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Conservation of momentum spaceship problem

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data
    A spaceship of mass 2.00×10^6 kg is cruising at a speed of 6.00×10^6 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 4.60×10^5 kg, is blown straight backward with a speed of 2.10×10^6 m/s. A second piece, with mass 8.30×10^5 kg, continues forward at 1.30×10^6 m/s.

    2. Relevant equations
    P1+P2+P3 = P(system)
    m1(v1)+m2(v2)+m3(v3) = m(total)v(total)

    3. The attempt at a solution

    So, ... I'm thinking this is a conservation of momentum problem... so at first I used the above equations. First off, I tried to find the mass of the third piece (since it's not given, its necessary) which is: M3 = M_t - (m_1 + m_2) = 7.1 x 10^5 kg.

    Then, I substituted all of the answers into the final equation, which gives me 1.4 x 10^7 m/s. However, this is wrong. Can anyone point me to what I'm missing?

  2. jcsd
  3. Nov 25, 2007 #2

    Doc Al

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    Staff: Mentor

    The mass looks right to me. Show exactly what you plugged in to get your answer for the speed.
  4. Nov 25, 2007 #3

    Thanks for responding!

    Here we go...

    m1(v1)+m2(v2)+m3(v3) = m(total)v(total)
    (4.6*10^5)(2.1*10^6) + (8.3*10^5)(1.3*10^6) + (7.1*10^5)(v3) = 1.2*10^13

    (1.2*10^13) - (9.66*10^11+1.079*10^12) = (7.1 *10^5)(v3)

    9.955*10^12 = (7.1 *10^5)(v3)

    (v3) = 1.4*10^7 :(

    I have to be missing something obvious... :(
  5. Nov 25, 2007 #4

    Doc Al

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    Staff: Mentor

    Careful with signs: The 4.60×10^5 kg piece moves backwards.
  6. Nov 25, 2007 #5
    Oh my goodness, yes, you are right.

    Thanks for the help... I got it now, and that makes sense. I was wondering how it would matter if the piece went forward and not backwards.

    Thanks again. :)
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