Lööpitystä, uutisia, ym. sälää / SYYSKAUSI 2015

[Örfil]

Haluan sanoa, että se törmäävien autojen (tai auton ja moottoripyörän) "suhteellinen nopeus toisiinsa nähden" ei suoraan kerro törmäyksen voimmakkuutta.

Siinä eilisillan väännössä selitin tuon ja tuo onkin sinällään totta. Absoluuttinen voimakkuus eli liike-energia määräytyy kummankin möhkäleen yhteenlasketusta liike-energiasta, mutta on helppo päätellä että mitä suurempi se suhteellinen nopeus on, sitä enemmän tuhovoimaa vapautuu kun verrataan samaa tilannetta eri nopeuksilla keskenään. Kuinka paljon tarkalleen vapautuu, riippuu tietenkin esim. siitä onko suurin osa siitä nopeudesta isomassaisella vai pieni-heh-massaisella klöntillä.

Tässä menee herkästi käsitteet sekaisin, aihe on helppo mutta toisaalta vaikea. Kun sotketaan samat massat, eri massat, joustamaton kappale jne. niin pian ollaan juurikin siinä tilanteessa että kaikki puhuu ihan just omaa sivupolkuaan asiassa.

 

[Örfil]

Nojuu. Lisäksi syön lihaa niin olen senkin vuoksi ärtsympi.

Olen rauhallinen kuin herne. Nenässä.

 

[Örfil]

Mulla on Voima-lehti, jolla voin läpsiä ja räpytellä.

Wiki:

Lehti on poliittiselta suuntaukseltaan viher-vasemmistolainen.

Eikö lehdellä ole muuta käyttöä?

 

[Antsa Rieppo]

Olen rauhallinen kuin herne. Nenässä.

Olen rento kuin retiisi. Sanoo yks asiakas.

 

[Dominic]

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Is two cars colliding at 50mph the same as one car colliding into a wall at 100 mph?
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I was watching a youtube video the other day where an economist said that he challenged his physics professor on this question back when he was in school. His professor said each scenario is the same, while he said that they are different, and he said he supplied a proof showing otherwise.

He didn't say whether or not the cars are the same mass, but I assumed they were. To state it more clearly, in the first instance each car is traveling at 50mph in the opposite direction and they collide with each other. In the second scenario, a car travels at 100 mph and crashes into a brick wall. Which one is "worse"?

When I first heard it, I thought, "of course they're the same!" But then I took a step back and thought about it again. It seems like in the first scenario the total energy of the system is the KE of the two cars, or 12mv2+12mv2=mv2. In the second scenario, it's the KE of the car plus wall, which is 12m(2v)2+0=2mv2. So the car crashing into the wall has to absorb (and dissipate via heat) twice as much energy, so crashing into the wall is in fact worse.

Is this correct?

To clarify, I'm not concerned with the difference between a wall and a car, and I don't think that's what the question is getting at. Imagine instead that in the second scenario, a car is crashing at 100mph into the same car sitting there at 0mph (with it's brakes on of course). First scenario is the same, two of the same cars going 50mph in opposite directions collide. Are those two situations identical?

newtonian-mechanics energy kinematics collision
shareciteimprove this question
edited Dec 1 '12 at 1:45

David Z
46.2k2185183
asked Dec 1 '12 at 0:30

smaccoun
197118
4
The "with its brakes on of course" is what makes both situations different, without the brakes they are the exact same, see my answer below. – Jaime Dec 1 '12 at 2:02
1
Possible duplicate of Train crash: are these situations alike? and Classical car collision – voix Dec 1 '12 at 9:18
2
I thought Myth busters tested this, and found that two cars at 50mph equals one car at 50mph hitting a wall. In essence the wall is like a plane of symmetry. – ja72 Dec 1 '12 at 14:55

As Jamie said, if the second car doesn't have brakes on, the situation are identical : one is viewed from the centre of mass frame, and the second is the same event viewed from a train moving in the same direction and at the same speed of one of the cars – Marco Aita Dec 1 '12 at 19:13

..but a wall is not like a car (i.e. it is like a very massive car), so hitting a wall at 100mph is not like hitting an equally massive car at 100mph (see the chosen answer to Classical car collision for a good explanation). – Marco Aita Dec 1 '12 at 19:35
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6 Answers
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23
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I don't think any of the other answers have made the following point clear enough, so I am going to give it a try. Both scenarios are very similar before the collision, but they differ greatly afterwards...

