Something I was taught in High School Driver Education class.

[Towlie]

I remembered this while reading the Toy Science thread:

In a Driver Education class I took way back in High School, there was a lot of effort made to scare us into being good drivers, including such extreme measures as showing us pictures of terrible accidents with lots of blood.

One of the things we were taught is that if you're driving at 50 MPH and you meet another car head-on that's also going 50 MPH, it would be like hitting a brick wall at 100 MPH. (The implied meaning of "brick wall" here is a wall that's unaffected by the collision.)

I've always been suspicious of that claim. It seems to me that it would be more like hitting a brick wall at 50 MPH, not 100 MPH. I suspect that either the teacher didn't understand the physics of it, or he felt it was okay to lie to us if it made us drive more carefully.

What do you guys think?

 

[athene]

The instructor's statement makes very little sense.

1. You are very unlikely to hit a car exactly head on, as our instinct is to swerve out of the way of certain death. If you don't hit the car head-on the physics I will attempt to apply is irrelevant.

2. Even if you do hit the car head on, the transfer of energy is different from hitting a brick wall. Unless we're dealing with really ancient cars, they will be designed to compress in the front. Unles the brick wall compresses also, the stopping distance is doubled in the first case. Deceleration will be slower.

3. One of the cars will most likely "win" - push the other car out of the way. This changes the transfer of energy again.

I may be totally wrong about the physics of the impact, as I've never studied the physics of a car crash. My guess is that the equivalent is somewhat higher than 50, but less than 100.

None of it changes the fact that hitting a brick wall isn't even scarier than hitting another car (though you would arguably have to be more drunk to manage it). It's just an excuse to change the speed to a higher number.

 

[GeeMack]

I remembered this while reading the Toy Science thread:

In a Driver Education class I took way back in High School, there was a lot of effort made to scare us into being good drivers, including such extreme measures as showing us pictures of terrible accidents with lots of blood.

One of the things we were taught is that if you're driving at 50 MPH and you meet another car head-on that's also going 50 MPH, it would be like hitting a brick wall at 100 MPH. (The implied meaning of "brick wall" here is a wall that's unaffected by the collision.)

I've always been suspicious of that claim. It seems to me that it would be more like hitting a brick wall at 50 MPH, not 100 MPH. I suspect that either the teacher didn't understand the physics of it, or he felt it was okay to lie to us if it made us drive more carefully.

What do you guys think?

What do you think your speed is relative to the location of the other vehicle?

 

[Monza]

What do you think your speed is relative to the location of the other vehicle?

Exactly. The point of the teacher's comment was simple vector addition. In that sense, the teacher was correct. The relative velocity of the other vehicle with respect to you is 100 MPH.

In actual practice, you wouldn't hit directly head on, not all of the energy will be directly transferred, etc. as previously mentioned.

 

[Bob Klase]

I've always been suspicious of that claim. It seems to me that it would be more like hitting a brick wall at 50 MPH, not 100 MPH.

I agree with athene. Ignoring the 'brick wall' part, if you're doing 50 MPH and you hit a a stationary car head on it would be like hitting a stationary car head on at 50 MPH. If you hit a car moving at 50 MPH directly at you head on I'm guessing the impact is going to be significantly more than when you hit the stationary car.

 

[Gate2501]

3. One of the cars will most likely "win" - push the other car out of the way. This changes the transfer of energy again.

I was in a head on collision in the "loser" car a few years ago. I was the passenger in a little Geo Metro, when we were hit head on by one of those old Monte Carlo's(I think that's the name of the car, not too sure). The other car pushed us a good 15 feet before we came to a halt. Basically, we slowed him down, while he threw us totally backwards.

The front end of the Geo looked like a crushed Dixie Cup, but thanks to our seat belts, and a little luck, we were both fine. The 3 passengers in the other car were all not wearing seat belts, and came out bloodied, one had what looked to be a severely broken nose.

 

[pgwenthold]

There is a huge difference in kinetic energy. The kinetic energy of a car moving 100 mph hitting a brick wall is 4 times that of 2 cars going 50 mph colliding.

Moreover, given that the brick wall is less forgiving, the car hitting the brick wall is going to be the object to dissipate that energy (basically by being blown to smithereens). OTOH, if it is two cars, they can each carry away that kinetic energy by blowing off pieces. The collision with the other car is far more elastic.

 

[Uncayimmy]

There is a huge difference in kinetic energy. The kinetic energy of a car moving 100 mph hitting a brick wall is 4 times that of 2 cars going 50 mph colliding.

I disagree. The two cars going 50mph are only going 50mph relative to some third point of reference (the ground). Relative to car A, car B is going 100mph and vice-versa.

 

[alexi_drago]

If two cars each travelling at 50mph hit exactly head on the front of each car will stop instantly and the impact on each car will be reduced by it's own crumpling, the same as one car hitting a solid wall at 50mph.

I think.

 

[lionking]

Exactly. The point of the teacher's comment was simple vector addition. In that sense, the teacher was correct. The relative velocity of the other vehicle with respect to you is 100 MPH.

In actual practice, you wouldn't hit directly head on, not all of the energy will be directly transferred, etc. as previously mentioned.

Absolutely, and I cannot understand why some here think otherwise.

 

[alexi_drago]

Exactly. The point of the teacher's comment was simple vector addition. In that sense, the teacher was correct. The relative velocity of the other vehicle with respect to you is 100 MPH.

