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Homework Help: What is the speed of a piece of a spaceship that blew up?

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A spaceship of mass 2.30×106kg is cruising at a speed of 5.50×106m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.20×105kg , is blown straight backward with a speed of 2.30×106m/s . A second piece, with mass 8.40×105kg , continues forward at 1.30×106m/s .

    2. Relevant equations

    3. The attempt at a solution

    Using mv=p, I calculated p of each piece like this:
    p1 = 1.196*10^12
    p2 = 1.092*10^12
    p_total = 1.265*10^13

    Then I found p3 by subtracting p2+p1 from p_total.

    I divided p3 by the mass of the mystery piece: p/940,000 = 1.0302*10^13

    But... the answer was wrong. I also tried subtracting: p1-p2 =pt and working from there. What am I doing wrong?

  2. jcsd
  3. Oct 8, 2013 #2
    Momentum is a vector. You need to add the three momenta (after explosion) as vectors. Here, in 1D, be careful with the signs. A diagram may help.
  4. Oct 8, 2013 #3
    But it's not 1D, right? Because one goes forward, one goes back, and the mystery piece I am solving for goes off in an unspecified direction...
  5. Oct 8, 2013 #4
    I got it, thanks for that tip.
  6. Oct 8, 2013 #5
    I believe you are right in using the conservation of momentum to tackle this problem. The initial momentum should equal the final momentum of the combined three pieces, such that

    pinitial = p1 + p2 + p3

    m(total)v = m1v1 + m2v2 + m3v3

    p(total) = (2.3x106kg)(5.50 x 106m/s) = (1.265x1013)

    p1 = (5.20×105kg)(-2.30×106m/s) = (-1.20x1012)

    p2 = (8.40×105kg)(1.30×106m/s) = (1.092x1012)

    p3 = (9.4x105kg) v3

    Because the first piece was blown backwards, it becomes important to apply the negative to its velocity component, and should look like,

    (1.265x1013) = (-1.20x1012) + (1.092x1012) + (9.4x105kg)(v3)

    It becomes simple to solve for v3 . I assume the sign was causing your problem in the calculations. The mass of the third piece was found by the missing mass, m(total) = m1 + m2 + m3

    Because the problem doesn't specify, i took the assumption that the collision only occurs in a single direction, along the x-axis, and that the velocity of each piece doesn't have to be broken down into individual x- and y-components.
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