# Head-on Collision Physics - revisited **URL:** \ **Category:** Factual Questions\ **Created:** [August 4, 2004, 5:34am UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252 "2004-08-04T05:34:32Z")\ **Posts on this page:** 11\ **Page:** 1
**Author:** ![astro](https://avatars.discourse-cdn.com/v4/letter/a/9dc877/32.png) [@astro](https://boards.straightdope.com/u/astro)\ **Post date:** [August 4, 2004, 5:34am UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/1 "2004-08-04T05:34:32Z")
I had some time this evening so I revisited an old thread I’d been meaning to review for some time, This is the thread below. Please do not re-open it. Post all answers here. [Head-on Collision Physics](http://boards.straightdope.com/sdmb/showthread.php?t=30041&page=1&pp=50&highlight=momentum+wall+impact+energy) Did we ever some to some final and absolute conclusion in this thread as to the OP’s question? Was he correct or was his friend correct re the physics of the collision? I’m assuming per the final comments that the OP was correct and his “knucklehead” friend was wrong. > [@](#): > > I have a knucklehead friend I am unable to convince of a fairly simple physics problem. He contends if two cars of equal mass (discounting crumple zones and the like) both traveling 35 mph meet head-on, the resulting collision is multiplied to a 70 mph crash. I told him it’s the speed of stopping that’s important, that this scenario is no different than had one car traveling 35 mph run into a immovable brick wall — the car goes from 35 to zero near-instantaneously. He disagrees, in the face of logic, and I can suffer his foolishness no longer. I’m long past my days of high school physics, and can’t quote the various laws which lay waste to his folly. Any takers? ---
**Author:** ![Richard\_Pearse](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/richard_pearse/32/13144_2.png) [@Richard\_Pearse](https://boards.straightdope.com/u/Richard_Pearse)\ **Post date:** [August 4, 2004, 5:58am UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/2 "2004-08-04T05:58:32Z")
It would seem to me that the effect would be the same as a 70mph crash between two vehicles, one stationary. It would probably also be the same as a 35mph crash into an immovable wall (half the speed but also only half the impact absorbtion). So they are both right, they just not talking about the same thing. ---
**Author:** ![SPOOFE](https://avatars.discourse-cdn.com/v4/letter/s/c5a1d2/32.png) [@SPOOFE](https://boards.straightdope.com/u/SPOOFE)\ **Post date:** [August 4, 2004, 6:03am UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/3 "2004-08-04T06:03:54Z")
When two objects collide, the kinetic energy of one object is imparted onto the other object. A stationary object has no (appreciable) kinetic energy. A moving object has plenty of the stuff, and will be only too happy to transfer it to its target. > [@](#): > > this scenario is no different than had one car traveling 35 mph run into a immovable brick wall — the car goes from 35 to zero near-instantaneously. The key word is “ **near** -instantaneously”. To our eyes, two 35-mph objects would decelerate about as fast as a 35-mph object hitting a stationary object. However, if you actually got down to the nitty-gritty and measured things in teensy fractions of seconds, you’d probably find that the two moving cars decelerate even MORE, uh, near-instantaneously. The collision would send a shockwave through the vehicle… it’s not just felt at the front end, at the point of impact, but throughout the whole thing. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 4, 2004, 6:39am UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/4 "2004-08-04T06:39:40Z")
Two cars hitting head-on at 35 MPH is equivalent to one car at 70 MPH hitting a parked car, or to one car at 35 MPH hitting a much more massive object (the proverbial brick wall). Or, for that matter, a car going 140 rear-ending a car going 70. Or a car going 70 rear-ending a brick wall going 35. ---