From a stationary reference, you see the cars driving towards each other at 50mph, but of course if you choose a reference frame moving with the first car, then the second will be headed toward it at 100 mph. How is this different from the wall scenario?

Well, from a stationary reference frame, after the crash both cars remain at rest, so the kinetic energy dissipated is 2×12mv2.

From the reference frame moving with the first car, the kinetic energy before the crash is 12m(2v)2=4×12mv2, but after the crash the cars do not remain at rest, but keep moving in the direction of the second car at half the speed. So of course the kinetic energy after the crash is 2×12mv2, and the total kinetic energy lost in the crash is the same as when considering a stationary reference frame.

In the car against a wall, you do have the full dissipation of a kinetic energy of 4×12mv2.

shareciteimprove this answer
answered Dec 1 '12 at 1:59

Jaime
3,056619
2
To put a finer still point on this, and to tie this in with my comment to the OP, the KE before the collision is not the whole story. In the frame in which the car is stationary while the wall is moving, the KE is enormous both before and after. It is the difference that tells the story. – Alfred Centauri Dec 1 '12 at 2:05

+1 for showing that the analysis works regardless of reference frame. When I discussed mine, I just assumed it would be easier to keep the reference frame the same in both situations: a 3rd person observation standing at zero mph relative to a wall. – smaccoun Dec 2 '12 at 19:54

> but after the crash the cars do not remain at rest, but keep moving in the direction of the second car If you consider the reference frame of the first car then isn't the first car always stationary and it is the second car that collides and slows down to 0 velocity? – 1110101001 Aug 15 at 21:05

Nope. The reference frame is not tied to the car, it is an inertial reference frame that happens to be one that the first car is stationary in at the starting time. A stationary car hit by another car will not stand still, but start moving backwards at half the speed of the incoming car. – Jaime Aug 15 at 21:57

Oh I see! Also, I assume the reason you can't use the reference frame of the car itself is that upon collision the car decelerates, hence the reference frame of the car is not strictly inertial. – 1110101001 Aug 15 at 22:55
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Actually, assuming that the oncoming car is the same mass as yours, colliding with an oncoming car at 50 MPH is equal to colliding with an ideal immovable wall at 50 MPH. Consider this:

I'm going to set up one of two experiments. I'm either going to ram car A into car B, both of them moving 50 MPH in opposite directions, or I'm going to ram car A into a solid wall at 50 MPH. However, I'm going to put up a shroud so that you can only see car A, you will be unable to see either car B or the wall, whichever one my coin-flip tells me to use.

Because you can now only see at car A and its contents, how would you tell which experiment I'd decided to do?

shareciteimprove this answer
answered Dec 1 '12 at 5:41

Aric TenEyck
1793

I would ask Schrödinger's Cat... – Mik Cox Dec 4 '12 at 0:23

I agree that the situations are identical, but fail to see how the shroud experiment proves that. It seems like you assumed they're the same at the beginning, and then concluded they're the same at the end based on your beginning assumption. Can you explain more how the shroud illustrates that? – smaccoun Dec 4 '12 at 0:25

Some people believe that head-on-at-50 is equivalent to brick-wall-at-100, while in fact it's equivalent to brick-wall-at-50. The shroud experiment encourages them to consider the different scenarios and figure out why head-on-at-50 is equivalent to brick-wall-at-50. – Aric TenEyck Dec 4 '12 at 2:19
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4
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Certainly they are not exactly the same - a wall is not the same thing as a car, and a crash is a very complicated physical event. Even if simple calculations involving momentum and energy or descriptions involving reference frames suggest that aspects of a car-car and car-wall collision are the same, the real collisions will be fairly different.

In this case, though, simple considerations do reveal that the car-car crash at 50 mph is almost certainly safer than crashing 100 mph into a wall. Your energy calculation is a fine way to see this.

