Elastic/Inelastic collision upon Traincrash

[TheChemist_]

Homework Statement

A railway waggon of mass m=20t is traveling with v=18km/h. It then hits 10 other stationary railway waggons, with each one having the same mass of m=20t . (Friction is not being considered)

Homework Equations

a) Assume the 10 railway waggons are connected through strong metal springs: Is it an elastic or inelastic collision. Describe what happens precisely and determine the velocity of the waggons after the impact.
b) The 10 waggons are connected through metal rods. Is it an elastic or inelastic collision? What are the collision "partners"? Determine the velocity of the waggons after the impact.
c) All 10 waggons have a automated connector socket, which upon impact connects each waggon with the next. Is it an elastic or inelastic collision. Determine the velocity of the waggons after the impact. How big is the loss of kinetic Energy?

The Attempt at a Solution

So let's start with what I have thought out for b)
All waggons being connected with a metal rod, we could also assume the other waggons act like one big waggon with 10x the mass of one waggon (m=100t). It's definately an elastic collision, so we can just work with the momentum formula (see attached files):

so v1' (of the lonely waggon) would be -8.18m/s (going in the opposite direction)
and for the other waggons v2' = 0.909m/s
Is this correct?
But what about a) and c)
I could think that for c) we have an elastic collision upon first impact and then it goes on to be a inelastic collision. Is that correct?

 

[BvU]

Hi,

b) no. The forces to reverse the direction of the single waggon would let it disintegrate.
What did you answer for a ) ?
c) the question is unclear. Any physical difference with b) ?

Is this in the context of an introductory mechanics physics course, or does reality play a meaningful role here ?

 

[TheChemist_]

Ok so could you explain a little further why b) is wrong? would the formula: v1' = ((m1-m2)/(m1+m2)*v1 be correct for the one waggon? and v2'= (2*m1/(m1+m2))*v1 for the rest of the waggons be correct?
to c) well I thought of it like this: The first waggon hits in an elastic collision waggon number 1 of 10. because the masses are equal, waggon 1 of 10 will travel with the same speed as the lonely waggon. When waggon #1 (of 10) hits #2 in an inelastic collision they connect and move on. Their velocity should be half of waggon #1 before impact (or equal to the starting velocity). On the next impact the same thing happens and the velocity should get even lower, is that correct?
For a) I thought of pure elastic collisions but what happens at the end? Can you explain it to me?
It's for an introductory mechanics curse for chemists

 

[TheChemist_]

and for c) it's not a connector socket (sry for my bad english), it's a coupler that goes between trains

 

[haruspex]

It's definately an elastic collision

Why?
As far as I can see, the question statement provides no basis for saying whether the collision is elastic or inelastic in a) or b). It only describes the connections between the 10 waggons. We need to know about the contact between the one waggon and the stationary waggon it hits.
In a), all we can say is there are elastic collisions between adjacent pairs of the 10.
As you say, in b) we can treat the 10 waggons as a single waggon, so there is no collision between them at all.

At a guess, you are supposed to assume the same kind of connection between the one and the 10 as the 10 have between themselvs, so elastic in a) and inelastic in b).

c) is also unclear. Are we to take the 10 as initially tightly coupled (as in b) or loosely coupled? If loosely, you have inelastic collisions between all adjacent pairs.

 

[TheChemist_]

Yeah for c) they are loosely coupled, so we have inelastic collsions between all pairs. But how will the velocity change there?
and for a) and b), I assume, regarding other examples we got for exercise, that they would have mentioned it if the first waggon would "stick" to the other 10 so that it would be in fact an inelastic collision. I assume for all 3 cases (a,b,c) that the first waggon just bounces of waggon #1 of the adjacent. (Originally it says "aufprallen" which is german for "to bounce/thud/dash against")

 

[haruspex]

an introductory mechanics curse for chemists

So it would seem

could you explain a little further why b) is wrong?

If the initial collision is elastic (have you left out part of the problem statement?) then your answer to b is correct.

to c) well I thought of it like this: The first waggon hits in an elastic collision waggon number 1 of 10.

Again, if that is known you omitted to mention it.

When waggon #1 (of 10) hits #2 in an inelastic collision they connect and move on. Their velocity should be half of waggon #1 before impact (or equal to the starting velocity). On the next impact the same thing happens and the velocity should get even lower, is that correct?

With the loosely coupled view, yes.

In a), all we can say is there are elastic collisions between adjacent pairs of the 10.

I should add that although the collisions would be elastic there, just as much energy ends up getting lost as if they were not. Do you see why?

