But 1,000 tons of dead flies will not act like a 1,000 ton cue ball, even in an idealised model. It will act like billions of cue balls. I originally asumed that the flies would be confined to the track, but of course, unless they are lined up in a single straight line, in practice the flies are going to scatter off in every direction.
And if they are in a straight line of perfectly elastic spheres, each weighing the same as a fly, I susspect thsat many flies would go all the way around the world. So, after the train hit the first one, a series of collisionsd would travel along the line of flies at the speed of sound in the material that the flies are made of, until the last fly caught up with the rear of the train.
Assuming that everything is confined to a great circle, the train would have slowed down becauise of repeated collisions with the first fly, while the last fly would be travelling at the original speed of the train.
At that time, a new series of collisions is going to propagate back from the train. Eventually, assuming that the system is damped, but on a fricyionless track, the whole system , it will be travelling at half the original speed of the train. If it’s not damped, I suspect that the train wll be travelling at approximately half ofits original speed, but the flies will be bouncing backand forth at speeds ranging from the original speed to the train to stationary.
True, because if the ball is rolling, it has both translational and rotational momentum. The collision only transfers the translational momentum to the other ball, and the rotational momentum keeps the cue ball moving. Back spin of exactly the right amount leaves the cue ball without rotation, so it stops.
Right, the rolling billiard balls aren’t quite the same as perfectly elastic particles colliding on a frictionless surface. But the velocity of the cue ball will be much closer to zero than to half it’s velocity before the collision.
That depends on how you idealize the model. If all of the flies strike the train simultaneously, it’s no different than a single particle of 1000 tons striking the train.
This is just handwaving unless you can account exactly for the 50% loss of kinetic energy in the system.
If the flies go all the way around the world, they’re not in a straight line, though. One could imagine some sort of frictionless track that constrained the flies to a circle, but that’s getting pretty far from our original scenario. Equivalently, you could imagine a linear track with periodic boundary conditions such that objects leaving one end instanteously reappeared at the other with the sam velocity. But suppose that when the train strikes them, our spherical flies are all in a line in direct contact with one another along the direction of travel. If the spheres are perfectly elastic, the speed of sound is infinite and the line of spheres acts as a single, solid body.
If it’s damped, the ultimate speed of the train and flies will be zero: damping takes energy out of the system.
In the undamped case, I don’t think the train will settle to a steady state speed, but I’ll have to think about it some more.
To the OP, as Giles has indicated previously, in the real world a coasting train would be stopped by friction and wind resistance before flies affected the velocity appreciably unless there were a tons of them.
Just for grins, we can calculate how many flies it would take to stop a train in an idealized model. Assume a frictionless system in a vacuum, a flat fronted train, spherical flies, and perfectly elastic collisions. The flies hit the train one at a time and do not interact with each other in any way. The train is travelling with ground velocity v[sub]TG[/sub],which is positive before any collisions. So all flies fly toward the train with a negative ground velocity v[sub]FG[/sub].
Let’s calculate how much the velocity of the train is changed by a single collision.It’s easiest to do this in the reference frame of the fly at the moment of collision, so
0 = v[sub]FG[/sub] - v[sub]FG[/sub] = velocity of the fly before collision
v[sub]0[/sub] = v[sub]TG[/sub] - v[sub]FG[/sub] = velocity of train before collision
v[sub]F[/sub] = velocity of fly after collision
v[sub]T[/sub] = velocity of train after collison
m = mass of fly
M = mass of train
u = m/M
Substitute for v[sub]F[/sub] from [2], solve for v[sub]T[/sub]
v[sub]T[/sub] = ((1 - u)/(1 + u))v[sub]0[/sub]
So we have the change of train velocity (in flying fly’s frame of reference) due to a single collision. After n collisions
v[sub]n[/sub] = ((1 - u)/(1 + u))[sup]n[/sup]v[sub]Tinitial[/sub]
Shifting back to the ground reference frame, we can solve for N, the number of flies it takes to stop the train
N = (log(v[sub]TGinitial[/sub] - v[sub]FG[/sub]) - log(0 - v[sub]FG[/sub]))/(log(1 + u) - log(1 - u))
for smaller than 0.1 or so, log(1 + x) = x, so for small u, train initally travelling 100 mph, flies travelling -10 mph
N = 1.2 * M/m
So if the train weighs 1000 metric tons and flies weigh one gram, it takes about 1.2e9 flies to stop the train. Or, to put it another way, the total mass of the flies needed is about 1.2 times the mass of the train for the given train and fly velocities.