**Author:** ![SPOOFE](https://avatars.discourse-cdn.com/v4/letter/s/c5a1d2/32.png) [@SPOOFE](https://boards.straightdope.com/u/SPOOFE)\ **Post date:** [August 4, 2004, 6:48am UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/5 "2004-08-04T06:48:48Z")
Here’s another way to think of things: Energy needs to come from somewhere, and it needs to go somewhere. In the two-car collision, we have enough energy to get two cars up to speed. In the car-hit-wall collision, we have enough energy to get one car up to speed. Twice as much energy means twice as much energy needs to go somewhere. It can’t just vanish into the cosmos (well, some of it does, in the form of heat, but that wouldn’t account for all of it). ---
**Author:** ![minor7flat5](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/minor7flat5/32/258_2.png) [@minor7flat5](https://boards.straightdope.com/u/minor7flat5)\ **Post date:** [August 4, 2004, 12:16pm UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/6 "2004-08-04T12:16:34Z")
How about if we consider a perfectly ideal experiment where we smash two mirror-image cars together, each going 35mph. Since the experiment is ideal, wherever there is a protrusion on car A, there is a matching protrusion on car B. Consequently, there would be no reason for any part of car A to extend past any part of car B. It would be as if there were an invisible plane separating the cars. If you had a sufficiently durable and thin material, such as Lexan, you could place a sheet of the stuff at the precise point where the invisible plane would be, and the two cars could smash into it, leaving it standing. Now what’s the difference between smashing into your sheet of Lexan while it is backed against a wall and smashing into it when there is a compensating smash coming from the other side? The divider never moves. ---
**Author:** ![minor7flat5](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/minor7flat5/32/258_2.png) [@minor7flat5](https://boards.straightdope.com/u/minor7flat5)\ **Post date:** [August 4, 2004, 12:17pm UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/7 "2004-08-04T12:17:49Z")
> [@SPOOFE](#): > > Twice as much energy means twice as much energy needs to go somewhere. There’s twice as much steel crumpling, absorbing twice as much energy. Each car is still crumpling to the tune of 35mph. ---
**Author:** ![Strainger](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/strainger/32/244_2.png) [@Strainger](https://boards.straightdope.com/u/Strainger)\ **Post date:** [August 4, 2004, 4:54pm UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/8 "2004-08-04T16:54:46Z")
I’ve taken the file I talked about [here](http://boards.straightdope.com/sdmb/showpost.php?p=570285&postcount=61), converted it to PDF, and posted it [here](http://fff.fathom.org/pages/strainger/collide.pdf) (WARNING: PDF, in case you didn’t notice that the first time). Don’t get too caught up in the “Toyota Camry” thing; it’s a joke. ---
**Author:** ![SCSimmons](https://avatars.discourse-cdn.com/v4/letter/s/e495f1/32.png) [@SCSimmons](https://boards.straightdope.com/u/SCSimmons)\ **Post date:** [August 4, 2004, 5:21pm UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/9 "2004-08-04T17:21:58Z")
> [@Chronos](#): > > Or a car going 70 rear-ending a brick wall going 35. And just try explaining _that_ one to the insurance company! ---
**Author:** ![Absolute](https://avatars.discourse-cdn.com/v4/letter/a/b2d939/32.png) [@Absolute](https://boards.straightdope.com/u/Absolute)\ **Post date:** [August 4, 2004, 6:50pm UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/10 "2004-08-04T18:50:09Z")
Wait a minute: energy increases with the square of velocity, right? So two cars hitting each other at 35 mph is not an equivalent impact to one car hitting a stationary car at 70. 2 \* 35 ^ 2 \< 70^2 - a car moving at 70 mph has more energy than two cars moving at 35. Or am I missing something… ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [August 4, 2004, 9:20pm UTC](https://boards.straightdope.com/t/head-on-collision-physics-revisited/258252/11 "2004-08-04T21:20:35Z")
Two cars at 35 have energy proportional to 2 \* 35^2 prior to collision, and 0 afterwards. One car at 70 and one at zero have 70^2 + 0 before collision, and 2 \* 35^2 after collision, and are both moving 35 MPH. They slow down and stop through friction with the ground, but that isn’t nearly as destructive, and so is being neglected. So the collision itself disipates energy proportional to 2 \* 35^2 in both cases.