A Speeding Train

No because of the enormous differential in their masses.

Let's be completely honest... I cant see a train barreling down a track in the wrong direction after being hit by a bullet of incomparable mass. So the train will just continue on it's merry way with no real regards to the hoodlums trying to stop it with their puny firepower. (A much better idea would have been for the hoodlum to stand on the track and hope the engineer has a conscience.)

From a mathematical point of view one might be able to say that if you can break up the time it takes the bullet to and from the train into infinitely small pieces you will find a small space in time where both the train would not be moving and the bullet would be at zero speed. But thats just because nothing can move at a given "per second" velocity if there is no change in the time. But thats just splitting hair about integrals and not really worth a mention, I just felt like rambling.

All things considered, the long and the short of it, is no.

It passes the missing time.

If both the bullet and the train are, as you say, "perfectly rigid" then there will not even be a moment when the bullet stops. If the bullet is malleable, which most real bullets are then there will be a point at which the back of the bullet goes through a stationary instant elative to the environment but not the front of the bullet (if the train is still perfectly rigid). If the train is also elastic yet the bullet will not penetrate, then there may be an instant when the spot where the bullet contacts the front of the train that that spot is motionless relative to the environment depending on numerous variables.

You could also say: There is no perfectly rigid material, or: Yes it could completely stop the train if the bullet's velocity is high enough, etc, etc, but I hope we agree in advance, that this is a thought experiment, to test a point: If when the bullet was at zero-speed (also non existent in the real world) and touched the train at that point, shouldn't the train (and earth for that matter) have a zero speed at the same time.

I hope I haven't blown my own argument here, with the introduction of earth...

With no elasticity, my presumption of the stated premise of this thought experiment, the instant the bullet touched the surface of the train, it would be traveling at the speed of the train, minus some miniscule diminishment of the train's velocity by impacting the bullet. However, considering further ramifications of the hypothetical zero elasticity precept would infer infinite deceleration of the bullet (dv/dt where dt=0) and as a consequence when plugging infinite deceleration into the formula for impact force the bullet would impart infinite force to the train. Infinite force would infer infinite power and so both the bullet and the train might indeed come to a dead stop right before they experienced total and cataclysmic annihilation. I mean a thought experiment is a thought experiment, right?

Enchanted by your logic. Really, what a post!

If the train stopped relative to the earth, then where would the force come from to start it again?

No. the train does not stop. The train's velocity is slightly reduced at the 'instant' of impact.

I agree. Unless the train also changed direction, there is no force applied to the train that would have re-accelerated it to whatever speed it ended up going. The total change in kinetic energy in the bullet would be the amount of energy that was lost in the train.

It could not have stopped. Not even for a split second. If it had it would no longer posses Kinetic energy and therefore could not continue without a new force to propel it.

What's the last thing to go through a fly's brain when it gets splattered against a car's windscreen?

Its arsehole.

Pure physics.

That would depend on the brain's elasticity otherwise it might bounce and go through the arsehole instead

I think we should drop the thought experiment and try this at home... now where did i leave those spare flies...

no...it would have to be a perfectly rigid brain and an elastic arsehole for that to work

Why would the train stop? The change in the bullet's momentum causes a change in the train's momentum, but it just has a "step" change in velocity and continues on in the same direction.

If the bullet was perfectly rigid, there would be no "moment" when the velocity would be zero, however the velocity of the bullet would pass through zero as it instantly changed velocity from +V to -V. The acceleration would be infinite. Given F = mA², then F = m( ∞ )² , hence the force exerted is infinite. The mass of the particle is irrelevant as long as it has some mass. It could be something as small as a neutron and the force would still be infinite. Of course the train's de-acceleration would also have to be infinite, so there would be an infinite force opposing an infinite force. One cannot say that they would cancel out as the term " ∞ - ∞ " is as undefined as 0/0.

...If the bullet was perfectly rigid, there would be no "moment" when the velocity would be zero...

True, the bullet seem to have no means to transfer it's momentum

This, however, does not answer the question.

F=ma not ma^2. Of course when a is infinite it doesn't change the outcome.

Also: "however the velocity of the bullet would pass through zero as it instantly changed velocity from +V to -V" Not so! It is not a continuous function. The transition time is zero. There is no opportunity for before-now-after. It is discontinuous. It jumps from +V to -V in zero time and hence never exists at 0V.

...It jumps from +V to -V in zero time and hence never exists at 0V...