In actual practice, you wouldn't hit directly head on, not all of the energy will be directly transferred, etc. as previously mentioned.

Absolutely, and I cannot understand why some here think otherwise.

I don't agree.

If two cars hit head on, each going at 50mph the closing speed is 100mph and would have the same effect as one car travelling at 100mph hitting a stationary car head on. In that case the stationary car will crumple and accelerate as much as the moving car will crumple and decelerate. In the case of hitting a solid wall (strong enough to resist the impact) it doesn't crumple or move.

 

[Towlie]

I just wrote almost the same thing you did, Alexi, but you posted your message first. I would just add that the parked car would be in neutral gear and have no brakes applied.

 

[Towlie]

In actual practice, you wouldn't hit directly head on, not all of the energy will be directly transferred, etc. as previously mentioned.

For the purposes of this discussion, let's assume that the two cars are identical in every respect, their speeds are the same, and they do hit directly head-on.

 

[BenBurch]

A single course brick wall is just not that solid, either...

 

[alexi_drago]

A single course brick wall is just not that solid, either...

I was assuming it was intended as a wall strong enough to resist the impact without much damage or movement, trees can be very bad also.

 

[Towlie]

Right. I defined "brick wall" in my initial post.

 

[Uncayimmy]

Towlie, I'm not clear on what your objection is. You said the teacher said "like" rather than "precisely the same as" - which is what my teacher said as well. I think the point the teacher was trying to make is that your energy and the energy from the other car are combined because of your relative speed (layman's terms 'cause I'm a layman). If you're going 50mph (relative the ground) and hit a car going 49mph in the same direction, that's like hitting a "brick wall" at 1mph.

Only it's not, really. Laymen and experts alike could fill page after page describing how a brick wall is not like a car. I don't even know if you could make a brick wall the same shape and size of a car yet keep it the same weight. Even if you could, it's a different material with a different distribution of mass. Furthermore, a car is on wheels whereas the brick wall is on the ground. There are countless other factors as well.

But let's take a simpler example. Suppose you have two soda cans. You roll A at 10mph towards B, which is stationary. Essentially, A stops and B moves in the direction A was going for a little bit. Next you roll A and B at 5mph towards each other. Essentially they hit and they each bounce back a little bit.

In a perfect world the two cans are going to end up the same distance apart each time. From our reference frame watching it happen, the two collisions look different. From the reference frame of the center of mass of the two cans (this frame will move), the collisions look the same.
You can check it out visually with this Java animation:
http://qbx6.ltu.edu/s_schneider/physlets/main/momenta4.shtml

If you forget "brick wall" and just think about two vehicles colliding head on, then the teacher is essentially right - relative velocity is what counts.

 

[GlennB]

I disagree. The two cars going 50mph are only going 50mph relative to some third point of reference (the ground). Relative to car A, car B is going 100mph and vice-versa.

True, but the kinetic energy is proportional to the square of v, which here is 100. Which makes 4x the energy of the 50mph / immovable-brick-wall scenario. But the damage is shared between 2 cars, so the head-on would seem to be doubly-damaging than the plain 50mph/wall impact.

 

[Uncayimmy]

True, but the kinetic energy is proportional to the square of v, which here is 100. Which makes 4x the energy of the 50mph / immovable-brick-wall scenario. But the damage is shared between 2 cars, so the head-on would seem to be doubly-damaging than the plain 50mph/wall impact.

Where did the 50mph brick wall impact come from? You wrote, "The kinetic energy of a car moving 100 mph hitting a brick wall is 4 times that of 2 cars going 50 mph colliding."

 

[alexi_drago]

Towlie, I'm not clear on what your objection is. You said the teacher said "like" rather than "precisely the same as" - which is what my teacher said as well. I think the point the teacher was trying to make is that your energy and the energy from the other car are combined because of your relative speed (layman's terms 'cause I'm a layman). If you're going 50mph (relative the ground) and hit a car going 49mph in the same direction, that's like hitting a "brick wall" at 1mph.

Only it's not, really. Laymen and experts alike could fill page after page describing how a brick wall is not like a car. I don't even know if you could make a brick wall the same shape and size of a car yet keep it the same weight. Even if you could, it's a different material with a different distribution of mass. Furthermore, a car is on wheels whereas the brick wall is on the ground. There are countless other factors as well.

But let's take a simpler example. Suppose you have two soda cans. You roll A at 10mph towards B, which is stationary. Essentially, A stops and B moves in the direction A was going for a little bit. Next you roll A and B at 5mph towards each other. Essentially they hit and they each bounce back a little bit.

In a perfect world the two cans are going to end up the same distance apart each time. From our reference frame watching it happen, the two collisions look different. From the reference frame of the center of mass of the two cans (this frame will move), the collisions look the same.
You can check it out visually with this Java animation:
http://qbx6.ltu.edu/s_schneider/physlets/main/momenta4.shtml

If you forget "brick wall" and just think about two vehicles colliding head on, then the teacher is essentially right - relative velocity is what counts.

I don't believe the relative velocity is what counts.

Consider poolballs colliding (ignoring spin/friction etc.):

Two poolballs moving towards each other at 50 mph in opposite directions hit head on, the effect each ball has on the other during the time of contact is as if each ball has hit a stationary solid immovable surface at 50 mph.

The relative velocity when two balls are involved is 100mph

The relative velocity when one ball and an immovable stationary surface is involved is 50mph

But the forces acting on each ball in each case is the same.