Something I was taught in High School Driver Education class.

[Uncayimmy]

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.

I'm not an expert in this field, but I'm sure the experts have my back, which is to say they will enthusiastically correct me when I'm wrong.

Stationary is relative. There is nothing that is absolutely stationary because there is no absolute. That "stationary immovable surface" is hurling through space on the surface of this rotating planet. We can "see" this depending on the reference frame we choose. We can also choose a reference frame in which it is stationary.

So, in your example you are actually working with relative velocities whether you realize it or not. Suppose we're looking at the pool table from the reference frame of the sun. The Earth is moving at about 67,000 mph relative to the sun, which means the pool table and everything on it is as well.

If we hit the cue ball so it goes 50mph in the opposite direction as the Earth relative to the sun, it is now moving 66,950mph relative to the sun (ignoring relativistic effects). When it impacts the cushion (your immovable "stationary" surface), the relative speed is only 50mph. It's the same relative speed we calculate if we're standing in the room and doing the math, only in that case we use 0mph and 50mph rather than 67,000mph and 66,950mph.

Thus with the two cars impacting each other, we can choose from an infinite number of reference frames, all of which are going to come up with a relative velocity of 100mph. From the reference frame of standing on the highway, we see each going 50mph. We could do the same calculation from one of the cars and just as accurately say that car is stationary and the other car is moving at 100mph.

 

[alexi_drago]

I'm not an expert in this field, but I'm sure the experts have my back, which is to say they will enthusiastically correct me when I'm wrong.

Stationary is relative. There is nothing that is absolutely stationary because there is no absolute. That "stationary immovable surface" is hurling through space on the surface of this rotating planet. We can "see" this depending on the reference frame we choose. We can also choose a reference frame in which it is stationary.

So, in your example you are actually working with relative velocities whether you realize it or not. Suppose we're looking at the pool table from the reference frame of the sun. The Earth is moving at about 67,000 mph relative to the sun, which means the pool table and everything on it is as well.

If we hit the cue ball so it goes 50mph in the opposite direction as the Earth relative to the sun, it is now moving 66,950mph relative to the sun (ignoring relativistic effects). When it impacts the cushion (your immovable "stationary" surface), the relative speed is only 50mph. It's the same relative speed we calculate if we're standing in the room and doing the math, only in that case we use 0mph and 50mph rather than 67,000mph and 66,950mph.

Thus with the two cars impacting each other, we can choose from an infinite number of reference frames, all of which are going to come up with a relative velocity of 100mph. From the reference frame of standing on the highway, we see each going 50mph. We could do the same calculation from one of the cars and just as accurately say that car is stationary and the other car is moving at 100mph.

Take a large sheet of paper and draw bricks on it and have it stationary and vertical to the ground and use that as the frame of reference.
Take 2 cars and drive them both at 50mph towards each other so that the collide exactly at the paper on opposite sides.
The effect on each car is as if it's just hit a solid immovable wall at 50mph.

 

[realpaladin]

Take a large sheet of paper and draw bricks on it and have it stationary and vertical to the ground and use that as the frame of reference.
Take 2 cars and drive them both at 50mph towards each other so that the collide exactly at the paper on opposite sides.
The effect on each car is as if it's just hit a solid immovable wall at 50mph.

If you ever watched MythBusters you know how nigh-on impossible it is to crash exactly at the same time in the same spot

 

[alexi_drago]

If you ever watched MythBusters you know how nigh-on impossible it is to crash exactly at the same time in the same spot

I'm sorry, I didn't make it clear that I wasn't actually expecting anyone to carry out such a test but more to imagine it and if you want do some calculations.

 

[Uncayimmy]

Take a large sheet of paper and draw bricks on it and have it stationary and vertical to the ground and use that as the frame of reference.
Take 2 cars and drive them both at 50mph towards each other so that the collide exactly at the paper on opposite sides.
The effect on each car is as if it's just hit a solid immovable wall at 50mph.

Hold the paper stationary relative to what?

Did you visit this link?
http://qbx6.ltu.edu/s_schneider/physlets/main/momenta4.shtml

It provides two reference frames. One is what they call the lab. The other is the reference frame of the center of mass. You can change the speeds all you want. The default has one object moving at 2 and the other at 0. Change it to 1 and -1. TheN change it to 0 and -2. Then try 2.5 and 0.5.

* From the reference frame of the lab, each collision will look different. Do you agree or disagree with the simulation?

* From the reference point of the center of mass all four collisions will look exactly the same. Do you agree or disagree the simulation?

I'm saying that 0mph vs 100mph is the same as 70-30, 50-50, 30-70, and 0-100. The two objects will end up the same distance apart. It is just as valid to claim that "A hit B, which resulted in A stopping and B moving distance d" as it is to say "A and B collided and both moved 1/2d from the impact point and thus ended up d apart." It's all a matter of your reference frame.

