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Homework Help: Linear momentum conservation - rocket

  1. Jul 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A two-ended "rocket" that is initally stationary on a frictionless floor, with its center a the origin of an x axis. The rocket consists of a central block C(of mass M = 6.00kg) and blocks L and R (each of mass m = 2.00kg) on the left and right sides. Small explosions can shoot either of teh side blocks away fro mblock C and along the x axis. Here is the sequence: (1) At time t = 0, block L is shot to the left with a speed of 3.00 m/s relative to the velocity that the explosion gives the rest of the rocket. (2) Next at the time t = .80 s, block R is shot to the right with a speed of 3m/s relative to the velocity that block C then has. At t = 2.80s, what are the (a) velocity of block C and (b) the position of its center?

    http://ready2goxtr.googlepages.com/Problem943.jpg

    2. Relevant equations

    Pi = Pf




    3. The attempt at a solution

    Pi = m1v1 + m2v2 + m3v3 Pf = m1v1f + m2v2f + m3v3f

    I dont really understand whats happening in the problem
     
  2. jcsd
  3. Jul 8, 2008 #2

    tiny-tim

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    Hi Ready2GoXtr! smile:

    The question is rather badly worded :frown:, but I think it means that the first explosion is between L and C, and the second between R and C, so that each explosion pushes both blocks away from each other. :smile:
     
  4. Jul 8, 2008 #3
    alright so should I set PI = to 0 seconds and PF = to .80? seconds
     
  5. Jul 8, 2008 #4

    tiny-tim

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    Not following you. :confused:

    Pi = 0, yes … but P is constant, so Pf = 0 also.

    Hint: after the first explosion, v2 = v3, and they are both 3 m/s different from v1 :smile:
     
  6. Jul 8, 2008 #5
    You mean at t = .80 seconds v2 = v3?
     
  7. Jul 8, 2008 #6
    Assuming that v2 is the intial velocity of R block and v3 is the initial velocity of the C block, that statement is true. (Note: Initial means before the R block is shot off).
     
  8. Jul 8, 2008 #7

    tiny-tim

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    No, I mean between t = 0 and .80 …

    the explosion is between blocks 1 and 2, so block 2 pushes block 3, and therefore blocks 2 and 3 have the same velocity (v2 = v3), until the second explosion. :smile:
     
  9. Jul 8, 2008 #8
    so could i say that m1v1i = 2kg * 3m/s?
     
  10. Jul 8, 2008 #9

    tiny-tim

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    Nooo …
    Going to bed now … :zzz:​
     
  11. Jul 8, 2008 #10
    Okay i think ima just sskip it, dunno how to do the question
     
  12. Jul 8, 2008 #11
    I decided to give it another try. since the momentums have to equal each other, I set m1v1 =- (m2+m3)v2 v2 = .75 m/s i

    then i just decided. well what would have it been if it was just m1 and m2

    m1v1 = m2v2 v2 = -1 then i took -1 + .75 = equal final velocity which is -.15 m/s

    For part b that was easy

    vit = .6m then vft = -.42 added them .18 m
     
  13. Jul 9, 2008 #12

    tiny-tim

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    Hi Ready2GoXtr! :smile:

    No, those aren't the results I get.

    It's difficult to tell unless you show your full working, but I think you subtracted 1 from 3 instead of adding.

    Hint: You need two positive speeds (one left, one right) which add up to 3, and one is 4 times the other. :smile:
     
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