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Homework Help: Conservation of Momentum in a Spaceship Explosion

  1. Jul 29, 2004 #1
    Hi all,

    I have become frusturated at a conservation of momentum problem. A space ship(ss) with a mass of 2.0*10^6 [kg] is crusing at a speed of 5.0*10^6 [m/s], when it blows up. One section(s1), with mass 5.0*10^5 [kg] is blown straight backwards at a speed of 2.0*10^6 [m/s]. The second section(s2), with mass 8.0*10^5 [kg], continues forwards at a speed of 1.0*10^6 [m/s]. A third section moves at a certain speed.

    Trying to find the speed of the third section, I initally approached the problem as a kinetic energy problem, considering that I might be able to use KE=(1/2)mv^2 and simply subtract s1+s2 from ss because the explosion is internal and the change in kinetic energy will be 0:

    First, the third section has a mass of 7.0*10^5[kg]

    KE(ss)=2.5*10^19 [J]
    KE(s1)= 1.0*10^18 [J]
    KE(s2)= 4.0*10^17 [J]

    KE(s1+s2)= 1.4*10^18[J]

    So I subtracted this from KE(ss):

    2.5*10^19 [J]-1.4*10^18[J]= 2.36*10^19 [J], which is the KE for the third section

    Then I set up 2.36*10^19 [J] = (1/2)*(7.0*10^5)(v_s3)^2, and I obtained a velocity for the third section of v_s3= 7.93*10^6 [m/s]! The solution in the back of the book reads 1.46*10^7 [m/s].

    So, ok, this is obviously not the correct way to set up this problem, so I did vector sum for momentum using p(vector)=m*v(vector).

    The space ship has a vector 9.45*10^12 [kg*m/s]
    Section1 has vector 1.144*10^12 [kg*m/s]
    Section2 has vector 1.105*10^12 [kg*m/s]

    Section 3 has mass of 7.5*10^5 [kg]. I subtracted section1+section2 vectors from the space ship vector, and the result was 7.201*10^12 [kg*m/s]. Then I inserted this back into the p(vector)=m*v(vector):

    7.201*10^12 [kg*m/s]= 7.5*10^5[kg] * v

    I solved for v and recieved 9.6*10^6[m/s].

    Im obviously not doing something right. I think I may not be factoring in that these are vectors? Thanks for any help.
  2. jcsd
  3. Jul 29, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Forget about KE. It's not conserved in an explosion.



    Right, but realize that vectors have direction. (Or at least a sign.)

    Why 9.45? 2.0 x 5.0 = 10. And in what direction? Call it the positive x-direction.
    Again, check your arithmetic. Section 1 moves backwards: that's the negative x-direction; Section 2 moves forward: the positive x-direction.

    Yet another arithmetic error!
    Careful! Signs matter.
    Two main mistakes: (1) arithmetic errors (2) not properly treating the sign (which represents the direction, in this case) of the vectors.
  4. Jul 29, 2004 #3
    Thanks for the help. Here is what I have done.

    The Ship has a vector of +1.0*10^13.
    Section 1 has a vector of -1.0*10^12.
    Section 2 has a vector of +8.0*10^11
    So, section 3 has a vector of 1.02*10^13, which I set up this way:

    1.02*10^13= 700000*v, which returns 1.457*10^7, which is most excellent.

    My mistakes are fairly embarassing, but not keeping things organized and thinking too far ahead are my achilles heel with physics.

    Thanks again.
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