Something I was taught in High School Driver Education class.

[sol invictus]

Huh?

"Huh?" indeed. Once again, I agree with everything in your post... and as far as I can tell, you agree with everything in mine. You don't seem to have read the thread very carefully.

Everyone agrees that the claim attributed to the teacher in the OP is incorrect. The more interesting question is what running into the wall actually is equivalent to. Again, my claim (and that of just about every other post) is that running into a solid wall at 50mph causes the same damage to the car as running head-on into an identical car, with both cars traveling at 50mph. Running into a wall at 100mph is much more damaging.

I can't tell whether you agree with that or not.

 

[Tim Thompson]

Kinetic Energy II

Tim, what we're really concerned with here is the impact on a single car, so your math appears to show a difference of four times, not just double. Am I right?

I don't think so, because you can't get rid of that factor (1/2), but I am unsure of the scenario. But whatever you have in mind, just go back to the basics.

If the only thing moving is the car, then the total energy available to cause ruin & havoc is the kinetic energy of the car, (1/2)M_carV_car². That total energy is independent of whether the car strikes an immovable wall or another moveable car. Clearly the amount of energy turned into damage depends on how inelastic the collision is. My little equations simply show that two cars slamming into each other head on have only half the kinetic energy of a single car slamming into the "wall" at double the speed of the car.

All you really need to do is construct your scenario, whatever it is and then calculate the total kinetic energy. Whatever it it, that's all you get. What happens to it depends on the details of the collision.

But I have ignored momentum and that is something else to consider (conservation of momentum). In a truly elastic collision the car just bounces off the wall with equal momentum backwards. But any real collision is doubtless far from elastic and the specifics of how the car & wall react to the collision will dictate what happens damage wise.

 

[alexi_drago]

I haven't wasted time looking at links since I checked out the first one (from the Univ of Illiois) which claimed they were the same based on the assumption (implicit in their scenerio) that one of the cars had an infinite effective mass, basically equal to a brick wall.

Where is that implied? (I'm not saying it's not, I just can't see where)

On the 'Brick wall' thing, I've always thought that running into a brick wall is used to describe encountering something (doesn't always have to be physical) that stops you dead in your tracks, could be something stopping you from solving a problem or completing work as well as a physical object so a car hitting a brick wall is stopped dead as would happen to a car hitting another head on travelling at the same speed and that's probably where the analogy comes from imo (both are stopped dead in their tracks) but some confusion has crept in along the way concerning the relative speeds.

 

[richardm]

What if the brick wall was on a conveyor belt moving at the same speed as the car?

 

[Ron Swanson]

I realize "Brick Wall" is just a phrase as metioned.

But I had to add this YouTube video of a poilce car "hitting a brick wall"

http://www.youtube.com/watch?v=BIigQwGtboQ

 

[alexi_drago]

What if the brick wall was on a conveyor belt moving at the same speed as the car?

Well, to me anyway, I would consider a stationary brick wall to be fixed firmly to the earth and it's speed relative to the earth before AND after the collision to be equal to zero.
Having a moving (relative to the earth) brick wall may imply that it's speed could change due to the collision so therefore the mass of the brick wall and possibly the conveyor and whatever was powering the conveyor would have to be taken into consideration.

 

[Towlie]

Don't take that seriously, Alexi. It's just a humorous allusion to the old airplane on a conveyor belt puzzle.

 

[alexi_drago]

I realize "Brick Wall" is just a phrase as metioned.

But I had to add this YouTube video of a poilce car "hitting a brick wall"

http://www.youtube.com/watch?v=BIigQwGtboQ

Yes, in reality if you have the choice of a brick wall or a tree, go for the wall.

 

[Mason]

Oh! Oh! I got one!!

What if one car were to hit a brick wall at the exact same moment as the second car were to hit the other side of that same brick wall?

Now remove the wall, and what has changed?

