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## 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 ...