# Simple Physics Question **URL:** \ **Category:** Factual Questions\ **Created:** [May 15, 2016, 3:28am UTC](https://boards.straightdope.com/t/simple-physics-question/754817 "2016-05-15T03:28:53Z")\ **Posts on this page:** 9\ **Page:** 3
**Author:** ![Francis\_Vaughan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/francis_vaughan/32/3093_2.png) [@Francis\_Vaughan](https://boards.straightdope.com/u/Francis_Vaughan)\ **Post date:** [May 28, 2016, 1:26pm UTC](https://boards.straightdope.com/t/simple-physics-question/754817/41 "2016-05-28T13:26:29Z")
> [@DanielB](#): > > Who’s right? However momentum must also be conserved. Momentum is mass x velocity. So here you get 60 mph. So clearly something has to give. The thing that gives is that if the two cars remain coupled together some of the kinetic energy in the first car gets converted to some other form of energy. Which is almost always eventually heat. (You even count the sound of the crash in this.) The crash where little to no energy is lost is an elastic collision, and here the only way to get both kinetic energy and momentum to remain conserved is that the cars do not end up in one lump, but rather they bounce apart, and you get two cars moving at different velocities. In reality a crash is often a mix of the two. The cars don’t merge, but a lot of energy does vanish into heat, and you end up with two wrecks moving at different velocities (at least for a while.) ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [May 28, 2016, 6:03pm UTC](https://boards.straightdope.com/t/simple-physics-question/754817/42 "2016-05-28T18:03:18Z")
Right. Momentum is always conserved in a collision. Energy is only conserved in a perfectly elastic collision. Cars are not elastic. ---
**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:** [May 28, 2016, 6:12pm UTC](https://boards.straightdope.com/t/simple-physics-question/754817/43 "2016-05-28T18:12:32Z")
> [@scr4](#): > > Right. Momentum is always conserved in a collision. Energy is only conserved in a perfectly elastic collision. Cars are not elastic. Close. Momentum is always conserved. Energy is also always conserved. However, in an elastic collision, _kinetic_ energy is conserved (as well as momentum). ---
**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:** [May 28, 2016, 6:16pm UTC](https://boards.straightdope.com/t/simple-physics-question/754817/44 "2016-05-28T18:16:34Z")
> [@scr4](#): > > Right. Momentum is always conserved in a collision. Energy is only conserved in a perfectly elastic collision. Cars are not elastic. > [@robby](#): > > Close. Momentum is always conserved. Energy is also always conserved. > > However, in an elastic collision, _kinetic_ energy is conserved (as well as momentum). To expand on this, by definition: [ul] [li]An elastic collision is one in which both momentum and kinetic energy are conserved.[/li][li]An inelastic collision is one in which momentum is conserved but kinetic energy is not.[/li][li]A perfectly inelastic collision is an inelastic collision in which the two objects stick together after the collision, so that their final velocities are the same and the momentum of the system is conserved.[/li][/ul] ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [May 28, 2016, 6:40pm UTC](https://boards.straightdope.com/t/simple-physics-question/754817/45 "2016-05-28T18:40:48Z")
More clarification - When a car collides, nowadays, it is designed with “crumple zones”. Assume the solid brick wall (or better yet, rock cliff). It’s not going anywhere, is barely dented or deformed. As each part of the car in turn is unable to proceed, the car bit by bit “crumples”. Modern cars have been designed so that this crumpling in the front takes energy away from the vehicle and converts it to heat and mechanical deformation, this ensuring too that the passenger compartment crumples as little as possible. As others mentioned, physics is simple - two equal weight cars collide going same speed, they both stop dead - net momentum is zero. A one-ton small car hits a five-ton truck, both going same speed - net momentum is 4 tons going in truck’s direction. Colliding car will crumple extra as it converts its speed from say, +60mph to -48mph. Hitting a parked car going 60mph, same weight, the final wreck will be going 30mph but there are two cars’ worth of crumple zone to absorb the impact. Of course, all these are like spherical cows and chickens - theoretical scenarios; the momentum will quickly drop to zero since secondary effects like friction and motor stops and air resistance, skidding tires, etc. will make the result spectacularly interesting provided you are not inside it. SO physics agrees with common sense. Hitting a brick wall is worse than hitting something crumply like a car, and hitting an oncoming moving car is worse than hitting a stationary car, and hitting a giant truck is worse than hitting a 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:** [May 28, 2016, 6:47pm UTC](https://boards.straightdope.com/t/simple-physics-question/754817/46 "2016-05-28T18:47:48Z")
