Free space motion I was going through Kleppner and got this interesting doubt. At page 138 there is an article named rocket in free space.So if there is no external force. The fuel can expand rapidly or slowly without affecting final velocity of rocket. Or we can say the velocity of rocket change by same amount whether we burn all the fuel in one go. or slowly. But at page 148, Question no. 3.14 if men jump from railway flatcar all at same time vs men jump one by one the speed don't remain same but this is similar case as the rocket one there is no friction involved.
 A: I believe that the rocket can burn the fuel "slowly" or "quickly", but cannot burn the fuel "instantaneously".  Whereas on the flatcar, all the people can jump at once.
The rocket acceleration is assuming that the fuel burn is a continuous process that you can divide as finely as you want.  The burn of some small fuel $dm$ results in momentum transfer to the exhaust in one direction and the rocket plus remaining fuel in the other.
The flatcar has an option that the rocket does not have, which is for all the "fuel" to be expended at once.  In this case, none of the energy goes into accelerating the remaining fuel forward.
If you put extra fuel on the rocket, you get some benefit, but it also increases the mass that the first burns have to accelerate.  You don't get a linear increase in the final speed.
If you put extra jumpers on the flatcar, there's no downside.  The extra mass of the jumper isn't acted on by the others, so there's a linear increase in the speed with their number.
A: The difference is that in one case, the mass being accelerated is varying, but in the other case the force being applied is varying.
So in the case of the rocket, the amount of energy input will be the same total, regardless of how we do it through time. Same energy in. Also, same mass being accelerated by that energy. Accelerating a mass takes energy. $E=\tfrac{1}{2}mv^2$. If we have a single value for $m$, and we end-up adding the same overall energy $E$, then the final velocity $v$ must be the same regardless.
In the case of the train-car, we are not varying the total energy put in. We are varying what masses get accelerated. So if a man stays awhile and gets accelerated, then that takes up some of the energy. But if he leaves early, then that energy will not go into accelerating him; it will all go into accelerating the car. We could think of two possibilities: 1. One guy jumps right at the beginning. Or 2. This one guy waits until the very end to jump. Either way engine has the same power output over time, and so it has the same total energy to put into accelerating things to velocity.

*

*Total energy $E$ goes into speeding up the train and appears as kinetic energy: $$E= \tfrac{1}{2}~m_{train}~v^2$$


*Total energy $E$ goes into speeding up the train and man and appears as kinetic energy: $$E= \tfrac{1}{2}~m_{train+man}~v^2$$
If the $E$ is the same in both 1 and 2 above, then the $v$ will be lower in number 2 since $m$ is higher.