 

[BvU]

For a physicist this problem statement is riddled with incompletenesses. If the strong springs are to absorb the kinetic energy of 20 t moving at 5 m/s, they should have some length and it will take a while before the bump has propagated to the farthest car. And back and so on. Very complicated. I'm pretty sure this wasn't meant by the exercise composer: he/she aimed at an elastic collision. The lonely car should bounce back with close to the original speed, but in the other direction. Your blue formula block comes out of the blue. If you can check what situation it describes, I bet it's not the situation here. Here you deal with momentum conservation and - for elastic collisions - conservation of kinetic energy.

b) is definitely inelastic in anything resembling the real thing. Lots of damage. But you never know in 'an introductory mechanics curse for chemists' (*)

In the c case you can assume the exercise composer envisioned an elastic collision between the lonely car and the first car alone. What happens ? And then a sequence where the connectors act. Then what happens ?(*) One cold look at this as follows: the exercise composer wants to make an interesting exercise where two scenarios have to be calculated through, and he/she adds a third scenario that requires some more thinking.

would the formula: v1' = ((m1-m2)/(m1+m2)*v1 be correct for the one waggon? and v2'= (2*m1/(m1+m2))*v1 for the rest of the waggons be correct

Don't see that. Where does it come from ?

 

[BvU]

(sry for my bad english)

No need to be sorry. Your english is a lot better than my chinese

 

[TheChemist_]

 

[BvU]

When waggon #1 (of 10) hits #2 in an inelastic collision they connect and move on

The problem statement specifically says that 10 cars have a coupler (not a word of english there -- unless you read the word 'eleven' and abbreviated to '10'). I supposed the lonely waggon comes from a neighboring country and doesn't have it.

 

[TheChemist_]

Ok, I just got to make a statement: I am not actually bad at physics, but some of the problems to me pose further problems on how to solve the question, just as you said, the incompleteness of the questions shows that they are not thought out well in my eyes. The example before this on my homework is just utter garbage, it doesn't even make a point or get you to understand something. That's why I tried it here in the forum.

 

[TheChemist_]

The problem statement specifically says that 10 cars have a coupler (not a word of english there -- unless you read the word 'eleven' and abbreviated to '10'). I supposed the lonely waggon comes from a neighboring country and doesn't have it.

yeah probably

 

[BvU]

We'll do what we can to help, but telepathy isn't usually a speciality of the helpers here

 

[TheChemist_]

Yeah well I get it ;) I am just going to solve it my way and then blame the professor or his students who made these examples...
And you have to consider that I have to translate the whole thing from German to English
Thx guys, really appreciate your help

 

[haruspex]

One thing I should add. I always get confused on the exact definitions of elastic and inelastic - they're not quite how I would have defined them.
"Elastic" is the same as "perfectly elastic", i.e. no KE loss. That is the circumstance required for the equations in the blue box in post #1. If we apply that to the spring connectors then the 10 waggons will oscillate forever.
That leaves "inelastic" to refer to every collision involving some KE loss, and every real world macroscopic collision falls into that class. The extreme case, maximum KE loss, is coalescence, where the participants end up with the same velocity as each other (in the direction of initial relative motion).

So the short answer is that all these collisions are inelastic, but perhaps to varying degrees. The collisions between the 10 waggons (in case b there are no such) are all effectively coalescence because any other residual KE takes the form of local oscillations, which are not mechanically useful. They will die down over time.

 

[BvU]

I assume for all 3 cases (a,b,c) that the first waggon just bounces of waggon #1 of the adjacent

You can practice this with identical coins sliding on a smooth table. Or google some Newton's cradle youtubes (however, that's NOT what your exercise describes)

So @haruspex and I wonder if you are supposed to think physically or just act out 'monkey see, monkey do'.

I still don't get where your blue formula block comes from.

I

 

[TheChemist_]

I thin they are more considered like "monkey see, monkey do" which is maddening. The blue block formula comes from the elastic equation ( where no KE is lost) (combined of momentum conservation and energy conservation)

 

[haruspex]

if you are supposed to think physically or just act out 'monkey see, monkey do'.

It would help if we knew how these concepts are used in the chemistry aspects of the course (assuming they are).
At the atomic level, collisions are much more nearly elastic, and the KE that is lost takes measurable forms, such as excitation.

 

[BvU]

No atoms here, just collision mechanics. Railroad cars. 0.5 MJ of kinetic energy for the lonely one !

 

[BvU]

I thin they are more considered like "monkey see, monkey do" which is maddening. The blue block formula comes from the elastic equation ( where no KE is lost) (combined of momentum conservation and energy conservation)

Better to write down the conservation laws, initial conditions and possible further assumptions, then go from there.

 

[haruspex]

No atoms here, just collision mechanics. Railroad cars. 0.5 MJ of kinetic energy for the lonely one !

Sure, but that is my point. If the chemistry applications are more in the elastic realm than are macroscopic rolling stock, the idealised elastic view in this problem is defensible for a mechanics-for-chemists course; it is just that the question set is not a realistic model for the behaviour to be considered.