MythBusters: Does two cars crashing at 50 mph each = a 100 mph crash into wall?
05-16-10, 01:31 AM
#16
Sorry guys I'm hijacking this thread with math.
Upon reviewing the previous equations, I think I'm pretty set on the following explanation.
2 cars, each with mass m, heading at each other at velocity v --> the total energy of the system is:
(1/2)(m)(v)^2 + (1/2)(m)(v)^2 = m(v)^2 = total Energy
We're assuming that 100% of this kinetic energy is transferred into the mechanical energy that is at work in disfiguring these cars up (i.e. total energy transferred into "wreckage" for both of these cars, equally). In truth, that's probably not a perfect assumption, I'm sure some portion of the energy is simply lost as heat to the surroundings, but it's an OK assumption.
So the question is, what is the equivalent one car-stationary wall situation to the one described above. That is, what mass would a single car have to be going at twice the velocity in order to yield the same results?
Dear lord, I've thoroughly just confused myself now. The equations say one thing, but just intuitively, look at what it is saying. a single car of half the mass going at double the speed will yield the same results? That sounds ludicrous and probably is.
I'm sure I'm making some fundamental basic errors in assumption here, which is resulting in these kooky results/equations.
Someone who knows physics please chime in and clarify haha.
Upon reviewing the previous equations, I think I'm pretty set on the following explanation.
2 cars, each with mass m, heading at each other at velocity v --> the total energy of the system is:
(1/2)(m)(v)^2 + (1/2)(m)(v)^2 = m(v)^2 = total Energy
We're assuming that 100% of this kinetic energy is transferred into the mechanical energy that is at work in disfiguring these cars up (i.e. total energy transferred into "wreckage" for both of these cars, equally). In truth, that's probably not a perfect assumption, I'm sure some portion of the energy is simply lost as heat to the surroundings, but it's an OK assumption.
So the question is, what is the equivalent one car-stationary wall situation to the one described above. That is, what mass would a single car have to be going at twice the velocity in order to yield the same results?
Dear lord, I've thoroughly just confused myself now. The equations say one thing, but just intuitively, look at what it is saying. a single car of half the mass going at double the speed will yield the same results? That sounds ludicrous and probably is.
I'm sure I'm making some fundamental basic errors in assumption here, which is resulting in these kooky results/equations.
Someone who knows physics please chime in and clarify haha.
05-16-10, 01:47 AM
#17
The cars going at each other at the same speed has twice the kinetic energy of that one car going into the wall at the same speed. The accident should have looked twice as bad but it doesn't because in the two car collision, the energy is spread out to both cars, going in opposite directions.
Mass and velocity are not linear with each other. Halving mass while doubling velocity will not give you the same kinetic energy. You'll have more energy since the velocity is squared.
Mass and velocity are not linear with each other. Halving mass while doubling velocity will not give you the same kinetic energy. You'll have more energy since the velocity is squared.
05-16-10, 08:14 PM
#23
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I thought it was pretty sobering to see the car hitting the wall at 100 mph. No survival space at all for front seat occupants. I wonder what a 2IS would look like under the same circumstances. I'm betting not a lot better.
05-17-10, 04:08 AM
#24
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Yeah, you wouldn't survive in any car in the same crash.
05-17-10, 06:48 AM
#25
Newtons third theory of velocity will always hold true in this case. No matter what the speed is 1 car hitting at even 100 mph will never be the same as 2 cars hitting each other at 100 mph. The reason why is because of the transferrance of energy theory.
1 car hitting a stationary object will absorb the TOTAL energy impact of that crash b/c the object that it hits is unyielding.
2 cars hitting each other at an identical speed will not absorb the TOTAL energy impact, but they will disperse the kinetic energy amongst the 2 objects because BOTH of the object are yielding to one another, being Newtons third theory into play (equal but opposite reaction).
Basicaly,, 1 car,,stationary object,, energy has nowhere to go but back to the car itself.
2 cars hitting each other, both NON-STATIONARY objects, energy will be equally displaced between cars 1 and 2.
1 car hitting a stationary object will absorb the TOTAL energy impact of that crash b/c the object that it hits is unyielding.
2 cars hitting each other at an identical speed will not absorb the TOTAL energy impact, but they will disperse the kinetic energy amongst the 2 objects because BOTH of the object are yielding to one another, being Newtons third theory into play (equal but opposite reaction).
Basicaly,, 1 car,,stationary object,, energy has nowhere to go but back to the car itself.
2 cars hitting each other, both NON-STATIONARY objects, energy will be equally displaced between cars 1 and 2.
05-17-10, 07:51 AM
#26
2 cars of mass M hitting each other at 50 MPH is, indeed, not the same as 1 car of mass M hitting a stationary wall at 50 MPH. That's the simple part.
(1/2)(m)(v)^2 + (1/2)(m)(v)^2 = total energy
mv^2 = (1/2)(M)(2v)^2
where m = mass of car 1 and car 2
M = equivalent mass of a single car necessary to simulate the same collision for a one car-one wall set up
2mv^2 = 4Mv^2
M = 2m
Edit: That's totally wrong. That's yielding a result of M = (1/2)m
Interesting. I think this result is implying that 2 cars of mass m hitting each other at velocity v is equivalent to 1 car of mass (1/2)m (not 2m like I stated previously) hitting a single stationary wall at velocity 2v
Someone help out and clarify the math please.
(1/2)(m)(v)^2 + (1/2)(m)(v)^2 = total energy
mv^2 = (1/2)(M)(2v)^2
where m = mass of car 1 and car 2
M = equivalent mass of a single car necessary to simulate the same collision for a one car-one wall set up
2mv^2 = 4Mv^2
M = 2m
Edit: That's totally wrong. That's yielding a result of M = (1/2)m
Interesting. I think this result is implying that 2 cars of mass m hitting each other at velocity v is equivalent to 1 car of mass (1/2)m (not 2m like I stated previously) hitting a single stationary wall at velocity 2v
Someone help out and clarify the math please.
Double the speed, you have 4x the energy involved, so you'd need 1/4 the mass of a single car, or in this incase, 1/2 of the 2 cars (1/4 x 2M = 1/2).
If you wanted to simulate the same scenario with one car crashing into a wall, it would be to have a velocity of sqrt(2) * 50 mph, or 70.7 mph.
05-17-10, 11:07 AM
#28
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How about if one car is stationary while the other is traveling 100mph? Would that be equivalent at each traveling at 50mph into each other? I know in theory the 100mph car has twice the energy of 2 cars at 50mph, but I also thought speed is relative.
05-17-10, 11:22 AM
#29
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I believe that one car going 100 into a parked car would result in worse damage than the two cars at 50. Though the damage definitely less than going into a solid wall because the parked car obviously absorbs a lot more energy.
05-17-10, 12:10 PM
#30
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The steel wall is the biggest variable because it is solid, does not move, does not crumple, all force is distributed to the crashing car where the car is not a solid stationary object, will move, and will crush meaning more energy reacting on both objects leading to a different result and the effects not as devastating.