So according to this, introducing elasticity into the collision, transition time from plus-V to minus-V, not being zero, then zero velocity being allowed, and would allow the train to stop?

With elasticity, the train, having orders of magnitude larger mass, would only experience a reduction in velocity, albeit miniscule, over the period that the bullet decelerates and then accelerates in the other direction. It is conceivable that the spot on the train could briefly stop if there were adequate elasticity in the location on the train of the bullet's impact, though it seems unlikely but certainly not impossible. To stop the train would require an amount of energy equal to the trains momentum (1/2mv^2) which in theory if the bullet went fast enough could impart but practically speaking other issues get involved like energy spent in elastic deformation and the non-uniformity of the distribution of the force and material strength limitations. If the bullet could hold together with that much force it would likely expend a great deal of energy in material deformation. In actuality, were it to go that fast it would likely burn up in the atmosphere before ever getting to the train. Reality places many limitations on our hypothetical conjecture.

...only experience a reduction in velocity...

This indeed is to be expected in the real world of mass times velocity, but how do we get rid of the daemon, expressing that, since the bullet touched the train while in zero velocity... well, you know the rest.

I recalled this item from New Scientist, because I read it so many years ago, pondered it's consequence many times, but never was able to rest assured of what to do with it

The logical trap here is clear and disturbing.

I had one answer to the effect of: We find it hard to conceive a system to "slice a fluid-movement into still-frames" so to speak, not unlike an explanation to analogue reality to a cinema movement comprised of a string of still frames, but I must admit, it didn't help me get rid of the mentioned disturbing daemon, because even if we quantise-slice reality into frames of, say, Planck-time frames, the disturbing notion, is still there, present to point a disturbing disability of visualisation.

After all, reality is not really a film-strip of still frames, only to say that a movie-type fluid-movement, is an optical illusion.

It is a thought experiment with many assumed data.

Consider the bullet, which has a physical length L. If the front atom of the bullet stops - and it is stated to be perfectly rigid - then the back atom must also stop at the same time.

The length L means that the information has passed from front to back of the bullet in no time - which contradicts Einsteins Laws - nothing exceeds the speed of light, due to the divide by zero issue!.

As a further challenge - how did the gun shoot the bullet, as if it was infinitely rigid the gunpowder charge accelerated the front atom of the bullet at exactly the same time as it accelerated the backmost atom - again contradicting Einstein...

Have fun...

Yes when tow bodies impacted -in opposite directions- one body can manage to stop the other, depending on its Kinematics & kinetic energy. The body with larger kinetic energy is cabable to reverse the direction of the other (i.e. stop the other for a briefest moment). Kinetic energy & Kinematics of any body depends on its mass, acceleration, velocity, position, gravity acceleration, air resistance, ... etc.

That to say that a bullet with a larger kinetic energy can manage to stop the train.

...a bullet with a larger kinetic energy can manage to stop the train...

Dear Abdel Halim,

We agreed in advance to neglect such notion, (which is known to be true nevertheless), as the thought-experiment is meant to examine the following:

If the bullet, while reversing it's direction, reached zero-speed and touching the train at the same time, caused the train to stop for a brief moment?

When I was studying Dynamics as a subject we were taught that ..

speed of a = speed of b + speed of a relative to b.

All this was so long ago (45 years) don't remember all the right symbols.

So one can certainly say that for a fraction of time this speed of the bullet relative to the speed of the train was zero. So relatively the train had stopped !

Dear Yuval

A. If we considered that the bullet have a higher kinetic energy (at the moment of impact), the train will reverse its direction (not the bullet), caused the train to stop for a brief moment and the bullet may be de-accelerates.

B. If the train have a higher kinetic energy (at the moment of impact), the bullet will reverse its direction (not the train), caused the bullet to stop for a brief moment and the train may be de-accelerates.

In this case (B), while the bullet reversing it's direction, reached zero-speed and touching the train at the same time, caused the train not to stop for a brief moment.

Note : De-acceleration means decreasing (changing) of speed (velocity, v) per time (t), but not means zero speed.

Acceleration, a = Δv/ Δt .........the only case which a= 0 is when v= constant

Dear Abdel Halim,

Your bottom note, seem to conclude the matter, I think.

The application of Newtonian physics, although being much more intuitive than other kinds (as QM or chemistry for instance), does not imply it is easy to visualise all aspects of it, this example given.

The first answer I got to this question, which I quoted from a magazine, was some years ago, saying: "it depends on the speed of the bullet, it just might throw the train back...

