Actual relative velocities and real world scenarios

[Eise]

If you were standing in the path between two cars going 50 mph you are being hit at 50mph by each of them; not 100mph. The cars will experience a collision of 100mph but not you. Your speed, relative to each car, and vice versa, is 50 mph. It's the same thing in your scenario but you have to make relativistic corrections,

 

Hmmm. That is ambiguous, or may be even wrong.

 

Imagine the following 3 situations:

1. 2 cars, each with 50 km/h frontally collide
2. 1 car, 50 km/h, collides with a 'perfect wall'
3. 2 cars, one is standing still, and the other collides with it with a velocity of 50 km/h

The relative velocities are clear (forgetting relativity for the moment). But a person in between will experience the same effect in 1 and 2, where the relative velocities differ, but not in 3.

 

But this also means that the single car in situation 2 will have the same effect as one car in situation 1, even if the relative velocity in situation 1 is 100 km/h and in 2 50 km/h.

 

Or am I totally confused?

[StringJunky]

 

Hmmm. That is ambiguous, or may be even wrong.

 

Imagine the following 3 situations:

1. 2 cars, each with 50 km/h frontally collide
2. 1 car, 50 km/h, collides with a 'perfect wall'
3. 2 cars, one is standing still, and the other collides with it with a velocity of 50 km/h

The relative velocities are clear (forgetting relativity for the moment). But a person in between will experience the same effect in 1 and 2, where the relative velocities differ, but not in 3.

 

But this also means that the single car in situation 2 will have the same effect as one car in even if the relative velocity in situation 1 is 100 km/h and in 2 50 km/h.

 

Or am I totally confused?

1. Both experience a 100km/h collision; measured from either frame

2.Car and wall experience a 50km/h collision; measured from either frame

3. Both cars experience a 50km/h collision; measured from either frame.

 

It follows that 2 and 3 will experience the same degree of collision.

 

Whichever frame you measure from you are the stationary observer.. what that means is, in scenario 1, for example, the car opposite the observer vehicle is measured travelling at the combined velocities of the two vehicles; velocities are additive in ron-relativistic physics

Edited June 8, 2016 by StringJunky

[Eise]

1. Both experience a 100km/h collision; measured from either frame

2.Car and wall experience a 50km/h collision; measured from either frame

3. Both cars experience a 50km/h collision; measured from either frame.

 

It follows that 2 and 3 will experience the same degree of collision.

 

Whichever frame you measure from you are the stationary observer.. what that means is, in scenario 1, for example, the car opposite the observer vehicle is measured travelling at the combined velocities of the two vehicles; velocities are additive in ron-relativistic physics

 

As I said, the relative velocities are clear. But what about the effect? Take the perspective of one single car: will it have the same damage in all three situations?

[robinpike]

 

Hmmm. That is ambiguous, or may be even wrong.

 

Imagine the following 3 situations:

1. 2 cars, each with 50 km/h frontally collide
2. 1 car, 50 km/h, collides with a 'perfect wall'
3. 2 cars, one is standing still, and the other collides with it with a velocity of 50 km/h

The relative velocities are clear (forgetting relativity for the moment). But a person in between will experience the same effect in 1 and 2, where the relative velocities differ, but not in 3.

 

But this also means that the single car in situation 2 will have the same effect as one car in situation 1, even if the relative velocity in situation 1 is 100 km/h and in 2 50 km/h.

 

Or am I totally confused?

 

The amount of deceleration needs to be considered in each case.

 

In scenario 1), both cars are decelerated from 50 km/h to 0 km/h.

 

Scenario 2), the car is decelerated from 50 km/h to 0 km/h.

 

And in scenario 3), one car is accelerated from 0 km/h to 25 km/h and the other car is decelerated from 50 km/h to 25 km/h.

[Eise]

 

The amount of deceleration needs to be considered in each case.

 

In scenario 1), both cars are decelerated from 50 km/h to 0 km/h.

 

Scenario 2), the car is decelerated from 50 km/h to 0 km/h.

 

And in scenario 3), one car is accelerated from 0 km/h to 25 km/h and the other car is decelerated from 50 km/h to 25 km/h.

 

First time I see the word 'deceleration'... (non-native English speaker). But it makes a simple explanation. If nobody objects... then thanks.

[imatfaal]

 

The amount of deceleration needs to be considered in each case.

 

In scenario 1), both cars are decelerated from 50 km/h to 0 km/h.

 

Scenario 2), the car is decelerated from 50 km/h to 0 km/h.

 

And in scenario 3), one car is accelerated from 0 km/h to 25 km/h and the other car is decelerated from 50 km/h to 25 km/h.

 

You are assuming a certain sort of collision and other things.

 

The deceleration is related to both the change in velocity [m/s] (which you have partially covered) and the time [s^-1]in which that change occurs giving an acceleration [m/s^-2]; this is one of the (lesser) reasons why we have crumple zones - you must lose velocity but it is safer to do so over a longer period of time and importantly with greater control.