Let's go back to your pool table. Ball A is stationary relative to the table. We roll a second ball at N meters per second relative to the table, and it impacts ball A. The two balls end up d distance apart. This is a fact.

Next we roll both balls at 1/2N meters per second relative to the table each so they have a relative speed of N meters per second. How far apart will they be after the collision? Will it be the same as in the first trial?

 

[alexi_drago]

Hold the paper stationary relative to what?

stationary and vertical to the ground and use that as the frame of reference.

I'm not disputing what the relative velocities are but rather what forces are exerted on an object depending on whether it collides with an object with equal mass and deformability to itself or a solid immovable surface.

It's those forces that determine the severity of a collision.

 

[GlennB]

Where did the 50mph brick wall impact come from?

The o/p

 

[Uncayimmy]

I'm not disputing what the relative velocities are but rather what forces are exerted on an object depending on whether it collides with an object with equal mass and deformability to itself or a solid immovable surface.

It's those forces that determine the severity of a collision.

You didn't answer any of my specific questions. Did you look at the simulation or try any of the tests I asked about? Holding everything else the same, do you think the individual speeds of the vehicles relative to anything besides each other makes any difference at all?

There is no such thing as an immovable object. All objects are moving. No object is perfecly rigid either. We're talking about two cars colliding at a relative speed of 100mph to each other. You want to change one of the cars into something that doesn't exist. How does that make any sense? You could change it into something that could exist, but we'd need at least two other pieces of data: What is the mass of this other object? What is its elasticity?

I am not arguing that it's impossible to choose a mass and elasticity in such a way as to make your numbers work. You could find an infinite number of solutions to make it work. But that's not what you seem to be claiming when you say that the "effect on each car is as if it's just hit a solid immovable wall at 50mph."

As far as the two cars are concerned, you could vary their speeds relative to the ground all you want so long as you keep their speed relative to each other the same. 100-0, 90-10, 50-50, 10-90, or 0-100: it doesn't make any difference so long as we're using the same two cars, thus holding the mass and elasticity the same.

Do you agree with the above paragraph or not?

 

[Uncayimmy]

The o/p

See my posts about relative velocities. There is no stationary wall. It's moving relative to something. We're just using a reference frame where it's stationary. Let's try this another way because I'm not following you.

Suppose the brick wall is moving at 50mph relative to the ground just like the second car would be. The car and will hit each other at a relative velocity of 100mph. Is that any different than the car going 100mph into a wall that is stationary relative to the ground? How about a stationary (relative to the ground) and a brick wall moving 100mph (relative to the ground/car)?

I say it's the same in all three events. Do you agree or disagree?

 

[alexi_drago]

I was going to try to explain again but I don't know how to explain the same thing any differently.

Here's a link with probably a better written explanation.

http://van.physics.illinois.edu/qa/listing.php?id=92

 

[GlennB]

See my posts about relative velocities. There is no stationary wall. It's moving relative to something.

The o/p states "It seems to me that it would be more like hitting a brick wall at 50 MPH, ", which is why people are mentioning it. It was the whole point of the discussion.

 

[alexi_drago]

The wall is stationary relative to the ground it is anchored to, the same ground that the observer is standing stationary on and the same ground that the car is moving 50mph relative to. I don't see the need to complicate things further.

 

[69dodge]

See my posts about relative velocities. There is no stationary wall. It's moving relative to something. We're just using a reference frame where it's stationary.

Yes, but the essential point about a brick wall is that it will remain stationary (or moving, depending on the reference frame you're using) after the collision as well as before it. Cars aren't like that; they change velocity in a collision.

Let's try this another way because I'm not following you.

Suppose the brick wall is moving at 50mph relative to the ground just like the second car would be. The car and will hit each other at a relative velocity of 100mph. Is that any different than the car going 100mph into a wall that is stationary relative to the ground? How about a stationary (relative to the ground) and a brick wall moving 100mph (relative to the ground/car)?

I say it's the same in all three events. Do you agree or disagree?

I agree. But that wasn't the original question.

In all three cases you're considering here, there is one brick wall and one car. So, the cases are all equivalent.

The original question asks whether a pair of cars is the same as a car and a brick wall. Those two cases are not equivalent.

You can think of "a brick wall" as meaning "an object of infinite mass." Brick walls are attached solidly to the Earth, and the mass of the Earth is effectively infinite compared to that of a car.

 

[brodski]

The wall is stationary relative to the ground it is anchored to, the same ground that the observer is standing stationary on and the same ground that the car is moving 50mph relative to. I don't see the need to complicate things further.

