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Momentum of cannon ball homework

  1. Apr 5, 2007 #1
    A cannon shoots a shell straight up. It reaches its maximum height, 875 feet, and splits into two pieces, one weighing 3 lb and the other 5 lb. The two pieces are observed to strike the ground simultaneously. The 5 lb piece hits the ground 1,659 feet away from the explosion (measured along the x axis). Find the magnitude of the velocity of the 5 lb piece jsut after the explosion.

    Work done before explosion = 8lb * 32 ft/s^2 * 875 ft = 224000 J

    work done before explosion = work done after explosion

    work done after explosion = work done by 3lb + work done by 5 lb = 224000 J

    work done by 3lb= 3lb * 32 ft/s^2 * 875 ft = 84000 J

    work done by 5 lb = 224000 - 84000 = 140000 J

    work = integral of force done
    work done by 5 lb = integral of force in x direction + integral of force done in y direction

    where do I go from here?
     
  2. jcsd
  3. Apr 5, 2007 #2

    Hootenanny

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    Convervation of energy is not required for this question. Have you met any kinematic equations before?
     
  4. Apr 5, 2007 #3
    yes i have
     
  5. Apr 5, 2007 #4

    Hootenanny

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    So, lets consider this problem in horizontal and vertical components. List the knowns and unknowns in each direction.
     
  6. Apr 5, 2007 #5
    xcomp of 3 lb
    a = 0
    v= Vo
    s=Vot + So = 0

    Y comp of 3 lb
    a= g
    v= gt + Vo
    s= gt^2/2 + Vot + So

    xcomp of 5 lb
    a = 0
    v= Vo
    s=Vot + So = 0= Vot + 1659

    Y comp of 5 lb
    a= g
    v= gt + Vo
    s= gt^2/2 + Vot + So = 0 =gt^2/2 + Vot + 875
     
  7. Apr 5, 2007 #6

    Hootenanny

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    Looks good to me :biggrin:, there's no need to consider the path of the 3lb particle since we are not asked to calculate it in the question. One small correction for your 5lb y component;

    s= -gt^2/2 + Vot + So = 0 =-gt^2/2 + Vot + 875

    So now you have two equations with two unknowns, I'm sure you know how to proceed from here.
     
  8. Apr 5, 2007 #7
    Actually as said this is a momentum problem; so need to do all that ugly stuff =p. You're already past the "grunge" way of doing the problems.

    It's like the idea of a ball striking another ball and both going off at different angles. This is a two dimensional collision essentially.

    One piece will go off at angle gamma, the other will go off at angle beta.
     
  9. Apr 5, 2007 #8

    Hootenanny

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    I disagree, the problem is solvable if we consider it as a purely projectile motion problem, the fragment will behave as if it was a projectile given some initial horizontal velocity at some height above the ground [ignoring air resistance of course].
     
  10. Apr 5, 2007 #9
    You can do it that way; however, it is a grunt method. The key thing was that it is momentum, so he/she should be learning it through momentum.

    Such as you could go about projectile motion in how far something will land. Set it up into the x and y vectors, say that there's no acceleration in the x and do a lot. However, you can accomplish it through energy.

    Momentum is the key in this and how to apply it is like with 2 dimensional collision. Same idea applies.
     
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