Another is to consider the car-car collision from a frame co-moving with the second car. In this frame, you're going 100 mph and crash into a stationary car. So the question is like asking whether it is worse to crash into a stationary wall or a stationary car when going 100 mph (apart from the fact that the movement relative to the road is a little different). Of course crashing into the car is less dangerous than crashing into the wall, confirming your earlier result.

I have often heard the same problem rephrased so that you consider crashing into a wall at 100 mph or crashing into a car when you're both going 100 mph. It may be that this was the original problem the physics professor mentioned, and it got distorted somewhere in the game of telephone it played since then.

In that scenario, some people say they are equally bad because the energy dissipated per car is the same. Personally, I would probably go for the wall because at least some of the car's energy should go into the wall, but here the details become important (e.g. what if I fly through the window and then hit the wall?), and the energy alone is not a strong enough difference to say what which is worse. I imagine that either crash is very likely to be fatal at that speed.

Addressing your new question, two cars crashing head-on each at 50 mph is essentially the same as one car going 100 mph and crashing into a stationary car, by the relativity principle. However, relativity is broken by the existence of the road, so to the extent that the cars interact with the road during the collision there may be some differences.

 

[Dominic]

shareciteimprove this answer
edited Dec 1 '12 at 2:32

answered Dec 1 '12 at 0:44

Mark Eichenlaub
30.2k574146
1
I think the idea here is to get at the fundamental physics concepts, not whether or not hitting a car or a wall is worse. For that reason, I don't think your second point is at all convincing, and it is the very reason why I originally thought the answer would be the same (and why I think most people would). For the second proposition, and that may very well have been the original question, I think that one is most certainly te cars. Using the same energy calculations, there is twice as much energy that needs to be dissipated for the cars colliding, so I think the cars are clearly worse there. – smaccoun Dec 1 '12 at 1:15

I don't know what you're talking about because phrases like "second such-and-such" are ambiguous; I don't know how you're counting. Please rephrase to say what specific concepts you are trying to refer to. – Mark Eichenlaub Dec 1 '12 at 1:28

By second proposition, I meant your rephrased version of the problem (your 3rd paragraph, which is a different problem) where the 2 cars colliding are each going a 100, and the second scenario the car is going 100 and crashes into the wall. For an analysis of that, I would say the two cars colliding is worse, because the total energy is twice that of a single car going 100 and crashing into the car. – smaccoun Dec 1 '12 at 1:37

I thin your third paragraph, where you use reference frames, is possibly the misconception that this problem is getting at. I originally thought that too, and it is true that relative to the other car the first car is going 100mph. This is why I had to revert to an energy argument to see a difference. You could just as easily make the problem be only walls crashing at each other, or only cars crashing at each other. Again, I'm not sure about this, but my suspicion is that the reference frame argument isn't sufficient in its own right. But, I posted it on here to see some other perspectives – smaccoun Dec 1 '12 at 1:40
1
Sorry man, in my 2nd comment I meant your 4th paragraph, LOL. Whoo, I need to learn how to count – smaccoun Dec 1 '12 at 1:46
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I think that it makes sense to move away from the specific walls and cars and consider simply an inelastic collision on of two masses, m1 and m2. Otherwise we get stuck in the details.

When two bodies collide, the devastating effect in collision depends only on their relative velocity v1−v2. Kinetic energy, which has the destructive effect is equal to

12m1m2m1+m2(v1−v2)2
The rest of the kinetic energy is associated with the movement of the center of mass of the system. This energy in the collision does not change, and has no effect of destruction.

In given case, if faced two identical cars moving toward each other with one and the same velocity v the energy of destruction is

12mmm+m(v+v)2=mv2
Now, consider the case where a car collides with a massive barrier at speed 2v.In this case m1=m, v1=2v, m2=∞, v2=0

The energy of destruction:

12m(2v)2=2mv2
I.e. the latter case is much more dangerous.

shareciteimprove this answer
edited Dec 1 '12 at 9:45

answered Dec 1 '12 at 9:24

Martin Gales
1,3891131
add a comment
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2
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The most straightforward way to see how different the two scenarios are is to:

(1) consider two cars crashing into each other from a reference frame in which one of the cars is stationary and the other has a speed of 100mph

(2) consider the one car crashing into the wall with a speed of 100mph.