How would you feel if the brick wall were not stationary, but was moving towards the car at an additional 50 MPH to the car which is already doing 50MPH*, do you still think that the forces involved would be identical?
If not, how would they change?




*This is just another way of describing a car hitting a wall at 100MPH

 

[pgwenthold]

Yes, but the essential point about a brick wall is that it will remain stationary (or moving, depending on the reference frame you're using) after the collision as well as before it. Cars aren't like that; they change velocity in a collision.



I agree. But that wasn't the original question.

In all three cases you're considering here, there is one brick wall and one car. So, the cases are all equivalent.

The original question asks whether a pair of cars is the same as a car and a brick wall. Those two cases are not equivalent.

You can think of "a brick wall" as meaning "an object of infinite mass." Brick walls are attached solidly to the Earth, and the mass of the Earth is effectively infinite compared to that of a car.

To put it another way, what are the reduced masses of the two situations?

This is the problem with the "center-of-mass" description. The argument that the velocity is relative is correct, but when that argument is made, the mass of the brick wall gets neglected. If you want to argue center-of-mass, then you have to talk about the reduced mass and not just the mass of the car.

It's like firing a ping-pong ball at a bowling ball. If you do that (and hit appropriately), the ping-pong ball is going to get smashed, and the bowling ball is unaffected. However, shoot two ping-pong balls at each other, and they distribute the damage between them.

For the folks arguing "center-of-mass" don't forget the superposition of the two objects. In the car/car scenerio, the damage must be distributed over the two cars, because they are indistinguishable. OTOH, yo can readily distinguish a car and a wall.

Even the link above makes the assumption that the cars hit each other and stop completely. That focuses only on the available energy of the system. It does not talk about the more important problem: energy dissapation. How much energy that goes into a collision is no more important than how energy gets OUT of the collision. German car makers recognize this, for example, and have designed their cars to crumple up (dissipate energy) in a prescribed, safer fashion (unfortunately, there's no safe way for the engine to rip through the driver's seat after a head on collision)

 

[Jeff Corey]

Car Talk discussed this a year ago (5/25/2008). One answer, "The impact forces are equal. Assuming the cars are identical (equal in mass) with crumple zones working as engineers intend; the head-on collision will result in the cars coming to rest without rebound at the point of impact in both scenarios. (inelastic collision)The occupant in the (intact) cab will go from 60mph to zero over a distance of perhaps three feet. ... So the accelerations will be equal. F=ma Therefore the forces felt inside the cab will be equal for both scenarios."

 

[Modified]

Obviously, hitting a car head on where both are going 50 mph is a lot like hitting a stationary car head on at 100 mph, which is not much like hitting an immovable wall.

 

[Towlie]

Modified - I agree that "hitting a car head on where both are going 50 mph is a lot like hitting a stationary car head on at 100 mph." That analogy would be much more useful in a Driver Education class and would be perfectly honest as well.

But I want to address 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).

Perhaps there's a basic misunderstanding here. Of course the head-on collision of two cars traveling at 50 MPH each will involve twice the energy of one car hitting a brick wall at 50 MPH, and therefore the collision will cause twice the damage, but that energy and damage will be split between the two cars, and what we're discussing here is the effect on one car, not both.

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.

No, not really. I thought so at first but then I had to edit this. It would actually be more like hitting a brick wall at 1/2 MPH. In both cases you would slow down by 1/2 MPH.

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 won't bother addressing this because neither of us believes a brick wall is like a car. The soda can analogy isn't very helpful either.

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.

But we're not going to forget about "brick wall" because that's the whole point of this discussion. Please read post#11 from alexi_drago again, because it's crucial to understanding this:

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]

Car Talk discussed this a year ago (5/25/2008). One answer, "The impact forces are equal. Assuming the cars are identical (equal in mass) with crumple zones working as engineers intend; the head-on collision will result in the cars coming to rest without rebound at the point of impact in both scenarios. (inelastic collision)The occupant in the (intact) cab will go from 60mph to zero over a distance of perhaps three feet. ... So the accelerations will be equal. F=ma Therefore the forces felt inside the cab will be equal for both scenarios."

That sounds like it's in agreement with the positions of alexi_drago, modified, myself, and possibly others, but you didn't give us enough information. What were the exact conditions under which they stated the impact forces were equal?

Is it on the Web? Do you have a link?

 

[Jeff Corey]

That sounds like it's in agreement with the positions of alexi_drago, modified, myself, and possibly others, but you didn't give us enough information. What were the exact conditions under which they stated the impact forces were equal?

Is it on the Web? Do you have a link?

It's discussed under Reviews, here.http://www.cartalk.com/ct/review/show.jsp?showid=200821