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Momentum and kinetic energy of pieces resulting from an explosion

  1. Dec 30, 2011 #1
    1. The problem statement, all variables and given/known data

    A body at rest contains an explosive device which when ignited fractures the body in two pieces, one of which is two times as massive as the other.

    i) Why must the pieces travel in opposite directions?

    ii) How much more momentum does the larger object have, than the smalller?

    iii) How much more kinetic energy does the larger object have, than the smaller?


    2. Relevant equations

    p = mv, p = momentum, m = mass and v = velocity

    KE = (1/2)mv^2, KE = kinetic energy

    3. The attempt at a solution

    i) The pieces must travel in opposite directions due to the conversation of momentum. As the body was at rest before the explosion, and the explosion should exert roughly equal force in every direction, there is no net external force on the object and so their momentum after the explosion should add up to 0. The only way for this to happen (excluding the possibility that the objects remain at rest) is if they travel in opposite directions.

    ii) As the individual momentums of the objects must equal 0, the larger object has the same amount of momentum as the smaller. This can be explained by the smaller object, which has a small mass, having a higher velocity to compensate, thereby giving the same momentum.

    If p = momentum, m = mass of smaller object, v = velocity of larger object and x is an unknown, then mxv=2mv.

    Rearranging this gives x = 2, therefore the smaller object will have a velocity twice that of the larger object.

    iii) KE = (1/2)mv^2,

    For the smaller object:

    KE = (1/2)m(2v)^2 = 2mv^2

    For the larger object:

    KE = (1/2)2mv^2 = mv^2

    Therefore the smaller object will have twice as much kinetic energy as the larger object.


    Have I done this correctly? I'm a bit unsure about iii)!
     
  2. jcsd
  3. Dec 30, 2011 #2
    i)correct
    ii)correct
    iii) and correct again !!!

    Nice explanation by the way, i cant write that good !!!

    PS: Are you phosgene from chemical forums???
     
  4. Dec 30, 2011 #3
    Oh, thank you :)! And no, I'm not, though I do have an account there :p
     
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