Something I was taught in High School Driver Education class.

[alexi_drago]

I'm not sure what's come up here but remember that if you use a moving car as your frame of reference, and the car changes speed (as would occur during a collision) your frame of reference does not change with the car.

 

[Modified]

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.

As long as you assume the collision is equally bouncy in both cases, the change in KE will be the same in both cases.

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

You have to choose one inertial reference frame for both the pre and post-collision measurements. I meant the frame in which one of the cars is stationary before the collision, of course.

 

[Uncayimmy]

As long as you assume the collision is equally bouncy in both cases, the change in KE will be the same in both cases.

What do you mean by both cases? Do you mean the two different reference frames?

You have to choose one inertial reference frame for both the pre and post-collision measurements. I meant the frame in which one of the cars is stationary before the collision, of course.

For example, I'm sitting on the roof of the car. As far as I am concerned the car is stationary. I see car B approaching at 100mph. The two cars collide. If they bounce off of each other perfectly, I will see car A moving away from me at 100mph. Car B will be underneath me and stationary. If they stick together perfectly, I'll see car A moving away from me at 50mph.

Is that what you're saying?

 

[alexi_drago]

What do you mean by both cases? Do you mean the two different reference frames?


For example, I'm sitting on the roof of the car. As far as I am concerned the car is stationary. I see car B approaching at 100mph. The two cars collide. If they bounce off of each other perfectly, I will see car A moving away from me at 100mph. Car B will be underneath me and stationary. If they stick together perfectly, I'll see car A moving away from me at 50mph.

Is that what you're saying?

Pretty much although car B will not be directly underneath you if it's left stationary.

 

[Uncayimmy]

I'm not sure what's come up here but remember that if you use a moving car as your frame of reference, and the car changes speed (as would occur during a collision) your frame of reference does not change with the car.

I don't understand what that means. I can imagine two frames of reference at the start of the experiment where car A is stationary. In one case I am stitting on the roof. In the other case my twin is strapped into the driver's seat with a five-point safety harness. In the imprecise language of English I can say that both frames of reference are where car A is stationary before the collision.

When the two cars collide, my twin and I can both accurately describe what happened. Again, using the imprecise language of English, my twin would be reporting from the "frame of reference of the car" and I would not. My twin would always describe car A as stationary. Immediately after impact I would report car A as moving. Neither one of us is wrong.

In the imprecise language of English only one of us is looking at things from the reference frame of car A. Modified said in the frame of one car which I took to mean my twin.

 

[Uncayimmy]

Pretty much although car B will not be directly underneath you if it's left stationary.

I was describing everything from my frame of reference. In that case, car B is underneath me and therefore stationary. We're ignoring, of course, all of the other factors that will cause us to slow down at different rates.

 

[Modified]

What do you mean by both cases? Do you mean the two different reference frames?

Yes, or any other inertial frame you like.

 

[Modified]

I was describing everything from my frame of reference. In that case, car B is underneath me and therefore stationary. We're ignoring, of course, all of the other factors that will cause us to slow down at different rates.

I think alexi meant that car B would be several feet in front of you and stationary.

 

[alexi_drago]

I don't understand what that means. I can imagine two frames of reference at the start of the experiment where car A is stationary. In one case I am stitting on the roof. In the other case my twin is strapped into the driver's seat with a five-point safety harness. In the imprecise language of English I can say that both frames of reference are where car A is stationary before the collision.

When the two cars collide, my twin and I can both accurately describe what happened. Again, using the imprecise language of English, my twin would be reporting from the "frame of reference of the car" and I would not. My twin would always describe car A as stationary. Immediately after impact I would report car A as moving. Neither one of us is wrong.

In the imprecise language of English only one of us is looking at things from the reference frame of car A. Modified said in the frame of one car which I took to mean my twin.

Your frame of reference shouldn't change, you got it right in your previous post.

Using the other example of this topic with a car hitting a wall, say you use the car's initial condition as your frame of reference then the wall is moving at 50mph relative to you, after the collision (if the car and wall collide with no bounce or destruction of the wall) then the car and the wall end up both moving at 50mph in the same direction the wall was initially moving.

 

[alexi_drago]

Sorry, mis-read this bit:

When the two cars collide, my twin and I can both accurately describe what happened. Again, using the imprecise language of English, my twin would be reporting from the "frame of reference of the car" and I would not. My twin would always describe car A as stationary. Immediately after impact I would report car A as moving. Neither one of us is wrong.

In that case your twins frame of reference has changed.

EDIT: this is why I suggested that the ground is assumed to be stationary and everything else moves relative to that.

 

[alexi_drago]

Consider a third car in the two car collision scenario that is travelling alongside you as you sit in one car, before the collision it is stationary relative to you but after the collision it is now moving away from you even though it didn't hit anything and no forces were applied to it.

 

[Uncayimmy]

I think alexi meant that car B would be several feet in front of you and stationary.

I disagree.

