t measured in “miles,” it’s in “miles per hour” or “miles per minute,” or other distance per unit of time thing.
The note by elenl looks right on for believable wisdom - confusing a wall with a stationary automobile probably is a lot of the cause of the confusion for the guy who first phoned this question in.
Crashing two cars together vs. crashing into a wall
First of all this is a stupid question!
Why would you choose to crash into a wall or another car? The safe driver ? obviously a consideration beyond the two (or three plus
counting the caller and family) ? would choose to avoid the collision. Regardless, lets investigate the question any way.
Two identical cars collide. Let?s make some assumptions. The both collisions are inelastic and do not bounce back. Do you
have enough sense to wear your seat belts, if not there is no sense in continuing, please skip to the good byes.
People get heart in a collision from the impulse. Impulse is change in momentum divided by time. dP/t now observe car one
from above ? as if you?re already dead and decided to leave your body a bit early (it
helps in rationalizing the poor decision). The car goes from 60 mph to 0 in t sec. t being the time it takes the car to crumple.
It makes no difference if you are stopped by an immobile wall or another car with the
opposite momentum mass times velocity (mV) you will experience the same impulse.
Kinetic energy is only conserved in an elastic collision. It is not conserved in an inelastic collision. Momentum is always conserved. The greater the impulse, the more trouble your in. The change in momentum divided by time is impulse. The shorter the time, the greater the force required to change the momentum (mass * velocity). Second law F=ma.
Last night I was driving home listening to a podcast from radio station Triple J in Australia. Dr Karl hosts a wacky science call in show every Thursday morning, which is podcast here: http://www.abc.net.au/science/k2/podcast/drk_rss.xml
On this week’s program (about 25 mins in), Angus from Brisbane (Queensland, Australia) phoned in with the exact same question! (except his car was traveling at 60 km/h not miles…) Assuming that not too many people from Brisbane Australia listen to NPR podcasts, I’m going on the premise this is entirely coincidental, which opens up a whole new series of questions about the mysterious ways of the universe…
No Marc, you’re wrong. Kinetic energy goes up as the square of the velocity. A car at 60 has .5m60^2 units of energy. Two cars have a total of 1m60^2. A car at 120 has .5m120^2 or .5m460^2 units, which is twice the energy of 2 cars @ 60. Two identical cars hitting at 60 each must each dissipate (absorb) .5m60^2 units, while one car hitting a wall at 120 must dissipate 2m*60^2 units, four times as much. This assumes that the wall is very rigid and unmovable, and absorbs very little of the energy.
In driver’s ed, they tell you that 2 cars going 60 in a head-on collision is the same as 1 car doing 120 hitting a parked car. This may strictly be true, so long as the parked car can be shoved backward without hindrance, but it can lead to fallacies such are being discussed here (that hitting the wall at 120 is the same as a head-on at 60).
Your 50 minutes is up and any hope of tenure should be up as well. You would not be welcome in our University Physics Department. Not at least until you can correctly understand frames of reference.
While not the first to post the CORRECT answer, I join the minority in providing the following FACTS:
Going from 60 mph to 0 mph almost instantly is where the harm is done. The two situations are equally
bad, because in both of them you slow down too quickly–with the same harm in both cases. END OF STORY.
For all of you attention-disorder types, this is not about side impacts, rollovers, or monster trucks vs
scooters. For you academic-wannabe types, remember the Three Doors from Let’s Make a Deal–you missed
that one, too, while spouting off probability terms. Velocity vectors, energy, heat dissipation, and
all sound nice–but it is the impulse (quick change in momentum) which hurts you. In both situations
you will experience the same sudden stop. We few waste our time repeating this to you many, but you
and the brothers are wrong!