Dear Yuval

"it depends on the speed of the bullet, it just might throw the train back..."

I think it depends not only on the speed, but also on the mass of both bullet & train, and in turn which of them has higher kinetic energy.

Right, my point was, that given the bullet existing mass, increasing it's speed, will still increase it total kinetics.

Due to the stated premise that the bodies are rigid, the bullet never passes through a moment when it is "stopped" and this is why it experiences infinite acceleration. Non-infinite acceleration infers that the velocity changes over a period of time and hence would go through zero when reversing. Yes of course this is quite impossible and a good thing otherwise any to colliding objects would result in the release of infinite energy, all quite absurd but none the less it is a thought experiment. Maybe that's what caused the big bang <LOL>.

Maybe that's what caused the big bang

I have a colleague that would say "they've crossed their beams"...referring to too many smart people letting loose...then crossing those thought paths in a similar result to "crossing the beams" in Ghostbusters (the end of the universe).

...just a funny way of explaining "an argument over theory vs reality".

Yeah, after a while it's like, just pick your set of constraints and create your own hypothetical universe. It's the Burger King of conjecture: "Have it your way." But sometimes fun, none the less.

I cannot imagine a perfectly ridgid object, I guess the closest would be a diamond which would probably imbed itself into or shatter the object which it struck. much like a Ramset gun which shoots steel pins into solid steel or hardened cement. It would entirely depend on the degree of rigidity and hardness of the colliding objects. With enough speed and hardness the cohesiveness of the atoms in the material may be split??

There is a moment where the kinetic energy of the bullet minus the kinetic energy of the train gives zero speed for the bullet but no the train with the enormous mass continue its way

Hi athmio,

But since this is a thought experiment, and given that both bodies are perfectly rigid, how is it that if the bullet, at the moment of impact (for one miniscule instant), has zero velocity relative to the earth, that the train does not also have zero velocity relative to the earth? (wow, that was a long sentence!) If the bullet changes direction 180 degrees then it had to have passed through zero velocity. At that moment did not the train also have zero velocity?

Somehow this question reminds me of the old "If an unstoppable object collides with an immovable wall, what happens?"

Cheers,

John

That's the problem when hypothesizing something that violates the laws of physics, which do you choose to violate and which do you choose to let operate. "Perfectly rigid" right off the bat infers communication at greater than the speed of light. Still, if we sum the force vectors of the two you find that the train's is much greater than the bullet's and normally that would infer that it is only slowed. However, when you plug in the numbers to calculate the amount of force created at the moment of contact then the result is infinity which of course must destroy everything in the universe and so there is no "momement" after the instant of contact when there any longer exists a train or a bullet or the earth on which they started their fateful journey. But if you overlook that little gotcha, then the summing of the force vectors only diminishes the train's slightly and another conundrum resulting from the inelastic premise results and that is that the bullet changes direction instantaneously. This seems to be causing some trouble with some's ability to imagine but it infers no intermediate moment during which the velocity of the bullet changes or can slow down and pass through zero, zero time, nada, it never passed through zero because the train never stopped and there is no compression, no bounce, no time. At the instant the touch the train looses momentum in the amount transferred to the bullet and the bullet experiences a jump from in the polarity of its vector in zero time. Sure it is impossible but no more so than the state of having no elasticity.

Yes Thats the problem ! In the same breath what will happen to apples in the trees when riped fully assuming gravitational force is zero

The centripital force from the Earth spinning tosses them into space. Assuming that it hasn't already tossed soil into space and of course without gravity there couln't have been rainfall and so on and so on. My how quickly things break down when you mess with just one little law.

...Perfectly rigid" right off the bat infers communication at greater than the speed of light...

You're basically right to point this out, rcapper, So I'm not arguing against. However:

I must admit, that when I quoted the item from "New Scientist", the additives of "perfect rigidity" and "symmetrical impact" were mine, in the framework of what I called in the OP "No escape bull", meaning, this is a thought-experiment, let's not twist and bent, It's merely a geometrical question.

Maybe not all-the-way to Topology, but somewhat of the "kinematic geometry" sort.

Then, Rigidity, became a deciding issue, so I tried to get rid of it in #25, to point out, that this does not kill the daemon-doubt of the paradox.

Then, we're off to relating to imaginary worlds, stressing the aspect that this is a thought experiment, as if not a real, concrete, issue.

Not a real, concrete world issue, that's for sure, as we can easily tell ourselves, from everyday experience. But it is a visualisation issue, thus, probably a geometrical or mathematical issue.

