Train a, traveling at 4 m/s hits train b, traveling at 2 m/s. What is the resulting velocity of the trains. I am assuming that train a overtakes train b?
Is the answer 3 m/s each, or train a 2 m/s, train b 4 m/s?
There is insufficient data for a meaningful answer.
Specifically, we need the masses of the two trains (at least we need the mass ratio).
Knowing how many cars in each train would be a bit of a help. Some of the energy will be absorbed by distortion of the colliding cars and reduce the final velocity.
Reframe the question lucidly, and return later.
Sorry, I am also assuming identical trains of identical mass.
Also, this is very elementary physics. We are assuming all energy is conserved in the velocity of the trains. No energy is absorbed in deformation, or anything else.
Is it a head-on collision or is it a rear-ender?
Rear ender.
Maybe this is a silly question, but it isn’t addressed. Were they traveling in the same or opposite directions?
Since the faster train hits the slower, they could be on the same track going the same direction, with train b ahead of train a.
Just trying to learn.
Yes, that is precisely the situation.
We used to do these problems in maths and physics, except with pool balls with a quoted coefficient of restitution and approach angle. Sometimes the balls would follow each other, sometimes bounce apart, and sometimes one would bounce off while the other would be stopped dead.
I don’t often play pool, but when I do I can work it out from these first principles. And generally lose to any player who practices more than I do.
What I mean is faster train a rear ends slower train b. After the collision are they both travelling 3 m/s, which is what I think, or is train b then traveling 4 m/s?
Oh bloody hell, I couldn’t help by work it out. But it’s 4am and IR tired, and my answer comes to 10^-2 m/s (about 3.16 m/s). This assumes the trains stick to each other and move off as one lump, and don’t deform and make no noise as they collide.
Train a 2 m/s, train b 4 m/s, assuming equal masses and an elastic collision. That’s the only way to conserve both kinetic energy and momentum.
You can think of it this way. From the point of view of the reference frame moving 2 m/s, a train moving 2 m/s hits a staionary train. The first train stops (in its tracks!) and the second starts moving at 2 m/s.
The closest you can come to an answer is:
momentum before collision = momentum after collision (lots of simplifying assumptions)
2m[sub]1[/sub] + 4m[sub]2[/sub] = v[sub]f[/sub](m[sub]1[/sub] + m[sub]2[/sub])
v[sub]f[/sub] = (2m[sub]1[/sub] + 4m[sub]2[/sub])/(m[sub]1[/sub] + m[sub]2[/sub])
Ummmm . . . no
You’re assuming that the two trains will have equal velocity after the collision which is almost never the case.
Knowing the two masses (m1 and m2) and the two starting velocities (v1 and v2), there are two variables to solve for: the two velocities after the collision (call them w1 and w2).
By conservation of momentum: m1v1 + m2v2 = m1w1 + m2w2
By conservation of energy (in this case kinetic): (1/2)m1(v1^2) + (1/2)m2(v2^2) = (1/2)m1(w1^2) + (1/2)m2(w2^2)
Two equations and two unknowns - have fun 
To work it backwards like the OP may be asking, assume v1=4, v2=2, w1=3, w2=3. this gives:
4m1 + 2m2 = 3m1 + 3m2
8m1 + 2m2 = (9/2)m1 + (9/2)m2
The first equation gives m1 = m2
Thus by the second equation 10m1 = 9m1
Therefore m1 = m2 = 0; the conditions are impossible :eek:
That’s right I am. And in actual cases they probably won’t but since the OP said basic momentum question I’m assuming that’s what was meant. It’s also easy to arrange that. Just have the fore and aft couplings on the trains engage upon colliding. And, of course ignore the inevitable jiggling in the couplings while things are settling down.
And anyway, energy isn’t conserved. You can hear the collision. 
If that wasn’t the intent then your solution can be used. But I don’t intend to touch that. I don’t have a lot to do, but even I can’t waste time on such things.
Sticking to my assumption, we can go further. Set the mass of the faster train equal to k times that of the other.
m[sub]2[/sub] = km[sub]1[/sub]
and set m[sub]1[/sub] = 1
In this case the equation reduces to:
f[sub]f[/sub] = 2(1 + 2k)/(1 + k)
A plot of the equation is here.
Obviously if 2 is very light compared to 1 the final velocity will be close to 2 m/s. And if it is very heavy the final velocity will approach 4 m/s.
But the OP wasn’t assuming that.
I probably went beyond the OP’s original question by solving backwards to find what masses would result in equal ending velocities but I was curious about the result myself.
Plus if your assuming an elastic collision, then how do you justify the conservation of momentum?
Momentum conservation and energy conservation are very different things. Well, not very different, actually, but you can cook up situations in which one is conserved and the other isn’t.
Assuming that the trains have equal mass, the answer still depends on whether the collision is elastic or inelastic. If you demand that the trains end up with the same velocity (say, their couplers engage when they collide), then the answer is that both will have a final velocity of 3 m/s. If you demand that energy is conserved, then they’ll essentially “switch velocities” like you suspected.
There’s actually a range of values for the final result, depending upon whether the collision is elastic, inelastic, or somewhere in between. In real life, things tend to be in between. In that case, you can defibne a “coefficient of restitution” that is the ratio of the velocities after collision (or use some other figure of merit). In the case of a truly elastic collision, both energy and momentum are conserved. In the case of a completely inelastic collision, momentum is conserved and the objects move off with the same velocity afterwards. For the intermediate case, momentum is conserved and you can solve for the final velocities using the momentum law and your coefficient of restitution.
But momentum is always conserved, unless some outside force acts on things.
Hold the phone. Assume the two trains meet and stick, and go 3 m/s. No deformation occurs, noise, or other nickel & dime energy loss. Assume each has mass m. Before collsion, kinetic energy is 8+2 = 10 kg*m[sup]2[/sup]/sec[sup]2[/sup] (joules). After collision, kinetic energy is 4.5+4.5 = 9. Where did the energy go?
I learned conservation of momentum and conservation of energy in physics but don’t remember the profs pointing out that you can’t have your joules and eat them too.