Conservation of Momentum in Multiple-Person Jumping Scenario

[AlexChandler]

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)v₁-m_p(v₁-u)=0

when the second person jumps we have

(m_c+(n-2)m_p)v₂-m_p(v₂-u)=(m_c+(n-1)m_p)v₁

third person jumps off

(m_c+(n-3)m_p)v₃-m_p(v₃-u)=(m_c+(n-2)m_p)v₂

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(v₁-u) from both sides of the 2nd equation we find:

(m_c+(n-2)m_p)v₂=m_p(v₁+v₂-2u)

continued use of this fact leads to the equation

m_cv_n=m_p(v₁+v₂+...+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!

 

[gneill]

Suppose we let ∆p = m_p*u be the impulse (momentum) that is contributed by the jumping person to the remainder of the people and the flatcar. The impulse is always the same with respect to the flatcar, because the velocity u is always the same with respect to the flatcar, and the people are all of the same mass.

When the first person jumps the initial velocity (v₀) is zero, and the "final" velocity v₁ is

v_1 = \frac{\Delta p}{m_c + n \; m_p}

When the next person jumps, the flatcar is already moving at v₁, so the new velocity is ... ?EDIT: I reconsidered the amount of mass that should be used in the denominator. I now think it should reflect the mass of the cart plus people before the jumper departs. So I've amended the (n - 1) term to be n in the formula.

 

[AlexChandler]

So the velocity after the second person jumps is given by

m₂v₂=m₁v₁ +∆p

(m_c+(n-2)m_p)]v₂=(m_c+(n-1)m_p)v₁+m_pu

so looking at what we had found for v1, this equation becomes

v₂=2m_pu /(m_c+(n-2)m_p)

and continued use of this leads to:

v_n=nu m_p/m_c

Is this what you are saying?

Thanks for the help!

 

[gneill]

So the velocity after the second person jumps is given by

m₂v₂=m₁v₁ +∆p

(m_c+(n-2)m_p)]v₂=(m_c+(n-1)m_p)v₁+m_pu

so looking at what we had found for v1, this equation becomes

v₂=2m_pu /(m_c+(n-2)m_p)

and continued use of this leads to:

v_n=nu m_p/m_c

Is this what you are saying?

Thanks for the help!

Not quite. It appears to me that the change in velocity for the flatcar for each person who jumps will be given by

\Delta V = \frac{\Delta p}{current\; total \; mass}

where ∆p is u*m_p. So that

v_0 = 0

v_1 = 0 + \frac{\Delta p}{m_c + n m_p}

v_2 = 0 + v_1 + \frac{\Delta p}{m_c + (n - 1) m_p}

v_3 = 0 + v_1 + v_2 + \frac{\Delta p}{m_c + (n - 2) m_p}

and so on. Casting this into summation notation, the velocity after the mth person has jumped is:

v_m = \Delta p \sum_{i = 1}^{m} \frac{1}{m_c + (n + 1 - i)m_p}

or slightly more elegantly,

v_m = \Delta p \sum_{i = 0}^{m - 1} \frac{1}{m_c + (n - i)m_p}

 

[AlexChandler]

Not quite. It appears to me that the change in velocity for the flatcar for each person who jumps will be given by

\Delta V = \frac{\Delta p}{current\; total \; mass}

where ∆p is u*m_p. So that

v_0 = 0

v_1 = 0 + \frac{\Delta p}{m_c + n m_p}

...

v_m = \Delta p \sum_{i = 0}^{m - 1} \frac{1}{m_c + (n - i)m_p}

Ahh I see where i was making my mistake now

However, shouldn't the velocity after one person jumps be:

v₁=m_pu/(m_c+(n-1)m_p) ?

And then the final velocity should be the sum from 1 to m instead of 0 to m-1?

 

[gneill]

Ahh I see where i was making my mistake now

However, shouldn't the velocity after one person jumps be:

v₁=m_pu/(m_c+(n-1)m_p) ?

And then the final velocity should be the sum from 1 to m instead of 0 to m-1?

It would be, if the velocity of the jumping person was taken to be with respect to the velocity of the cart immediately before he jumped, rather than the 'new' velocity that the cart has after he departs. I followed your lead in that respect.

 

[AlexChandler]

Aha! I see. I think I understand it now. I tend to confuse myself quite often.
Thank you very much you have been extremely helpful!