In #31, Johnjohn spots my doubt, back to the OP: "...If the bullet changes direction 180 degrees then it had to have passed through zero velocity. At that moment did not the train also have zero velocity?..."

Now, I'm nor trying to correlate this fairy tale to the concrete world, only to show this paradox of visualisation, and ask the participants here to find a way to cast-off this, with a reasonable explanation, which I never could.

Well all artificial constraints aside, the kinetic energy of the train would have to experience an equal and opposite reaction in order to be stopped. Simple Newtonian physics and there is no ambiguity whatsoever. Since the bullet does not have an equal and opposite momentum to that of the train, it will, in a proper world, regardless of how rigid it can be made in reality, impart its energy to the train, slowing it ever so slightly, and then receive energy from the train, which continues to move in its original direction, reverse its direction and yes the bullet will, in this real world, experience a moment when it goes through zero velocity as it reverses direction. I think I did state this in one scenario. However, I don't presume to be able to know a posters intent other than by what is said and if I missed your amending that premise in some of your subsequent posts it is because try as I might I found some of your wording confusing to me and perhaps I should have inquired as to what exactly you were intending to say. In order to make it clear what I am saying I do make an effort to state the conditions governing the thought that I am trying to convey and thereby another may see where we differ in preconceptions or frames of reference.

I have no questions concerning the Newtonian reality of the matter. It's clear and taken for granted. We can all very easily visualise a bullet hitting the train, and it's real-world consequence.

My "geometrical" problem is as presented in the eighties magazine: with that brief moment when the bullet pauses in zero velocity (whatever that means), while touching the train:

At that moment precisely, what would the velocity of the train?

- If the velocity of the train is not zero, and it touches the bullet, than the bullet never paused.

- If the velocity of the bullet is zero, and it touched the train, then the train paused.

So, which is it?

Or if it's neither, please explain.

It is a matter of where you place your frame of reference. Since all the materials involved are not only elastic but also malleable, let's consider for a moment the front surface only of the bullet. When it contacts the train, the energy imparted to the front of the train will cause elastic deformation of that front surface of the train until the point at which the energy absorbed by the elasticity of the front of the train is equal and opposite to the remaining energy of the bullet. Prior to this instant, the bullet will have been slowing. At this instant the front of the bullet and the surface of the train will be "stopped" relative to the boundary between them. Depending on the amount of energy transferred at this boundary, how much goes into elastic deformation, how much goes into actual deformation (where the modulus of elasticity has been exceeded) and the relative speed of the two entities, there will be some point when the boundary between them is "stopped" relative to the external reference frame of the earth. It may be before at or after the point that the relative boundary between the two entities passes through stationary, we can't know unless we take all the variables into consideration but it will occur. The bullet, being the obvious looser in this pushing contest and evidenced by its reversal of direction will pass through a stationary moment with respect to the earth reference frame. It may actually be more correct to say that, due to the fact that it will deform and that the compression wave will propagate through the bullet from front to back, the stationary moment will actually pass through the bullet, from front to back. Since it possess, as does all matter, elasticity and has length then the stationary moment does not occur for the entire bullet at the same instant. As I mentioned in an earlier post, it is not inconceivable that the spot on the train where the bullet strikes may also experience such a moment, depending on the elasticity/malleability of that spot, but not the whole train.

...the stationary moment will actually pass through the bullet, from front to back...

And:

...it is not inconceivable that the spot on the train where the bullet strikes may also experience such a momen...

So, in principle, this may also occur in the train, as in:

...the stationary moment will actually pass through the train, from front to back...

(if ignoring the mass difference for a moment).

- - - - - - - - - - - - - -

Well, I must admit this must be the best explanation I got so far, for this annoying paradox.

If you ignore the mass differential then you ignore reality and are back to the fantasy world I thought you were objecting to earlier. The mass differential is real and as a consequence, Newtonian physics dictates that excepting possibly a tiny spot on the train where the bullet hits, the train will not experience any stationary moment. When you subtract a very small number from a very large number the sum does not momentarily go through zero!

No,no, I was only ignoring to visualise, relax, man, don't destroy what you just carefully balanced in my mind here.

It's all agreed in advance, what happens in the real world.

The imaginary is just to conceive/contain the paradox into a reasonably coherent visualisation.

Right on YUVAL,

Many of us realize that this is just an exercise in visualization. Let's all relax and take it as that.

