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Explosion Question did I do this correctly?

  1. Jul 4, 2004 #1
    Hey all... I got an answer to this, but I'm not sure if I did it correctly.

    q: A body of mass 8.0kg is travelling at 2.0 m/s under the influence of no external agency. At a certain instant an internal explosion occurs, splitting the body into two chunks of 4.0 kg mass each; 16 joules of translational kinetic energy are imparted to the two chunk system by the explosion. Neither chunk leaves the line of the original motion. Determine the speed and direction of motion of each of the chunks after the explosion.

    a: I assumed the original block was just 2 4kg blocks with the same kinetic energy travelling in the same direction. (to the right)

    Then, I subtracted 16J from the 8J of kinetic energy that the left most piece had because that's the way the 16J would be directed from the explosion.

    When I put that into the calculation, I got 2.0 m/s (to the left) for the left block, and when I put 24 J (to the right) into the kinetic energy equation, I got the right block travelling 3.46 m/s (to the right).

    Did I do this correctly??

    Steve
     
    Last edited: Jul 4, 2004
  2. jcsd
  3. Jul 4, 2004 #2

    AKG

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    Sorry, but I don't understand what you did here at all. What you need to do is use the idea of conservation of momentum:

    [tex]\mathbf{p}_{tot} = \mathbf{p}'_{tot}[/tex]

    where, if x is an initial quantity, then x' is the value after the explosion.

    [tex]\mathbf{p}_{tot} = \mathbf{p}'_{tot}[/tex]

    [tex]8.0kg \times 2.0m/s [forward] = 4.0kg(v'_1 + v'_2)[/tex]

    [tex]4.0 m/s [forward] = (v'_1 + v'_2)\ \dots \ (1)[/tex]

    Note: [itex]v'_1[/itex] and [itex]v'_2[/itex] are the final velocities of chunk 1 and chunk 2, respectively (although it doesn't really matter which chunk we call chunk 1 or chunk 2).

    Also, from the law of conservation of energy:

    [tex]\frac{1/2}(4.0kg)[(v'_1)^2 + (v'_2)^2] = 16 J[/tex]

    [tex](v'_1)^2 + (v'_2)^2 = 8m^2/s^2\ \dots \ (2)[/tex]

    Now, solve for your velocities using equations (1) and (2). I would suspect you might get two possible answers, one that [itex]v'_1 = v'_2 = 2.0 m/s[/itex] meaning that the explosion has no effect, or another solution which is the one you'd be looking for. Note, you can and should use the law of conservation of momentum because there are no external forces acting on the system, I believe. If this is wrong, let me know.
     
  4. Jul 7, 2004 #3
    yea AKG i tried it and got 2m/s, therefore it had no effect, and they both had the same speeds.Btw how did you get 3.46m/s??? Maybe you used another method...so either that or 2.82m/s of the sqrt of 8..
     
    Last edited: Jul 7, 2004
  5. Jul 7, 2004 #4

    Gokul43201

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    I think that should be :

    [tex]\frac{1}{2}(4.0kg)[(v'_1)^2 + (v'_2)^2] = 16J + \frac {1}{2}(8.0kg)(2.0m/s)^2[/tex]

    The way you solved it, the explosion adds no energy to the system, so the final velocities will be the same as the initial velocities.

    However, you are actually getting an extra 16J from some form of chemical energy.

    Answer v1' = 0 m/s and v2' = 4 m/s.
     
    Last edited: Jul 7, 2004
  6. Jul 7, 2004 #5

    AKG

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    Yeah, that makes more sense.
     
  7. Jul 12, 2004 #6
    thanks a million for the help
     
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