Assuming the wall is substantial enough that its mass and physical strength far exceeds that of the stationary car in (1), it's clear that the two scenarios significantly differ.

In (2), the car crashes into a stationary and effectively immovable, indestructible object at 100mph while in (1), the moving car crashes, at 100mph, into a stationary but otherwise identical object that importantly, can both move and deform.

shareciteimprove this answer
answered Dec 1 '12 at 1:24

Alfred Centauri
32k22384

Please see my edit. Do you think it would be the same if it was just walls crashing into walls, or just cars crashing into cars? Is my energy argument incorrect? Good points otherwise, but I don't think that's what the question is getting at. – smaccoun Dec 1 '12 at 1:50

@smaccoun, the total KE of the system is a minimum in the COM reference frame. Consider, for example, the frame in which the car is stationary and the wall is moving at 100mph. The KE in that frame is enormously larger than in the frame in which the wall is stationary. Yet, the damage to the car (and perhaps the wall) does not depend on the reference frame. Your argument cannot depend on just the KE before the crash. – Alfred Centauri Dec 1 '12 at 1:59
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1
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Damage should be the same if two cars colliding at 50mph and if a car travelling (50∗2√) mph crashes into a wall.

The energy of destruction is an internal energy so:

First case
Equation of energy conservation

mv22+mv22=T
where T - internal energy, m - car mass

so energy of destruction in first case:

T=mv2
Second case
Equation of energy conservation

m(v2√)22=(m+M)u22+T
where T - internal energy, m - car mass, M - wall mass

Equation of momentum conservation

mv2√=(m+M)u
so internal energy T=m(v2√)22(1−mm+M)
M>>m so energy of destruction in second case:

T≈mv2
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answered Dec 1 '12 at 11:12

voix
1,3761921

The calculations are correct but they solve the problem for the total energy of the system. If you are only interested in the damage caused on one car (because in the second case there is no "other car"), the same calculations show that to obtain the same damage (=change in internal energy=deformation+heat) you need to have the same speed v (again considering the wall very massive, M>>m so that u≈0). – Marco Aita Dec 1 '12 at 19:08

@Marco Aita, you are not right. Imagine a car standing before the wall, so damage will be equal for two cars in the second case. – voix Dec 2 '12 at 14:38

In the fist case each car dissipates 12mv2. In the case of an impact of a car with a wall of a very large (infinite) mass, the final speed is zero and therefore the car dissipates again an energy of 12mv2. If all this energy goes into deformation of the car you will get the same damage to the car as in the first case. The problem is that we can't really tell how much of the dissipated energy goes into deforming the car, so my reasoning is partial.. ..but I don't understand your point with the car in front of the wall, can you elaborate? – Marco Aita Dec 3 '12 at 0:10

@Marco Aita, if in the second case we put another car in front of the wall we will get the same damage to the car as in the first case. – voix Dec 3 '12 at 4:54

..I think we are talking about different things.. Anyway, the part with the wall is practically unsolvable unless we specify more about the wall characteristics.. Imagine the difference in hitting a massive solid steel wall and an equally massive marshmallow wall.. . It all depends on how much energy the wall absorbs.. – Marco Aita Dec 3 '12 at 13:09
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rev 2015.9.22.2837

 

[Dominic]

 

[Albert Oil]

Luin kokonaan ku oli niin hyvä.

 

[Örfil]

Referaatti kiitos, omin sanoin jotta tietää että puhutaanko nyt samasta asiasta.

 

[Örfil]

Aikaa on, sillä palaan vasta huomenna aamulla.

 

[Albert Oil]

Dominic tuossa todistelee laveasanaisesti, että Al-setä on hyvä mies ja kaikkien tulisi välittömästi höllätä kukkaronnyörejään miehen suuntaan.

 

[markoj]

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Komp

 

[Renessanssinero]

Wiki:


Eikö lehdellä ole muuta käyttöä?

 

[PekPoy]

Koitan arvata mistä on kyse osaltasi, ehkä tuo Lunssilan video tarkoitti sitä?