Remember, stars are only pinholes in a shell...

John

I think the part I don't get is whatever it is that you see as a paradox. To me it just looks like an application of Newtonian physics and material science but I would be interested to know what part appears to you to be a paradox.

To simplify the original problem slightly: we should realise that "stopped" is just an arbitrary speed when you look at things in a Newtonian relativistic sense. So to re- state Yuval's paradox he's saying that if you drop an incompressible ball on an immovable incompressible floor, then at "some time" between the ball falling and stopping the whole earth is clearly moving at say half the falling speed of the ball downward. The paradox clearly resolves itself when you realise that with the given constraints the ball stops instantly and there is no time between moving and stopping.

Uh...what? Why not then the train was never moving and everything was moving with respect to the train so nothing happened to the train when the bullet hit it because it was the only stationary entity. Since in reality, there really is no absolute reference frame more valid then some other and it's all relative, yeah, that works. But once again, I have no idea what is the paradox? I never understood what part that Yuval considered a paradox, nor do I understand what you are seeing as a paradox. Please point out what is contradictory to what you would expect to happen?

"When reversing it's direction there is a very short moment in which it is stationary, at "zero" speed. "

I'm with you on this, there is no paradox for me either: I think the problem lies in trying to reconcile the imaginary world where things can happen instantly and the real world where they can't. Clearly: in the real world there would be a "very short moment", whereas, in the imaginary world it is instantaneous.

Seems plausible to me. Very good explanation.

John

The moment that the bullet has zero velocity is the moment that almost touch the train or to be more specific just on its impact.

Consider the following experiment to check the movement of the train:

On the instant the train stop the first coach will keep on moving until it took up the slack in the coupling. (nothing to do with elasticity or -bullshit)

The impact will bump the loco into some motion and with no load on it the loco will accelerate.

In similar manner the rest of the coaches will come to stop and bump the previous one into motion . which will in turn bump the one in front to go faster. etc until it reaches the loco.

In my imaginary world the train will actually move faster.

The idea of standing in front of a speeding train and expect it to stop long enough to get out of the way is to scary for me.

I rather will go for: What will happen if you shoot a speeding train from the side?

The

What you just described is not-too-far from the "zero velocity point, passing through the object" described by rcapper

We in fact can regard the loosely-bound train cars banging each other in turns, as a form of "total system elasticity"

Could you please explain: "...What will happen if you shoot a speeding train from the side?..." ?

Yes it is also a form of elasticity. just trying to introduce some reality.

A bullet coming from the side will also come to an abrupt stop but there are no forces working on the train to stop it or even reduce speed. With no elasticity involved the bullet will be suddenly traveling at t mph.

A court case were lost because of the assumption that both objects stops. I do not know all the detail because I was only ±10. The person was alleged of not stopping at an intersection. The state used a professor to explain momentums, velocities etc. He told the court repeatedly that the cars actually stopped.

...A court case were lost because of the assumption...

I would have never expected such an aspect, to jump into the discussion .

So, rcapper gave me the best description yet, to visualise this imaginary state, which led me to the conclusion that I can never assume absolute rigidity, being fictional, even with such illusive visualisation.

A 'train' is a series of coupled units. 'Locomotive' would have been a better choice of word .

The average speed for a bullet is 1000m/sec A speed train has 220Km/h that mean the bullet approaches the train as the train moves 0.061Km/sec so when the bullet touches the train with all our assumptions it changes its direction and for a moment of time has zero velocity as the train has no contribution in the equation of velocity.

It would seem to me that one law of physics that must be obeyed in this hypothetical collision is the law of conservation of momentum.
As momentum is Mass X Velocity then we must concur that the bigger of these two quantities will win in a collision.
I.E the trains momentum is several orders of magnitude greater than the bullet so while the bullet will bounce off the train
with a velocity which equals it's own incoming velocity plus the train's velocity the train on the other hand will simply have it's velocity
decreased by a very small amount in order to conserve the overall momentum of the colliding objects.
Here is my attempt at the math:-

Assume Mass of Bullet = 10g = 0.01Kg and Velocity of bullet = 1000M/S
Assume Mass of Train = 10 tonne = 10,000Kg and Velocity of Train = 30 M/S (108Km/Hour)

Now, Momentum of Bullet (Before impact) = 0.01 X 1000 = 10 Kg.M/S
Momentum of Bullet (After impact) = 0.01 X 1030 = 10.3 Kg.M/S