 

The collision could also be elastic or inelastic or anywhere in between - momentum is always conserved without an outside force - but in some collisions (perfectly elastic) all kinetic energy is conserved, in others no kinetic energy is conserved (perfectly inelastic). You have to consider vectors and scalars (at odds to recent posting in speculations) momenta (as vectors) can cancel but energy has to go somewhere. Perfectly elastic collisions will have very different accelerations (final velocities will not be zero)

[Eise]

You are assuming a certain sort of collision and other things.

 

Well, I assume that frontal colliding real life cars are pretty close too perfect inelastic collisions, no?

 

I think the central lesson is that the frontal collision of two cars, each driving 50 km/h, even if they are approaching each other with a velocity of 100 km/h, cause the same damage to one car driving at 50 km/h as when this car collides with a perfect wall.

[imatfaal]

 

Well, I assume that frontal colliding real life cars are pretty close too perfect inelastic collisions, no?

 

I think the central lesson is that the frontal collision of two cars, each driving 50 km/h, even if they are approaching each other with a velocity of 100 km/h, cause the same damage to one car driving at 50 km/h as when this car collides with a perfect wall.

 

On the collision type - Perhaps; definitely more inelastic than elastic

 

On the central lesson - The acceleration may be similar but if you look at the problem from the frame of the point of impact then in the first case there is twice as much kinetic energy than in the second. In an inelastic collision this is all converted to other forms of energy - heat basically - which is manifest in a deformation of the structure of the car/cars. But conversely in the second case there are two cars which evens up the amount of energy transformed within each car. Seems wrong intuitively - but I would have to agree

[Eise]

Seems wrong intuitively

 

Yeah. That's why I found it interesting.

[swansont]

 

On the collision type - Perhaps; definitely more inelastic than elastic

 

On the central lesson - The acceleration may be similar but if you look at the problem from the frame of the point of impact then in the first case there is twice as much kinetic energy than in the second. In an inelastic collision this is all converted to other forms of energy - heat basically - which is manifest in a deformation of the structure of the car/cars. But conversely in the second case there are two cars which evens up the amount of energy transformed within each car. Seems wrong intuitively - but I would have to agree

 

 

Would a sensor array inside the car, measuring only the car's parameters (not seeing outside), be able to tell if you hit another car traveling at exactly your speed, or a perfect wall?

[imatfaal]

Would a sensor array inside the car, measuring only the car's parameters (not seeing outside), be able to tell if you hit another car traveling at exactly your speed, or a perfect wall?

 

No we could not tell - but that still goes against my instinct (which incidentally is why science works - finding out stuff that common sense says is the other way around).

 

Some things follow how we guess they should - others, even very simple mechanics, do not; fat tyres do not help you stop quicker if you are riding on a normal road - this is a favourite of mine that very very few people believe even when they are shown data

[swansont]

I would think as you had more fat tires your stopping time would get progressively worse.

 

Some part of this might be that this is a highly idealized situation, as the wall must be ideal, the masses and speeds must match exactly, and the collision must be perfectly inelastic. But an inelastic collision with the wall that isn't perfect elastic might mimic hitting a more massive or higher-speed car. Once you get away from the perfect case, there's a whole lot of phase space for what happens.

 

Another part is that the two-car collision has twice the energy as one car; a common lay misperception is that this must be like a 100 mph collision with a wall, since that's the closing speed, but anyone versed in basic physics knows that doubling the speed quadruples the energy, and shouldn't fall for that one.

[imatfaal]

I would think as you had more fat tires your stopping time would get progressively worse.

 

Some part of this might be that this is a highly idealized situation, as the wall must be ideal, the masses and speeds must match exactly, and the collision must be perfectly inelastic. But an inelastic collision with the wall that isn't perfect elastic might mimic hitting a more massive or higher-speed car. Once you get away from the perfect case, there's a whole lot of phase space for what happens.

 

Another part is that the two-car collision has twice the energy as one car; a common lay misperception is that this must be like a 100 mph collision with a wall, since that's the closing speed, but anyone versed in basic physics knows that doubling the speed quadruples the energy, and shouldn't fall for that one.

 

Belgian Beer and cycling is a marriage made in heaven and consummated every weekend in the low countries. I think it is the fact that we know it will not be ideal is the reason that it is difficult to grok the correct solution

[AbnormallyHonest]

 

Hmmm. That is ambiguous, or may be even wrong.

 

Imagine the following 3 situations:

1. 2 cars, each with 50 km/h frontally collide
2. 1 car, 50 km/h, collides with a 'perfect wall'
3. 2 cars, one is standing still, and the other collides with it with a velocity of 50 km/h

The relative velocities are clear (forgetting relativity for the moment). But a person in between will experience the same effect in 1 and 2, where the relative velocities differ, but not in 3.

 

But this also means that the single car in situation 2 will have the same effect as one car in situation 1, even if the relative velocity in situation 1 is 100 km/h and in 2 50 km/h.

 

Or am I totally confused?

 

I would say you might be looking at it in the wrong way. The relative "velocities" may all be different in relation to one another, but what can you say about their change in momentum?