That’s easy. It all turns into heat. A perfectly elastic collision, where all mechanical energy is conserved, is one idealization, but a perfectly inelastic collision, where all particles end in the center of mass frame (that is, they’re all stuck together), is another perfectly good idealization. And from my experience of flies hitting windshields, perfectly inelastic is a much better approximation than perfectly elastic.
1000 tons of dead flies aren’t going to just sit there on the tracks for your experiment, now will they? You’ll need something to make the entire group of flies stay together-I’d suggest ointment. As such, the model would have to include the weight of sufficient quantity of ointment to bind all the little buggers together, along with the effect of dislodged-but-still-stuck flies on velocity, and so forth.
Plus the energy necessary to break up a fly into thin goo, spread into a thin layer on a windshield.
If you allow perfectly inelastic collisions, all you need to stop a train is a fly with a momentum exactly opposite of the train’s own momentum. Which would be a really freaking fast fly. And in reality, the window would probably break if hit by a fly going that fast. Plus it also requires a fly traveling at relativistic speeds (going by the 1000 metric ton train and a 1 g fly).
Is perfect elasticity the same as perfect rigidity? If the motionless body of the fly cannot deform to accommodate the motion of the train, and the material of the moving train cannot deform to permit the brief period of acceleration that the fly must experience, then there would have to be a moment at which the fly stopped the train. Of course it could never happen in reality, because there’s no such thing as perfect rigidity.
Deformation doesn’t take place to allow the fly to change velocity, does it? Hypothetically, what would happen if a substance with absolute rigidity was discovered and used to make a small bb and a train? The bb thrown at an oncoming train would still have to come to a complete stop and reverse direction.
What I’m really asking is, does all this talk of rigidity really have anything to do with answering the OP? Even if the windshield becomes slightly deformed and the fly becomes mangled, each individual molecule of the fly must still come to a stop and reverse direction.
Yes, it must. Actually, that doesn’t mean any of the train’s atoms have to stop, since they’re never in absolute direct contact with each other anyway - they bounce off each others’ electromagnetic fields, so a region of the train’s material may slow down a bit while the fly slows down and reverses bit by bit.
But if the train and fly were truly rigid and capable of coming into absolute direct contact, then the train would have to be considered stopped while the fly was at the point of reversing direction and having stopped, there would be no reason for the train to start again (and we’d have to find an explanation for where all that energy went). That obviously can’t happen.
Why? Is there some law of physics that says this must be so? As a thought experiment, I have no problem imagining that the train chugs along at the same speed, while the truly rigid bb is the only object that stops and reverses direction.
Is the crux of the problem that if there is a point that the bb is motionless, then the train must also be motionless? If so, it does sound like this question is more like a Zeno’s paradox type problem as Thudlow Boink suggested.
With a truly rigid fly and a truly rigid train, it’s not correct to consider the fly stopped at the point/time of contact. Its velocity is undefined. Strictly speaking, the velocity of the train would be undefined also, since its velocity also undergoes an instantaneous change, although the magnitude of the undefinability, to make up a term, would be very small.
Take the fly and train moving towards each other at 100 m/s, and colliding at t=0, x=0, then the fly’s position as a function of time is
Xfly = 100 * abs(t)
dXfly/dt isn’t 0 at t=0, it’s just undefined.
The train’s position function is
Xtrain = (100 - d) * t - d * abs(t)
where d is very very small. Physicists and engineers will probably let you get away with saying dXtrain/dt is approximately 100 at t = 0 (and the mathematicians will probably cringe), but it doesn’t really make any sense to call it zero.
Let’s have a reality check here. Lets postulate that the fly and train are perfectly rigid. Are you going to tell me that the train’s momentum would be cancelled by a rigid fly ? We could cancel the momentum of a rigid train instantaneously though. With another identical rigid train in the opposote direction.
In short, the Law of Conservation of Momentum, (mass x velocity), is completely independant of the properties of material .
