# Tale of two cars (Physics question) **URL:** \ **Category:** Factual Questions\ **Created:** [March 2, 2012, 1:50pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543 "2012-03-02T13:50:04Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![Blaster\_Master](https://avatars.discourse-cdn.com/v4/letter/b/cab0a1/32.png) [@Blaster\_Master](https://boards.straightdope.com/u/Blaster_Master)\ **Post date:** [March 2, 2012, 4:08pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/21 "2012-03-02T16:08:07Z")
> [@Corcaigh](#): > > I’m pretty sure they did this on Mythbusters and #2 would cause more damage Yes, they did do this on Mythbusters. Jamie had previously incorrectly stated that two vehicles going 50 would be the same as one going 100 into a brick wall. They tested it by having one car crash into a wall at 50, an identical one into a wall at 100 and two more identical cars hitting head on both going 50. They took numbers, but the visual was enough to show that the head-on collision looked a lot like the first example and not at all like the latter. The math behind why was well explained upthread. I’d find the episode name but, well, the only thing harder to find than a specific Mythbusters myth involving car crashes is one involving explosions. ---
**Author:** ![westom](https://avatars.discourse-cdn.com/v4/letter/w/bbe5ce/32.png) [@westom](https://boards.straightdope.com/u/westom)\ **Post date:** [March 2, 2012, 4:09pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/22 "2012-03-02T16:09:28Z")
> [@Chessic\_Sense](#): > > But it’s not. Detriment is determined by change in energy over time, i.e. impulse. The OP did not define ‘detriment’. So every post is making assumptions. Recall my first post. > [@westom](#): > > Both cars are now scrap metal. Damage to each car is only relevant to energy [that] is imparted inside. Dogma’s assumption is detriment defined by the amount of energy absorbed entirely by each car. Acceleration and time has no relevance there. Acceleration to a human head? That is completely different. But again, too many hypotheticals exist to make a reliable reply to the OP’s question. ---
**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\ **Post date:** [March 2, 2012, 4:11pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/23 "2012-03-02T16:11:45Z")
> [@Blaster\_Master](#): > > Yes, they did do this on Mythbusters. Doesn’t look like the same thing, actually. There’s a difference between a brick wall and a car. The brick wall crumples differently, for one. And a car isn’t made of bricks. So, what happens when a car at 100mph crashes into a parked car? ---
**Author:** ![robby](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robby/32/11048_2.png) [@robby](https://boards.straightdope.com/u/robby)\ **Post date:** [March 2, 2012, 4:14pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/24 "2012-03-02T16:14:04Z")
> [@Alka\_Seltzer](#): > > I think we need a reply from an actual physicist. Kinetic energy is given by the formula 1/2 mv[sup]2[/sup], so the collision in scenario 2 is a lot more energetic. Well, for what it’s worth, I used to teach physics. 🙂 > [@Dog80](#): > > Kinetic energy is 1/2 m u[sup]2[/sup] > > Total kinetic energy in the first case is 2\*(1/2 m 50[sup]2[/sup]) = 2500m > > In the second case the energy is 1/2 m 100[sup]2[/sup] = 5000M > [@westom](#): > > If detriment is determined by energy, then Dog80 has struck the correct answer. > > A 100 Km/hr car striking a parked car is energy equivalent to two cars hitting each other at 70.7 Km/hr. All of these posts are incorrect, because you have to look at the change in kinetic energy dissipated in the collision. In the first case, the initial total KE of the two vehicles is 2500_m_. The final KE is zero, so the KE dissipated in the actual collision is 2500_m_. In the second case, the initial KE of car A is 5000_m_, but the KE of the two mangled vehicles _immediately after the collision_ is 2500_m_. (The rest of the KE will then be dissipated as friction as the two mangled vehicles slide to a stop). So the KE dissipated in the collision will again be 2500_m_. > [@Baracus](#): > > In the first case, all that kinetics energy is dissipated in the collision. > > In the second case, immediate post crash when the two cars are traveling a 50 km/h, their kinetic energy will be (1/2)_2m_50[sup]2[/sup] = 2500m. Thus there will again be 2500m of kinetic energy dissipated during the collision. This is correct. ---