I keep picturing those metal balls suspended by two strings in a row of five, and when you lift one and let it fall back in place it makes the one at the opposite end pop off, and they go back and forth for like 5 minutes.

Anyway, if the wall absorbs none of the energy, each car would hit the wall at 50 mph, and it would be just like a car hitting a wall at 50 mph. Remove the wall and assume that neither car absorbs the other car's energy (just like the wall didn't), and it would be the exact same situation, except for the lack of bricks.

Clack-clack-clack-clack-clack

 

[GlennB]

Consider this:

1) Two cars of identical mass m collide head on with velocity v for each car, relative to the center of mass. The total kinetic energy of the collision is simply the sum of the kinetic energies of the two cars ...

(1/2)mv² + (1/2)mv² = mv²
2) One car slams into a stationary immovable object ("wall") at double the speed from the case above. Now the kinetic energy of the car is ...

(1/2)m(2v)² = (1/2)m*4v² = 2mv²
Now, unless I am making some elementary mistake, the kinetic energy of the car hitting the wall is double the kinetic energy of the two cars colliding. This implies to me that the two scenarios are not at all equivalent, with the latter being in fact far more damaging.

Excellent, and I think I agreed with this much earlier in the thread confused

But it raises an interesting theoretical point. Suppose we change our frame of reference to that of one of the cars in the 50-50 headon collision. Now the kinetic energy of the 'moving' car is the same as the 100mph car in the 'wall collision, and has somehow doubled?! It's 100 v 0, therefore 2mv². Just by changing one's frame of reference ??

Doesn't seem right at all. My only guess is that kinetic energy can only reasonably be calculated through the absolute relative velocity (100) rather than the sum of the energies of separate velocities (50+50). Or I could be talking tosh.

 

[Modified]

But it raises an interesting theoretical point. Suppose we change our frame of reference to that of one of the cars in the 50-50 headon collision. Now the kinetic energy of the 'moving' car is the same as the 100mph car in the 'wall collision, and has somehow doubled?! It's 100 v 0, therefore 2mv². Just by changing one's frame of reference ??

The change in KE will be the same in both cases. The absolute value from any particular reference frame doesn't matter. After all, you could measure KE from an airplane passing overhead.

 

[Myriad]

Three cases:

1. 100 MPH car crashes head-on into stationary car.
2. 50 MPH car crashes head-in into another 50 MPH car.
3. 100 MPH car crashes into a solid immovable wall.

It's true that the cars in case 1 have twice the total kinetic energy of the cars in case 2. However, the same amount of kinetic energy gets converted into deformation of the cars in both cases. In case 1, at least half the kinetic energy going in still exists as kinetic energy after the crash. That is, both cars will end up moving at 50 MPH in the direction the 100 MPH car was originally going.

That's in the case where the collision is completely plastic, meaning that in case 2, the cars come to a stop when they collide without bouncing off.

So in case 1, there's twice as much kinetic energy, but half of it remains kinetic energy after the crash. In case 2, there is no kinetic energy after the crash. The same amount of energy deforms both cars in both cases.

If there's some elasticity involved, then the cars in case 2 will bounce off each other to some degree, so they will have some kinetic energy afterward. That means that in case 1, the 100 MPH car after the crash will be going at a little slower than 50 MPH, and the stationary car will be going a little faster than 50 MPH. The difference between total kinetic energies before and after the crash, and the amount of kinetic energy converted in crumpling the cars, will still be the same for both cases.

In the opposite extreme of a totally elastic collision, in case 1 the 100 MPH car will be left stationary and the stationary car will be moving at 100 MPH after the collision, and in case 2 both cars will be moving 50 MPH in the directions opposite to their original movements. The difference in total KE before and after will still be the same in both cases -- that is, zero in both cases. (Note that a more elastic collision is not better for the drivers; unless the cars are very long and made of atypical materials, the accelerations involved will be greater the more elastic the collision is. However, for a totally elastic collision we must of course have totally elastic drivers, who would therefore be unharmed.)

