How Does Jumping Men Affect the Velocity of a Flatcar?

[geoffrey159]

Homework Statement

N men, each with mass m, stand on a railway flatcar of mass M.
They jump off one end of the flatcar with velocity u relative to the
car. The car rolls in the opposite direction without friction.
(a) What is the final velocity of the flatcar if all the men jump
off at the same time?
(b) What is the final velocity of the flatcar if they jump off one
at a time? (The answer can be left in the form of a sum of terms.)
(c) Does case (a) or case (b) yield the larger final velocity of
the flatcar? Can you give a simple physical explanation for your
answer?

Homework Equations

Conservation of momentum

The Attempt at a Solution

Hello dear forumers, I tried this exercise but I have the feeling I have been walking on eggs all along. I need your aproval if you don't mind reading my solution. Thanks!Solution:

Rk 1: Since there are no horizontal forces, the momentum is conserved in this direction.
Rk 2: Since the men jump at speed u relative to the car, they jump at speed v-u relative to an inertial frame fixed on the track

(a) Just after the jump, the momentum is $P = M v + mN (v - u)$. By conservation of momentum, it is constant over time so $P = P(0) = (M+mN)\times 0 = 0$, and the final speed is $v = \frac{mNu}{M+mN}$

(b) Let ${(v_k)}_{k = 1...N}$ be the speed of the car after jump $k$.
Man $k$ jumps when all men who have jumped before him are at rest on the ground, so the momentum just after his jump is

$P = (M+m(N-k)) v_k + m (v_k - u)$

On the other hand, momentum after previous man jumped was

$P = (M+m(N-k+1)) v_{k-1} + m (v_{k-1} - u )$,

which by the time current man jumps, goes to

$P = (M+m(N-k+1)) v_{k-1}$

because previous man is at rest, so there is a relation:

$(M+m(N-k+1)) v_{k-1} =(M+m(N-k)) v_k + m (v_k - u)$

So that $v_k = v_{k-1} + \frac{mu}{M+m(N-k+1)} = v_1 + \sum_{i=1}^{k-1}\frac{mu}{M+m(N-i)}$

By conservation of momentum, as in question (a): $v_1 = \frac{mu}{M+mN}$

So that at the end:

$v_N =\sum_{i=0}^{N-1}\frac{mu}{M+m(N-i)}$

(c) If $v_N^{(a)}$ is the final speed of the car at question (a), and $v_N^{(b)}$ the final speed of the car at question (b),

Since $M+mN \ge M+m(N-i)$ for all $i$ between 0 and N-1,

$v_N^{(b)} = \sum_{i=0}^{N-1}\frac{mu}{M+m(N-i)} \ge \sum_{i=0}^{N-1}\frac{mu}{M+mN} = \frac{mNu}{M+mN} = v_N^{(a)}$

So the car will be faster if men jump one after the other. I can't explain why ...

 

[Chestermiller]

Part A is not done correctly.

Final momentum = -umN+Mv = Initial momentum = 0

Chet

 

[geoffrey159]

Hello, thanks for the reply.
I am not sure because speed u is a relative speed.

 

[Chestermiller]

Hello, thanks for the reply.
I am not sure because speed u is a relative speed.

They almost certainly meant it to be relative to the speed before the guy jumps off. But I can see where there can be some ambiguity.

Chet

 

[BvU]

Interesting ! I was siding with Chet at first, but Geoff opened my mind to a much more sensible alternative. After all, if you jump from a skateboard, you don't jump far. Unless there are still a number of folks left standing (crowded!) on the thing :)

 

[geoffrey159]

They almost certainly meant it to be relative to the speed before the guy jumps off. But I can see where there can be some ambiguity.

Hello !
I don't understand where the ambiguity was. We know from start that the speed of the train is not constant, so we cannot take the train as the origin of the frame. If we want to use momentum conservation, we have to find an inertial frame, don't we?

Interesting ! I was siding with Chet at first, but Geoff opened my mind to a much more sensible alternative. After all, if you jump from a skateboard, you don't jump far. Unless there are still a number of folks left standing (crowded!) on the thing :)

That's a Christmas comment ! :-) Thanks ! Do you find that solution reasonable ?

 

[Chestermiller]

Hello !
I don't understand where the ambiguity was. We know from start that the speed of the train is not constant, so we cannot take the train as the origin of the frame. If we want to use momentum conservation, we have to find an inertial frame, don't we?

Hu? If the guy jumps off with a velocity u relative to train velocity that existed just before he jumped, and the train was traveling at v just before he jumped, then his velocity relative to the ground after he jumps will be v - u. This is certainly different than your interpretation. Your interpretation is that he has jumped off with a velocity u relative to the train velocity that existed just after he jumped. I call that ambiguity when there are two possible interpretations to the problem statement. Another equally good interpretation would be that he jumped with a velocity u relative to the average of the train velocities that existed just before and just after he jumped.

Chet

 

[geoffrey159]

Ok, so forgive me but I want to understand a little bit more.
I'm trying to connect your previous post with the one where you said that 'final momentum = -umN+Mv'.
I don't get it yet. Why did you write '-umN' ?

 

[Chestermiller]

Ok, so forgive me but I want to understand a little bit more.
I'm trying to connect your previous post with the one where you said that 'final momentum =-umN+Mv.
I don't get it yet. Why did you write '-umN' ?

The initial momentum is zero, because the men and the flat car are both stationary to begin with. If now all the men jump off the rear of the flat car at the same time with a speed u (in the negative x direction) relative to both the ground and the flat car (whose velocity before they jump is zero), then the final momentum of the men in the positive x direction will be -umN. If the final velocity of the flat car is Mv, the new final momentum in the positive x direction of the men plus the flat car will be -umN+Mv=0.

Chet

 

[geoffrey159]

I understand, but it seems a little strange to use the flat car's velocity before jumping to conclude about momentum after jumping.
I am not 100% convinced.
Why not just stick to the basic definition of momentum and relative velocity ?
Momentum after jumping is a mass of mN kilograms traveling at a relative speed u from the car, plus a M kilogram mass traveling at speed v.
Using the definition, P = mN(v-u) + Mv, where (v-u) is just a replacement of relative velocity from the car to frame coordinates velocity.
To this point I make no particular assumption about speed.
For once, I tend to agree with myself !

 

[Chestermiller]

I understand, but it seems a little strange to use the flat car's velocity before jumping to conclude about momentum after jumping.
I am not 100% convinced.
Why not just stick to the basic definition of momentum and relative velocity ?
Momentum after jumping is a mass of mN kilograms traveling at a relative speed u from the car, plus a M kilogram mass traveling at speed v.
Using the definition, P = mN(v-u) + Mv, where (v-u) is just a replacement of relative velocity from the car to frame coordinates velocity.
To this point I make no particular assumption about speed.
For once, I tend to agree with myself !

I don't have any problem with this interpretation. As I said in post #7, this is the the same as assuming that the guys jump off the flat car with a velocity u relative to the flat car velocity v that exists just after they have jumped. From the problem statement, you could make a case for either interpretation. I'm just guessing that they meant for you to use the interpretation that I indicated.

Maybe you want to try the problem with both interpretations?

Chet