**Author:** ![DCnDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dcndc/32/2842_2.png) [@DCnDC](https://boards.straightdope.com/u/DCnDC)\ **Post date:** [March 2, 2012, 4:18pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/25 "2012-03-02T16:18:23Z")
> [@westom](#): > > The OP did not define ‘detriment’. So every post is making assumptions. Yes. Are we talking force or energy? The force of the collision in (2) is double that of (1). The energy OTOH is the same in both. ---
**Author:** ![robby](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robby/32/11048_2.png) [@robby](https://boards.straightdope.com/u/robby)\ **Post date:** [March 2, 2012, 4:23pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/26 "2012-03-02T16:23:27Z")
> [@Chessic\_Sense](#): > > But it’s not. Detriment is determined by change in energy over time, i.e. impulse. Nitpick: _Impulse_ is equal to the change in momentum of an object to which a force is applied. And as I stated above, the detriment to the people in the vehicle would be due to the peak forces experienced by the occupants. It’s not actually relevant in an inelastic collision to consider energy (including kinetic energy), because kinetic energy in an inelastic collision is not conserved. Instead, it is dissipated as friction, noise, and the crumpling up of the vehicles in question. ---
**Author:** ![Terr](https://avatars.discourse-cdn.com/v4/letter/t/839c29/32.png) [@Terr](https://boards.straightdope.com/u/Terr)\ **Post date:** [March 2, 2012, 4:24pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/27 "2012-03-02T16:24:04Z")
> [@Corcaigh](#): > > I’m pretty sure they did this on Mythbusters and #2 would cause more damage No. On Mythbusters the second scenario was crashing into a wall. Very different. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [March 2, 2012, 4:25pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/28 "2012-03-02T16:25:05Z")
> [@robby](#): > > Well, for what it’s worth, I used to teach physics. 🙂 > > All of these posts are incorrect, because you have to look at the change in kinetic energy dissipated in the collision. In the first case, the initial total KE of the two vehicles is 2500_m_. The final KE is zero, so the KE dissipated in the actual collision is 2500_m_. > > In the second case, the initial KE of car A is 5000_m_, but the KE of the two mangled vehicles _immediately after the collision_ is 2500_m_. (The rest of the KE will then be dissipated as friction as the two mangled vehicles slide to a stop). So the KE dissipated in the collision will again be 2500_m_. > > This is correct. **robby** , I understand and agree with your sentiment but I think your math is flawed. First, KE has a vector, does it not? I understand it doesn’t _have to_, in an absolute sense, but in the context of summing energies, it has to. So in the first scenario, it’s really 2500m for the first car and -2500m for the second car, summing to zero. After collision, it’s still zero, but some negative KE and some positive KE has been converted to damage, friction, heat, etc. In the second scenario, yes, we’d end with 2500m, but that’s still “little m”, i.e. a variable. But that little m is equal to 2 big-Ms, i.e. the fixed mass of an individual car. So we go from 5000M and 0M to 2500(2M), and energy is conserved. Of course, in the real world, we won’t come out of the collision at that “conserved energy” speed, and the difference will be the damage to the two cars. So my point is, the scenarios are equal as far as we can tell without knowing exactly how the cars are built and what safety features they have. In practice it may make a difference, but in theory, there is no difference. The situations are the same. ---
**Author:** ![westom](https://avatars.discourse-cdn.com/v4/letter/w/bbe5ce/32.png) [@westom](https://boards.straightdope.com/u/westom)\ **Post date:** [March 2, 2012, 4:25pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/29 "2012-03-02T16:25:35Z")
> [@robby](#): > > Well, for what it’s worth, I used to teach physics. The energy content of one car at 100 km/hr is two times larger than the energy content of two cars doing 50 km/hr. A car at 100 km/hr has four times the energy of a car at 50 km/hr. Again, if “detriment” is defined by energy dissipated, then a 100 km/hr car is a much more violent event. To have same energy, two cars must be doing 70.7 km/hr; not 50 km/hr. As Dog80 posted. I used to be a student. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [March 2, 2012, 4:28pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/30 "2012-03-02T16:28:41Z")
