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Conservation of Momentum involving a space ship blowing up into three pieces

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data
    A spaceship of mass 2.30×10^6kg is cruising at a speed of 5.40×10^6m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.20×10^5kg, is blown straight backward with a speed of 2.20×10^6m/s. A second piece, with mass 7.80×10^5kg, continues forward at 1.20×10^6m/s.

    What is the speed of the third piece? in m/s


    2. Relevant equations
    M3=m3=M-m1+m2

    Conservation of Momentum says:

    m1v1+m2v2+m3v3=MV


    3. The attempt at a solution

    To get the weight of the third piece i used
    M3=m3=M-m1+m2=
    2.30*10^6kg - 5.20*10^5 kg + 7.80*10^5kg= 2.56*10^6 kg

    Then
    m1v1+m2v2+m3v3=MV

    (5.20*10^5kg)(2.20*10^6)+(7.80*10^5kg)(1.20*10^6 m/s)+(2.56*10^6 kg) V3 = (2.30*10^6kg)(5.40*10^6ms)=

    This may sound dumb but I don't know how to do this equation

    I have 1.44*10^12 + 9.36*10^11 + 2.50*10^6

    2.0800025*10^12 = 1.242*10^13

    Then do I divide those 2, if so I get 1.67*10^-1 and it's wrong

    What am I doing wrong?

    Thank you
     
  2. jcsd
  3. Oct 22, 2011 #2

    Doc Al

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    Staff: Mentor

    Shouldn't that be: m3 = M - m1 - m2 ?
     
  4. Oct 22, 2011 #3
    I don't know, that piece is going to the right which is to the east and i'm calling that way positive?
     
  5. Oct 22, 2011 #4
    It is positive just a mistake, I added it in the equation.
     
  6. Oct 22, 2011 #5

    Doc Al

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    Staff: Mentor

    The direction of motion doesn't affect the mass. (It affects momentum, of course.)
     
  7. Oct 22, 2011 #6
    I tried subtracting that and I have
    1*10^6kg for mass 3

    When i plugged it into the conservation I still get 1.67*10^-1 ?
     
  8. Oct 22, 2011 #7

    Doc Al

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    Staff: Mentor

    Good.

    How can you get the same answer with different numbers?

    Two problems:
    (1) As already pointed out, you have the wrong value for the mass of the third piece.
    (2) You didn't incorporate the direction of motion. Things that go forward should have + velocity; things that go backward should have -.
     
  9. Oct 22, 2011 #8
    Ok m1 is positive and v1 is negative because it's going backwards?
    m2 positive because it's weight, and m2 is positive because it's going forwards?
    m3 doesn't matter because it's weight....then solve to get v3?
     
  10. Oct 22, 2011 #9

    Doc Al

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    Almost: Masses are always positive, regardless of direction.

    v1 is negative, v2 is positive. v3 you will solve for.
     
  11. Oct 22, 2011 #10
    Ok good, the only value i had to change was v1 to negative because it is going to the left or to the west.

    it made V1 value negative

    (5.20*10^5)(-1.144*10^12)+(9.36*10^11)+(1*10^6)=(2.30*10^6kg)(5.40*10^6m/s)

    i get -1.69*10^-2, is this correct?
     
  12. Oct 22, 2011 #11
    oopse disregard the (5.20*10^5), my mistake
     
  13. Oct 23, 2011 #12
    I was able to use a more simple way and now it's correct, thanks!
     
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