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## Homework Statement

I am having trouble with part B of this problem

N people, each of mass mp, stand on a railway flatcar of mass m

_{c}. They

jump off of 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 car if all the people jump at the same time?

(b) What is the final velocity of the car if the people jump one at a time? Leave

the answer as a sum of terms.

(c) Does case 1a or 1b yield the largest final velocity of the car?

## Homework Equations

P

_{i}=P

_{f}

## The Attempt at a Solution

When the first person jumps we have

(m

_{c}+(n-1)m

_{p})v

_{1}-m

_{p}(v

_{1}-u)=0

when the second person jumps we have

(m

_{c}+(n-2)m

_{p})v

_{2}-m

_{p}(v

_{2}-u)=(m

_{c}+(n-1)m

_{p})v

_{1}

third person jumps off

(m

_{c}+(n-3)m

_{p})v

_{3}-m

_{p}(v

_{3}-u)=(m

_{c}+(n-2)m

_{p})v

_{2}

We notice in each equation that we can subtract the momentum of the last person that jumped off from both sides, and we start to see a pattern. For example if we subtract m

_{p}(v

_{1}-u) from both sides of the 2nd equation we find:

(m

_{c}+(n-2)m

_{p})v

_{2}=m

_{p}(v

_{1}+v

_{2}-2u)

continued use of this fact leads to the equation

m

_{c}v

_{n}=m

_{p}(v

_{1}+v

_{2}+...+v

_{n}- nu)

however this is as far as i can get without things getting

*extremely complicated*. And i know that he does not want the answer in terms of all of the velocities that the car traveled on the way. There must be a more direct way to solve this. Does anybody have any ideas? Thank you!