> [@DCnDC](#): > > The force of the collision in (2) is double that of (1) No, the force is equal, too. I mean, at least in a scalar sense. A car going from rest to 50 km/h is the same as a car going from -50 km/h to 0. If it takes the same time and the cars have the same mass between scenarios (and why wouldn’t they?), and F=M\*dV/t, then the force is the same, except maybe for the sign depending on how you draw the coordinate system. ---
**Author:** ![robby](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robby/32/11048_2.png) [@robby](https://boards.straightdope.com/u/robby)\ **Post date:** [March 2, 2012, 4:28pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/31 "2012-03-02T16:28:51Z")
> [@DCnDC](#): > > …The force of the collision in (2) is double that of (1). No, it isn’t. See my Post #10 above. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [March 2, 2012, 4:30pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/32 "2012-03-02T16:30:37Z")
> [@westom](#): > > The energy content of one car at 100 km/hr is two times larger than the energy content of two cars doing 50 km/hr. > > A car at 100 km/hr has four times the energy of a car at 50 km/hr. > > Again, if “detriment” is defined by energy dissipated, then a 100 km/hr car is a much more violent event. > > To have same energy, two cars must be doing 70.7 km/hr; not 50 km/hr. As Dog80 posted. > > I used to be a student. Then you get half credit for stopping halfway through the problem. Yes, a car at 100 km/hr has four times the energy as a car at 50 km/hr. You don’t get to just stop right there, put down the pencil, and say you have an answer. Now find the energy post-collision and find the “energy dissipated,” which you admit is what is at issue here. ETA: Also note that only one car is going at 100 km/hr. The other is at rest. If you intend to damage two cars, then you have to average the energy across both cars. If you consider only one car to be changing energy, then you can’t claim the second car would be damaged. It’d have to be built of unobtanium. ---
**Author:** ![DCnDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dcndc/32/2842_2.png) [@DCnDC](https://boards.straightdope.com/u/DCnDC)\ **Post date:** [March 2, 2012, 4:33pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/33 "2012-03-02T16:33:56Z")
I see. My physics (and math) are rusty. ---
**Author:** ![Baracus](https://avatars.discourse-cdn.com/v4/letter/b/2bfe46/32.png) [@Baracus](https://boards.straightdope.com/u/Baracus)\ **Post date:** [March 2, 2012, 4:35pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/34 "2012-03-02T16:35:25Z")
> [@Chessic\_Sense](#): > > **robby** , I understand and agree with your sentiment but I think your math is flawed. First, KE has a vector, does it not? I understand it doesn’t _have to_, in an absolute sense, but in the context of summing energies, it has to. So in the first scenario, it’s really 2500m for the first car and -2500m for the second car, summing to zero. After collision, it’s still zero, but some negative KE and some positive KE has been converted to damage, friction, heat, etc. No, KE is not a vector. Consider a rubber ball fired at a brick wall. All of the KE is pointed at the wall prior to impact and _most_ of it is pointed away from the wall afterwards. ---
**Author:** ![robby](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robby/32/11048_2.png) [@robby](https://boards.straightdope.com/u/robby)\ **Post date:** [March 2, 2012, 4:35pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/35 "2012-03-02T16:35:45Z")
> [@Chessic\_Sense](#): > > **robby** , I understand and agree with your sentiment but I think your math is flawed. First, KE has a vector, does it not? I understand it doesn’t _have to_, in an absolute sense, but in the context of summing energies, it has to. > > …After collision, it’s still zero, but some negative KE and some positive KE has been converted to damage, friction, heat, etc. Well, your starting point here is incorrect. Kinetic energy is a scalar quantity. Note that you are squaring the magnitude of the velocity. There is no such thing as “negative” kinetic energy. ---