In case 3, if the wall suffers no displacement or damage, and the car ends up stationary at the wall (no significant bounce), then all of its kinetic energy has been converted into deformation of the car. That's four times as much energy of deformation as either car suffers in case 1 or 2. So the 100 MPH crash into a hypothetical invincible wall is much worse than the two 50 MPH cars crashing head-on. In fact it would be basically the same as two 100 MPH cars crashing head-on.

Crashing into the brick wall at 100 MPH would be comparable to crashing head-on into another car each going 50 MPH, if the result of the brick wall crash were that the car smashes through the wall and keeps going at 50 MPH afterward.

ETA: Crashing into an invulnerable wall at 100 MPH would also be comparable to crashing head-on with another vehicle, both going 50 MPH, if the other vehicle were much more massive such as a locomotive or loaded semi truck.

Respectfully,
Myriad

 

[edd]

Consider this:

1) Two cars of identical mass m collide head on with velocity v for each car, relative to the center of mass. The total kinetic energy of the collision is simply the sum of the kinetic energies of the two cars ...

(1/2)mv² + (1/2)mv² = mv²
2) One car slams into a stationary immovable object ("wall") at double the speed from the case above. Now the kinetic energy of the car is ...

(1/2)m(2v)² = (1/2)m*4v² = 2mv²
Now, unless I am making some elementary mistake, the kinetic energy of the car hitting the wall is double the kinetic energy of the two cars colliding. This implies to me that the two scenarios are not at all equivalent, with the latter being in fact far more damaging.

What you're missing is the conservation of momentum as well as energy. With two cars at 50mph all the kinetic energy can go into the squishing and mangling of things you'd not want squished and mangled. With one at 100 and a stationary object you aren't allowed to have all the energy disappear into squishing and mangling. Some has to be left over to provide the velocity of the final system - as conservation of momentum requires that the car+wall lump is still going a certain velocity in the same direction as the car was.

 

[GlennB]

The change in KE will be the same in both cases. The absolute value from any particular reference frame doesn't matter. After all, you could measure KE from an airplane passing overhead.

Agreed. But Tim's maths look conclusive, yet counter-intuitive. While I agree that the 50-50 head on should involve the same total KE as single 100-wall , Tim's figures seem to be out by a factor of 2. Because of the v² factor. Basically, (10/2)² + (10/2)² is not the same as 10²

 

[marting]

What you're missing is the conservation of momentum as well as energy. With two cars at 50mph all the kinetic energy can go into the squishing and mangling of things you'd not want squished and mangled. With one at 100 and a stationary object you aren't allowed to have all the energy disappear into squishing and mangling. Some has to be left over to provide the velocity of the final system - as conservation of momentum requires that the car+wall lump is still going a certain velocity in the same direction as the car was.

The momemtum lost by a car crashing into the wall would be transferred to the wall and earth it is connected to. In an fully inelastic collision all the momentum would be transferred. However, the energy pickup of the earth/wall system would be infintessimal. It is quite similar to the problem of a person jumping up in the air. Just before they jump their momentum is zero, Upon jumping their momentum is now mv. The earth moves in the opposite direction at v/A where A is an extremely large number, the ratio of the Earth's mass to the person's mass. Thus the Earth's momentum would be (v/A)*(A*m) which matches the person's momentum, mv.

 

[JoeTheJuggler]

When I heard my teacher say essentially the same thing, the above was the lesson I got out of it. In my mind I know that "brick wall" is vague. How big is it? Is it attached to a building? Is it just sitting on the ground or sunk into it? So I figured the teacher meant for us to imagine a brick wall built to be a reasonable but dramatic equivalent to a car. I figured the teacher's goal was relative velocity, not mass, since we're probably never going to find ourselves driving into brick walls.

That's what I was thinking too--that the point was mostly about relative velocity.