I stipulated that the collision was perfectly elastic. The vehicles bounce off of each other. Look at this simulation:
http://qbx6.ltu.edu/s_schneider/physlets/main/momenta4.shtml

Use the default values. A (red block) is moving at 2 while B (blue block) is stationary. After the collision, B is moving at 2 and A is stationary. This is a description from the frame of reference of the ground (they call it the lab).

My frame of reference is immediately above A. Before the collision A will be stationary relative to me. B will be moving towards us at 2. After the collision, A will be moving away from me at -2 because B doesn't touch me. From my frame of reference, B is now stationary What I measure is that B ran into A and accelerated it. B decelerated by the same amount and stopped right underneath me.

Now let's look at it from the frame of reference of A. In this case A measures B as approaching at 2 before the collision. After the collision, A measures B as moving away at -2.

All three are accurate descriptons of the same event. Are we in agreement?

 

[Uncayimmy]

Consider a third car in the two car collision scenario that is travelling alongside you as you sit in one car, before the collision it is stationary relative to you but after the collision it is now moving away from you even though it didn't hit anything and no forces were applied to it.

Exactly. It's the same as me riding on top of the car (ignoring friction, etc). This third car is now moving parallel to the second car (I stipulated a perfectly elastic collision). As far as the third car is concerned, the second car is now stationary and the first car is moving away. If you introduce their ground speeds into this, you are introducing another frame of reference, so don't do that.

 

[Towlie]

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?

That works for me!

 

[Uncayimmy]

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'm not sure I get your point. What changes is the total mass in the system and the elasticity of what the two vehicles are impacting. Replace the brick wall with a giant marshmallow.

 

[alexi_drago]

That works for me!

And me, kind of what I was saying about putting a paper wall in the middle of the collision.

 

[Modified]

All three are accurate descriptons of the same event. Are we in agreement?

If the collision were essentially instantaneous, like two hardened steel balls colliding, then car B would reverse direction almost instantaneously, and would end up several feet in front of you moving at your speed. If the collision took longer, you could end up over car B, or well in front of it, and moving at the same speed. When I imagine an almost perfectly elastic collision between vehicles, I naturally imagine the closest large-scale cases I'm familiar with, which are hard objects colliding.

 

[Uncayimmy]

If the collision were essentially instantaneous, like two hardened steel balls colliding, then car B would reverse direction almost instantaneously, and would end up several feet in front of you moving at your speed. If the collision took longer, you could end up over car B, or well in front of it, and moving at the same speed. When I imagine an almost perfectly elastic collision between vehicles, I naturally imagine the closest large-scale cases I'm familiar with, which are hard objects colliding.

I asked a very specific question, but you did not give me a specific answer. I described the same event using three reference frames. Are those three descriptions accurate or not given the stipulations (perfectly elastic, no friction, no air resistance)?

In order for you to say that car B would reverse direction, you'll need a reference frame in which B is moving. You can specify one of the reference frames I used or define your own. In my example the third reference frame (car A) has B reversing direction. I'm not in that reference frame, so you can't describe anything relevant to me.

Issues with reference frames aside, I think what you are driving at here is that when our two objects collide, they take time to do whatever it is they do (reverse directions, accelerate). I agree. That's why I chose to use the simulation with a coefficient of restitution of 1, which doesn't actually exist in the real world.

None of us has even attempted to calculate otherwise. Doing so would result in a loss of relative velocity as kinetic energy is lost. The collision would also, I assume, take the period of time I think you're talking about. This is a separate issue from trying to find out if we agree on the frames of reference.

I will say this, though. If as you say car B is moving the same speed as me after the collision (remember, I wasn't involved in the collision), then by definition the coefficient of restitution was 1. If the collision is going to take time, which it will in the real world, then the relative velocity will be lower as well. You can't have it both ways.

Or at least that's how this layman sees it.

 

[marting]

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

It is absolutely true that the average "brick wall" might not put up all that much resistance to a 50 MPH car. I think that when complexities aren't stated one is expected to use the most simple interpretation which means (to most) a fixed, inelastic, largely undamaged wall. Also, the purpose of the instructor in the OP was to dramatize the hazards of driving speed and intended the audience to think of the "brick wall" as an immovable object that was going to inflict most all the damage onto the car.

That said I concur and believe that actual single layer red brick walls would probably both have less mass across the car's cross section and tend to crack rather than crumple which would reduce the degree the car was slowed. But even though it might be reality I'd take bets that wasn't what the instructor intended.

Perhaps a better example would have been driving it into a concrete bridge support structure but it doesn't roll of the tongue like "brick wall."

 

[rsaavedra]

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,

I disagree with that. Hitting a stationary car at 100mph even if both cars have the same weight will result in the stationary car being pushed backwards, and the hitting car slowing down quickly while pushing it, but certainly not ending up at the collision point. However, when the two cars of the same weight collide head-on both going at 50 mph, assuming sufficient alignment of the impact, both wrecked cars will end up pretty much at the collision point. So the experienced deccelaration is much worse in the latter scenario.

I think the sudden decceleration and end result of two cars colliding head-on both going at 50 in sufficiently unlucky alignment would be much closer to the end result of one car going at 100 and hitting a stationary car that has right behind a solid unmovable titanium wall.