1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear momentum conservation - rocket
Loading...