**Author:** ![TripleFail](https://avatars.discourse-cdn.com/v4/letter/t/b9e5f3/32.png) [@TripleFail](https://boards.straightdope.com/u/TripleFail)\ **Post date:** [March 2, 2012, 4:37pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/36 "2012-03-02T16:37:27Z")
> [@Chessic\_Sense](#): > > Again, unless you’re expecting the second car to come out immaculate, you have to average the energy. It’s the same energy _per car_. > > In scenario 1, the delta-V is +50 for car 1 and -50 for car 2. > In scenario 2, the delta-V is -50 for car 1 and -50 for car 2. > > Each car changes velocity the same amount. You’re either not averaging across the cars or you’re not considering the end scenario. It’s the same error either way, but you have to do one or the other. humm Not sure what you trying to say. Are you saying in both scenario have the same amount of energy dissipated? I think it very clear that case 2 carries more energy. How the energy is dissipated is another question Of course I am don’t think the 2nd car will come out immaculate in either case. The real questions are in case 2 how much of that extra energy would be converted back to kinetic energy then come to a rest due to friction with road, and would the materials crumple zone material act the same way in both scenarios. As collision time would be different the material might handle the different intensity differently. ---
**Author:** ![JimboJamesJimmersonIII\_Esq](https://avatars.discourse-cdn.com/v4/letter/j/54ee81/32.png) [@JimboJamesJimmersonIII\_Esq](https://boards.straightdope.com/u/JimboJamesJimmersonIII_Esq)\ **Post date:** [March 2, 2012, 4:41pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/37 "2012-03-02T16:41:01Z")
**Robby** has been correct throughout the thread. The two cases are physically equivalent to each other. ---
**Author:** ![westom](https://avatars.discourse-cdn.com/v4/letter/w/bbe5ce/32.png) [@westom](https://boards.straightdope.com/u/westom)\ **Post date:** [March 2, 2012, 4:44pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/38 "2012-03-02T16:44:00Z")
> [@Chessic\_Sense](#): > > ``` > Yes, a car at 100 km/hr has four times the energy as a car at 50 km/hr. ... Now find the energy post-collision and find the "energy dissipated," which you admit is what is at issue here. > > ``` Energy dissipated is 1) energy of a car at 100 km/hr or 2) the energy of two cars at 50 km/hr. Conservation of energy. In both cases, all energy is absorbed equally both cars. But 1) means double the energy. Again, I keep saying this. Otherwise junk science results. Define “detriment”. a) Detriment is defined by the acceleration of a brain inside the skull? Or b) defined by the total amount of energy dissipated? b) is total energy dissipated in each car. a) defines detriment in terms of _where_ energy dissipates throughout each car. Why are so many answering a question before first defining the question? A major mistake. Define your interpretation of detriment in every answer. Otherwise any answer is possible. Meanwhile, the problem is framed by one constant. The amount of energy in the collision. One 100 km/hr car means double energy dissipates in both cars. ---
**Author:** ![robby](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robby/32/11048_2.png) [@robby](https://boards.straightdope.com/u/robby)\ **Post date:** [March 2, 2012, 4:50pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/39 "2012-03-02T16:50:55Z")
**Westom** , I have addressed all of your points in my earlier posts. ---
**Author:** ![TonySinclair](https://avatars.discourse-cdn.com/v4/letter/t/49beb7/32.png) [@TonySinclair](https://boards.straightdope.com/u/TonySinclair)\ **Post date:** [March 2, 2012, 4:51pm UTC](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543/40 "2012-03-02T16:51:25Z")
Wow. And here I always liked physics because, unlike politics, there was a definite right answer for elementary problems. The guys who say the two collisions are equivalent are right, and you shouldn’t even have to do any arithmetic to see it. You only have to remember that experiments done in inertial frames have identical outcomes (except for the infinitesimal relativistic effects that occur at such low speeds). [Previous page](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543.md?page=1) [Next page](https://boards.straightdope.com/t/tale-of-two-cars-physics-question/614543.md?page=3)