The Psychotic Train Riders Puzzle
Here is a somewhat cruel physics puzzle which illustrates the idea of momentum conservation.
Consider a train full of passengers moving along on a frictionless rail at a constant speed, v. In order to increase its speed, the passengers decide to throw themselves off the back end of the train, making use of the principle of conservation of momentum - much like a rocket expels fuel to propel it forwards.
Putting aside the question of their mental sanity, you, as a physicist, must now advise them on the best method of action. Seeing that there is only a limited number of people on board, should they (in order to achieve the greatest increase in speed):
- Hurl the people one at a time, or
- Throw themselves all at once?
Make the assumptions that all passengers have the same mass and throw themselves off the rear end of the train with the same velocity (relative to the train) equal to, say, -vf.
Let’s say that relative velocity is v_0.If they all throw themselves all at once,they’ll jump with relative velocity v_0.
Now by one.If first jumps with relative velocity v_0.Then train will have some velocity v_1.For second relative velocity will be v_0+v_1.
So by one v_f will be greater.
Ah, but you are dealing with CHANGES in momentum, Delta P, and not P.
When you think about it, the speed of the train is irrelevant! Let’s look at things from a frame at rest. Suppose the train is moving forwards with velocity v, and some guy jumps off with (relative!) velocity -vf. His velocity BEFORE jumping is:
v
And AFTER jumping it is:
v-v0
Therefore the change in momentum of the train system is:
-Delta_P = Delta_v_guy/mass_of_guy = (v-v0-v)/mass_of_guy = -v0/mass_of_guy
Independent of v!
So it’s back to the drawing board :).
So, a few days later, here is the solution (don’t read on if you still want to think about the problem!):
The trains riders should all throw themselves at once. Why is that? Each passenger imparts exactly the same momentum when he jumps off: his mass times the relative velocity. However, to whom exactly does he impart it to? If there are still passengers left on board, THEY get some of the momentum, so they “rob” the train of its momentum. So jumping off together would ensure that no momentum is “wasted” on other passengers - it would all transfer directly to the train.
This is why space shuttles burn their fuel rapidly when lifting off - they want to get rid of their momentum all “at once”, or at least as rapidly as possible. Hence the spectacular liftoffs seen whenever a shuttle is launched.
I believe your solution is incorrect. The burn rate for propellant (whether rocket fuel or people throwing themselves off the back of a train) only affects the ΔV in a gravity-dependent situation such as the vertical takeoff of the shuttle. However, for gravity-independent motion, such as in deep space or on a horizontal track, the burn rate does not affect the ΔV.
You can use the rocket equation to solve this problem:
ΔV = Ve ln{ (M+m) / m } - gM/c
where c is the burn rate, M is the fuel mass, m is the vehicle mass. You see that the burn rate only matters when gravity (g) affects the motion. As long as each person leaps off the train with the same relative velocity as every other person, the net effect is the same no matter when they jump.
Hi Matt,
We can “prove” my solution by examining a simple limiting case.
Let’s take a simple example of a train of mass M, with two people of mass m each standing on top of it. Let +v be the initial velocity of the train+people, and let v0 be the relative velocity with which they jump off. Let’s examine the two possible solutions from a frame at rest.
1.) Both passengers jump off at once.
In this case, conservation of momentum tells us:
Momentum of (people+train) = constant = (M+2m)*v.
The velocity of the people after jumping will be (v-v0), and so their momentum will be (v-v0)*2m. The train’s momentum will be M*vf, where vf is the final velocity. Hence:
M*vf + (v-v0)*2m = (M+2m)*v
which we can solve for vf (I’ll spare you the algebra):
vf = final velocity of train after the two passengers jumped off simultaneously = v + 2*m*v0/M
2.) In the second scenario we let person A jump off, followed by person B. Here we need to break up the solution into two stages, obviously.
Stage #1: let vf1 be the velocity of person B and the train after person A has jumped off. Person A’s momentum after jumping off is m*(v-v0), while person B+train’s momentum is (M+m)*vf1. Hence we must solve:
(M+2m)*v = (M+m)*vf1 + (v-v0)*m
Stage #2: Now let’s use conservation of momentum for the system consisting of person B and the train. Their “initial” momentum - that is, right after A jumped off - will be (M+m)*vf1. Conservation of momentum dictates:
(M+m)*vf1 = (vf1-v0)*m + M*vf2
where vf2 is the final velocity of the train and (vf1-v0)*m is the momentum of person B after jumping off.
So we have here two equations with two unknowns (vf1, vf2), which you can solve and obtain:
vf2 = final velocity of train after A & B have jumped off one by one = v + v0*[(2m+M)*m/((m+M)*M)]
I’ll leave it up to you to plug in numerical values and verify that the second case results in a lower final velocity.
This naturally leads the question as to how can we reconcile this with your continuous equation (of liftoff). If you agree with my analysis up to this point try & think about it, and if you feel you cannot reconcile the two come back & I’ll show you.
Of course, if you disagree with my above solution please point out the flaw and we’ll take it from there :).