Change in momentum of bullet = 20.3 Kg.M/S

This momentum change is absorbed by the train as follows:-

Momentum of Train (Before impact) = 10,000 X 30 = 300,000 Kg.M/S
Momentum of Train (After impact) = 300,000 - 20.3 = 299,979.7Kg.M/S

Thus Velocity of Train ( After impact) = 299,979.7 / 10,000 = 29.99797 M/S

Now for those of you who want to talk about infinite forces, I argue that whatever infinite quantity of force changed
the direction of the bullet's velocity from 1000M/S towards the train to 1030M/S away from the train is equal to
the same infinite quantity of force which reduced the velocity of the train from 30M/S to 29.99797 M/S and
all this occurred in zero time.

Aside from not accounting for the unknown amount of energy lost to heat as a result of metal deformation, I would concur with your description.

I'm just nit picking here; because it's pretty irrelevant for this discussion, but, if you're talking about a perfectly elastic collision, then the bullet and train start off approaching each other at 1030 M/s: and should finish up travelling apart at 1030 M/s. So the final velocity of the bullet should be 1060 M/s. And of course this only holds when the more massive object can effectively be taken as infinitely greater than the smaller (reasonable in this case).

Yes Randall, I would actually agree with your final bullet speed, if we assume a perfectly elastic collision, but now let me do a little more nit-picking. The original hypothesis stated and I quote from the first posting:-

"Imagine, an ideal, perfectly rigid and symmetrical impact: no elasticity, no time or space-warp, no escape-bullshit:

A speeding train goes one way, and a speeding bullet rushes to clash it's front.

At the moment of impact the bullet which naturally bounces off the train's front, reverses it's direction."

My interpretation of this is that there is an inherent paradox in this hypothesis. I believe that if we assume perfect rigidity, no elasticity, then the bullet does not deform on the front of the train and neither does it bounce off the front of the train. In this hypothetical environment the bullet will simply join to the front of the train and will go from 1000 Km/S in one direction to an ever so slightly less than 30Km/S in the opposite direction.

Well, we may as well do the mathematics under this assumption:-

Assume Mass of Bullet = 10g = 0.01Kg and Velocity of bullet = 1000M/S
Assume Mass of Train = 10 tonne = 10,000Kg and Velocity of Train = 30 M/S (108Km/Hour)
Assume train's direction of travel is positive momentum.

Momentum of Train (Before impact) = 10,000 X 30 = 300,000 Kg.M/S
Momentum of Bullet (Before impact) = 0.01 X 1000 = - 10 Kg.M/S

Total momentum of both objects = 290,990 Kg.M/S

Final velocity of Train and Bullet = 299,990 / (10,000 + 0.01) = 29.99897 M/S

As you can see there is not much difference to the train's velocity, as we would expect (29.99897 M/S with bullet sticking and 29.99797 M/S {OK, slight error here due to Randall's observation} with bullet bouncing perfectly).

I agree if we also choose to ignore the energy of impact since unless we do that, there is no longer a bullet nor possibly a train after the instant of impact, if neither is elastic. Yet another paradox in the hypothesis.

Have you seen some of the photos from NASA's high velocity impact experiments? Really tremendous fireballs. Great guy stuff. That's probably about a close as we will get to what a non-elastic collision would be like.

"...the bullet will simply join to the front of the train and will go from 1000 Km/S in one direction to an ever so slightly less than 30Km/S in the opposite direction..."

Perfectly intuitive for my vision, alas as rcapper stated earlier, it would only make sense given elasticity into the frame, otherwise it's all about fantasy, and for my money it was the first observation which made sense so far

Yes rcapper, I agree with you. I was assuming no heat loss or metal deformation as this is what the original hypothetical situation described. I also agree with your previous posts that in the real world the bullet would have to go from a plus velocity to minus velocity thus traversing through an instant of time where the bullet's velocity is zero. In the hypothetical case of no elasticity, metal deformation etc the function is as you rightly pointed out a discontinuos function. Therfore in that scenario the bullet's velocity changes from +V(bullet) to -V(bullet +train) at the point in time of impact, which is of zero duration but can still be defined on the real number line. Thus at that point the bullet's velocity changes from plus to minus but does not go through zero (Discontinuity of the function at this point). The train however, also goes through a step discontinuity. I.E. It goes from 30 metres per second to 29.99797 metres per second in zero time duration at the moment of impact. The upshot is that in the real or hypothetical world, the train does not go through zero and hence does not momentarily or otherwise stop.