Also, I just recently built a bit of brick wall just one red clay brick thick with type "N" mortar and no other support. A wall of this construction the size of the cross section of a car (the rest of the mass of the wall extending to the sides is mostly irrelevant) would be considerably less massive than a car and would be reduced to dust if a car going 50 mph hit it head on. The car would be damaged, no doubt, but I think it would make it through the wall.

It's a different story if you assume the brick wall is infinitely massive and can stop the car without sustaining any damage to itself.

I wasn't imagining a magical brick wall.

 

[JoeTheJuggler]

The momemtum lost by a car crashing into the wall would be transferred to the wall and earth it is connected to.

How's the wall connected to the earth? Are we talking about something other than a wall composed of nothing but bricks and mortar?

ETA: I once lived in an apartment upstairs of a small bar in a brick building, and one day a school bus drove into the bar (actually took out the entire corner of the bar).

I just googled for photos of cars crashing into brick buildings, and these are typical of what you get:

http://www.citynews.ca/itsyourstory...ategoryid=42&text=&imageid=2358&box=&shownew=

http://www.dailymail.co.uk/news/art...-kills-sports-car-crashes-police-pursuit.html

http://www.nj.com/news/index.ssf/2007/10/a_union_township_man_lost.html

And these structures are quite a bit sturdier than a free-standing brick wall.

 

[Modified]

Agreed. But Tim's maths look conclusive, yet counter-intuitive. While I agree that the 50-50 head on should involve the same total KE as single 100-wall , Tim's figures seem to be out by a factor of 2. Because of the v² factor. Basically, (10/2)² + (10/2)² is not the same as 10²

In the frame of the road, you start with two cars moving at 50 mph and end up with two cars moving at zero.

In the frame of one car, you start with one car moving at 100 mph and one at zero, and end up with two cars moving at 50 mph.

(50² + 50²) - 0 = 100² - (50² + 50²)

 

[Uncayimmy]

Excellent, and I think I agreed with this much earlier in the thread confused

But it raises an interesting theoretical point. Suppose we change our frame of reference to that of one of the cars in the 50-50 headon collision. Now the kinetic energy of the 'moving' car is the same as the 100mph car in the 'wall collision, and has somehow doubled?! It's 100 v 0, therefore 2mv². Just by changing one's frame of reference ??

Doesn't seem right at all. My only guess is that kinetic energy can only reasonably be calculated through the absolute relative velocity (100) rather than the sum of the energies of separate velocities (50+50). Or I could be talking tosh.

I'm with you. I think the problem here is Car A is not going 50mph relative to what it impacts (car B). It is going 100mph relative to B. And car A is going 100mph relative to car B. Tim measured two speeds relative to the center of mass. What we really have is one event with two masses striking each other at a relative speed of 100mph. Neither vehicle is striking the center of mass since that's just a concept - a location.

What Tim is describing is sort of like having two separate events. Each event has a mass moving at 50mph and striking a second stationary mass. His math works to describe adding together the kinetic energy of each of the two events. But that's not what's happening when two cars collide at a closing speed of 100mph.

The kinetic energy formula he gave is
E=(1/2)mv2.

To arrive at the correct answer, I say you need to do something more like this:
* Add the masses together: m1 + m2 = m
* Use their relative velocity for v

All the usual disclaimers apply - I'm just a curious layman.

 

[Uncayimmy]

In the frame of the road, you start with two cars moving at 50 mph and end up with two cars moving at zero.

Won't that depend on how elastic the collision is? Two bowling balls, for example, will bounce off each other. Well, maybe not at 50mph, but certainly at a lower speed they will hit and bounce apart.

In the frame of one car, you start with one car moving at 100 mph and one at zero, and end up with two cars moving at 50 mph.

(50² + 50²) - 0 = 100² - (50² + 50²)

How can that be? I thought that by definition if you choose the frame of one car, that car is always stationary.