Oleellista törmäyksen tuhojen kannalta on vapautuva liike-energian määrä. Ja se on suoraan verrannollinen siihen nopeuden neliöön (0.5*mv2)

Tuossa Lunssilan esimerkissä pohditttiin liikemäärän säilymistä, mutta oleellista törmäyksen tuhojen kannalta on se paljonko vapautuu liike-energiaa, ei se mihin suuntaan ja millä nopeudella rojut sinkoutuvat törmäyksen jälkeen. Pahin on jo tapahtunut, kun tuli kontaktia.

On siis eri juttu paikallaan olevan auton tapauksessa kuin kahden auton nokkakolarin yhteydessä, mihin suuntaan pääosa kappaleista tällin jälkeen liikkuu.

Sivulisäjarna: autossa ei ole muuten nopeusmittaria, se on oikeammin nimeltään vauhtimittari. Miksi?
Tuo liittyy myös yllä olevaan

Edit:
V: vauhdilla ei ole suuntaa.

Edit 2: taidan nyt arvata että pekkis keksi vielä erään kombinaation tästä hässäkästä, mutta annetaanpas sen vielä olla

Häh.

En mä edes tsiigannut mitään videota, kunhan kysyin selvyyden vuoksi. Jotta voidaan arvioida vastaako törmäys a törmäystä b, pitää tietää mitä verrataan. Mä en ainakaan tiedä mitä tässä verrataan. Autojen välimatkan hupenemistahti on molemmissa skenaarioissa sama, siitä ei liene epäselvyyttä (tämä on se suhteellinen nopeus). Muiden suureiden osalta kai asia ei ole niin selvä.

Mutta kai se nyt on päivänselvää, että joku vertailtava arvo on osoitettava ennen kuin voidaan vertailla? Mä en ole selvillä mikä se nyt on.

Sen verran kyllä fysiikantunneilta muistan (ja uskon että kaikki tähän keskusteluun osallistuneet tietävät) että nopeus on vektorisuure (tms.) ja vauhti ei.

 

[Dominic]

Jos auto on tullut käännökseen esim. 30 km/h ja mopo 50 km/h = 80 km/h nokkapokka.

En suosittele ajamaan tiilisenään sillä nopeudella.

Muutenhan tuo 80 km/h nokkapokka olisi vallan hyväksyttävä kommentti, mutta homma murenee siihen että tehdään vertaus tiiliseinään ajamisesta sillä nopeudella ja se kun nyt vaan ei ole sama asia.

 

[ProjectMaster]

IS uutisoi miehestä, joka heräsi koomasta 11-vuoden jälkeen ja ällistyi, kun idolinsa Federer oli vielä huipulla. Jos kuka tahansa meistä olisi vajonnut koomaan, mitä miettisitte ykkösistä?

Ekana tulee mieleen, että "Vieläkö Schumi ja Ferrari on mestaruusputkessa?" Varmana ällistyisit kun kuulisit ettei ole voittanut ainuttakaan enää 2004 jälkeen. Ehkä olisit tyytyväinen, mutta mieli muuttuu kun kuulet Schumin olleen jo kaksi vuotta vihannes.

Mitä muuta keksitte? On se aika iso menetys 11-vuotta.

 

[Tsuhna]

Ällistyisin varmaan siitä että Jaromir Jagr pelaa edelleen NHLssa. Ja sitten kun kuulisin että on välillä käynyt jossain ryssän liigassa jäähdyttelemässä ja sen perään tehnyt vielä comebackin..

 

[PuoliPöhkö]

IS uutisoi miehestä, joka heräsi koomasta 11-vuoden jälkeen ja ällistyi, kun idolinsa Federer oli vielä huipulla. Jos kuka tahansa meistä olisi vajonnut koomaan, mitä miettisitte ykkösistä?

Ekana tulee mieleen, että "Vieläkö Schumi ja Ferrari on mestaruusputkessa?" Varmana ällistyisit kun kuulisit ettei ole voittanut ainuttakaan enää 2004 jälkeen. Ehkä olisit tyytyväinen, mutta mieli muuttuu kun kuulet Schumin olleen jo kaksi vuotta vihannes.

Mitä muuta keksitte? On se aika iso menetys 11-vuotta.

Heh, Massa.

 

[Fagerholm]

Varmaankin päällimmäinen ihmetyksen aihe olis, miten Eccelstone voi elää vieläkin.