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Calculating the collusion and loss of energy?

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations
    m1v1i+m2v2i=m1v1f+m2v2f


    3. The attempt at a solution
    I tried doing MV=m1v1f+m2v2f

    I know that the sum of the momentum of the two parts is equal to the the momentum of the the full shell. Since I have two unknowns, I used the conservation of energy. I set used KE=KE1+KE2. The kinetic of the full shell equals to the sum of the kinetic energy of the two parts. Since I have two unknowns here too, I used a system of equations solve for v. Am I on the right path? I found two "ms" didn't add up to 5kg. Thank you in advance.
     

    Attached Files:

  2. jcsd
  3. Oct 22, 2013 #2

    rude man

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    It would be helpful if you stated the problem - preferably in its original version. If it's not in English, someone may be able to translate it accurately. Supplementally, give your own translation as best you can.
     
  4. Oct 22, 2013 #3

    collinsmark

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    Don't use conservation of energy. Energy isn't conserved in this problem.

    Yes, you will come back to that later when you find the difference in kinetic energies such that you can calculate the energy released. But that comes later.

    So go back to your first equation regarding conservation of momentum. That's one of the two equations you will need.

    The second equation that you'll need is actually quite simple. What is the sum of the two masses after the explosion? [Hint: it's the same as the shell's mass before the explosion.] Put that in equation form. :wink:

    That's not true for this problem. Additional energy is released as part of the explosion.

    But what you can do is find the kinetic energy of the shell before the explosion and compare that to the sum of kinetic energies of the two parts.

    According to your attachment, the two masses should add up to 4.5 kg. (not 5 kg).
     
  5. Oct 22, 2013 #4
    O-oh.. ohhh. Wow I feel slow. Thank you very. So it's 410m1+215m2=1440 and m1+m2=5. I understand this part now(although I'm 85% sure on the former's signage, I'm still loose on signage in physics).

    After this to find kinetic energy, I'm thinking of subtracting the kinetic energy of the full shell to the sum of the kinetic energies of the two parts using the new found masses, is that correct? Basically ΔKE=KEShell-(KE1+KE2).
     
  6. Oct 22, 2013 #5

    collinsmark

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    That looks right to me :smile:

    (Except for the lack of units. But perhaps I'm being nit-picky.)

    I think you mean 4.5 kg.

    Everything is going in the same direction for this problem, so the signs look good to me the way you have them.

    Almost right. You can usually expect the energy of the explosion to cause the kinetic energy of the subsequent pieces to be greater overall, compared to the original kinetic energy. And that holds true for this problem too. So what does that tell you about what you need to subtract from the other?
     
  7. Oct 22, 2013 #6
    Oh sorry about the unit thing. So I have it backwards? It should be the sum of the potential energy of the two parts minus the kinetic energy of the shell. Could I think of it as the absolute value of the change in kinetic energy, although there's no such convention? Thank you again.
     
  8. Oct 22, 2013 #7

    collinsmark

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    I don't think potential energy plays a role in this problem.

    What I meant in my last post is that explosions tend to increase the kinetic energy of the things around them. So if you want to find the energy released by the explosion, and assuming all of the energy released by the explosion went into increasing the kinetic energy of the two pieces (in addition to what they already had before the explosion, back when the two pieces were a single shell), then... :wink:
     
  9. Oct 22, 2013 #8
    Sorry I miswrote that, I meant Kinetic energy. As in ƩKEParts-KEShell. Thank you for everything by the way.
     
  10. Oct 22, 2013 #9

    collinsmark

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    That looks like the correct idea to me. :approve:
     
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