> [@DanielB](#): > > Hey maybe I missed it among the posts but a car at 120mph hitting an identical stationary car would result in them both continuing at around 85mph according to me, NOT 60mph as I saw mentioned a couple of times. There’s just not energy in two cars moving at half the speed compared to one car moving at double the speed. > > My brain sees it like this… > > 120mph squared = 120 X 120 = 14400 ‘units’ of energy in the moving car. > > After the collision you claim 2 cars at 60mph so we have 2 x 60mph squared = 2 X 60 X 60 = 7200 ‘units’ of energy only half of what we need. > > I say it’s 2 cars x 84mph squared = 2 X 85 X 85 = 14450 ‘units’ of energy which (the slight excess because I rounded the speed to 85mph) is what the original 120mph car had. > > Who’s right? > [@DanielB](#): > > And for clarity to my post above, the 85mph assumes the stationary car is in neutral with no brakes. > [@Francis\_Vaughan](#): > > However momentum must also be conserved. Momentum is mass x velocity. So here you get 60 mph. So clearly something has to give. > > The thing that gives is that if the two cars remain coupled together some of the kinetic energy in the first car gets converted to some other form of energy. Which is almost always eventually heat. (You even count the sound of the crash in this.) > > The crash where little to no energy is lost is an elastic collision, and here the only way to get both kinetic energy and momentum to remain conserved is that the cars do not end up in one lump, but rather they bounce apart, and you get two cars moving at different velocities. In reality a crash is often a mix of the two. The cars don’t merge, but a lot of energy does vanish into heat, and you end up with two wrecks moving at different velocities (at least for a while.) **Francis Vaughan** has the correct answer here, with a couple of nitpicks. A lot of the energy of the collision of two cars will go into the energy required to crumple the cars up, in addition to heat and noise. Note that modern cars are specifically designed with energy-absorbing crumple zones. Also, by definition, an elastic collision is one in which kinetic energy is conserved. In other words, it is not correct to state that an elastic collision is one in which “little to no energy is lost.” If _any_ kinetic energy is lost in the collision, it is not an elastic collision. For this reason, real-world macroscopic elastic collisions are rare. Real-world elastic collisions tend to be things like the collision of gas molecules. To go back to **DanielB** ’s question, consider the example from the OP of two identical cars each traveling in opposite directions at 60 mph and colliding head on. Because cars are designed to crumple up, they will likely end up sticking together, resulting in a perfectly inelastic collision. In order for momentum to be conserved, the resulting velocity of the crumpled-up cars after the collision will be zero. The kinetic energy of the system before the collision was the sum of two vehicles’ kinetic energies, which was some large number. The kinetic energy of the system after the collision is **zero** (because their final velocity is zero). (This means that the kinetic energy of the system was _not_ conserved, which is indicative of an inelastic collision.) Where did all of the kinetic energy go? It went into crumpling the two cars up, as well as the noise and heat generated as a result of the collision. ETA: And I see on reviewing that **md2000** covered some of the same points while I was typing this up, such as the idea of “crumple zones.” ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [May 28, 2016, 7:01pm UTC](https://boards.straightdope.com/t/simple-physics-question/754817/47 "2016-05-28T19:01:55Z")
It’s been years since I did any ral physics, but - the most interesting pretty much elastic collisions are the sort where you drop a solid steel ball bearing onto a thick steel plate. It bounces pretty spectacularly, but some of the energy is obviously converted to sound, and as anyone using a hammer can attest, steel on steel also produces heat. Many science museums demonstrate this with the bouncing ball bearing… OTOH, a rubber ball bouncing works similarly but more obviously - the energy is converted to deformation during the collision with the ground, but being elastic (Hooke spring principle) the deformed ball “un-deforms” and that energy is converted to kinetic energy -again, with some loss to heat and noise. Cars in collision crumple and the energy is absorbed and converted to heat and mechanical deformation, reducing the amount that remains to be used as momentum. Deformation also reduces the “rebound”, the car body is not springy. ---
**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:** [May 28, 2016, 8:23pm UTC](https://boards.straightdope.com/t/simple-physics-question/754817/48 "2016-05-28T20:23:32Z")
> [@](#): > > Quoth **robby** : > > Francis Vaughan has the correct answer here, with a couple of nitpicks. A lot of the energy of the collision of two cars will go into the energy required to crumple the cars up, in addition to heat and noise. Note that modern cars are specifically designed with energy-absorbing crumple zones. The energy required to crumple the cars up _is_ energy lost to heat. A permanently-bent piece of steel has the same energy content as a straight piece of steel. **md2000** , don’t think of “energy being used as momentum”. Energy and momentum are two different things, and momentum is absolutely conserved. The total momentum the cars have after the collision will be exactly equal to the total momentum before, regardless of what happens to any or all of the energy. ---
**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:** [May 29, 2016, 4:09am UTC](https://boards.straightdope.com/t/simple-physics-question/754817/49 "2016-05-29T04:09:01Z")
> [@Chronos](#): > > The energy required to crumple the cars up _is_ energy lost to heat. A permanently-bent piece of steel has the same energy content as a straight piece of steel. :smack: You’re right, of course. Not sure what I was thinking here. I withdraw my incorrect nitpick on this point, **Francis Vaughan** …my bad. ☹ [Previous page](https://boards.straightdope.com/t/simple-physics-question/754817.md?page=2)