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Momentum Conservation

  • #1
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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 mc. 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



Pi=Pf

The Attempt at a Solution



When the first person jumps we have

(mc+(n-1)mp)v1-mp(v1-u)=0

when the second person jumps we have

(mc+(n-2)mp)v2-mp(v2-u)=(mc+(n-1)mp)v1

third person jumps off

(mc+(n-3)mp)v3-mp(v3-u)=(mc+(n-2)mp)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 mp(v1-u) from both sides of the 2nd equation we find:

(mc+(n-2)mp)v2=mp(v1+v2-2u)

continued use of this fact leads to the equation

mcvn=mp(v1+v2+...+vn - 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!
 

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773
Suppose we let ∆p = mp*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

[tex] v_1 = \frac{\Delta p}{m_c + n \; m_p} [/tex]

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.
 
Last edited:
  • #3
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So the velocity after the second person jumps is given by

m2v2=m1v1 +∆p

(mc+(n-2)mp)]v2=(mc+(n-1)mp)v1+mpu

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

v2=2mpu /(mc+(n-2)mp)

and continued use of this leads to:

vn=nu mp/mc

Is this what you are saying?

Thanks for the help!
 
  • #4
gneill
Mentor
20,793
2,773
So the velocity after the second person jumps is given by

m2v2=m1v1 +∆p

(mc+(n-2)mp)]v2=(mc+(n-1)mp)v1+mpu

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

v2=2mpu /(mc+(n-2)mp)

and continued use of this leads to:

vn=nu mp/mc

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

[tex] \Delta V = \frac{\Delta p}{current\; total \; mass} [/tex]

where ∆p is u*mp. So that

[tex] v_0 = 0 [/tex]

[tex] v_1 = 0 + \frac{\Delta p}{m_c + n m_p} [/tex]

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

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

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

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

or slightly more elegantly,

[tex] v_m = \Delta p \sum_{i = 0}^{m - 1} \frac{1}{m_c + (n - i)m_p} [/tex]
 
  • #5
283
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Not quite. It appears to me that the change in velocity for the flatcar for each person who jumps will be given by

[tex] \Delta V = \frac{\Delta p}{current\; total \; mass} [/tex]

where ∆p is u*mp. So that

[tex] v_0 = 0 [/tex]

[tex] v_1 = 0 + \frac{\Delta p}{m_c + n m_p} [/tex]

...

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

Ahh I see where i was making my mistake now

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

v1=mpu/(mc+(n-1)mp) ?

And then the final velocity should be the sum from 1 to m instead of 0 to m-1?
 
  • #6
gneill
Mentor
20,793
2,773
Ahh I see where i was making my mistake now

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

v1=mpu/(mc+(n-1)mp) ?

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.
 
  • #7
283
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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!
 

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