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Exploding projectile

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    A 12 kg shell is launched at an angle of 64 degrees above the horizontal with an initial speed of 120 m/s. When it is at its highest point, the shell exploded into two fragments, one three times the mass of the other. The two fragments reach the ground at the same time. You can ignore air resistance. The heavier fragment lands back at the original launching point.
    How far from the launch does the lighter fragment land?
    How much energy is released in the explosion?


    2. Relevant equations

    p=mv
    k=.5mv^2
    x=vt
    vf=vi+at

    3. The attempt at a solution

    V(horizontal)=52.604m/s
    V(Vertical)=107.885m/s

    KE(Vertical)= .5mv(vertical)= 69796.576
    KE(Horizontal)= .5mv(horizontal)= 16603.424

    KE(start) = PE(max height)
    69796.576 = mgh
    h = 593.5

    vf=vi + at
    t=11.006

    x = vt
    x = 578.941

    If b = mass of small piece
    b + 3b = 12
    b = 3
    small mass = 3
    big mass = 9

    KE(Horizontal) = .5mv^2
    16603.424=.5(3)v^2
    v=105.21

    x = vt
    x = 1157.93

    total x = 1157.93 + 578.941 = 1736.87m

    annnd its wrong :confused:
     
  2. jcsd
  3. Nov 6, 2011 #2
    The horizontal component of the velocity of the larger mass is equal in magnitude to the horizontal component of the velocity of the entire shell at launch. Since neither of the two fragments has any vertical component of velocity and momentum is conserved you can figure out the horizontal component of the smaller fragment's velocity.
     
  4. Nov 6, 2011 #3

    PeterO

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    The bits I highlighted in red could be the source of your error. For a projectile, the KE at the top is not zero, that is only for things fired vertically. The KE at the top is a minimum, but not zero. KE is not a vector, so does not have vertical and horizontal components. It is a scalar, so just has magnitude

    The PE at maximum height will be the difference between the KE at the start and the KE at the top.

    I have not really followed your calculations exactly, just the overlying assumtions you have made.

    EDIT: Note; once you have the right answer, I will show you the overview method I would have used to get it.
     
  5. Nov 6, 2011 #4
    i think i got the same answer from what you are saying

    you are saying that the larger mass, which is 9kg, have velocity of horizontal launch = 52.604

    so since the conservation of momentum is

    m1v1 + m2v2 = 0
    9(52.604) +3v2 = 0
    v2 = 157.812

    then to get the distance

    x = vt
    x = 157.812 (11.006)
    x = 1736.879m

    and is wrong :confused:
     
  6. Nov 6, 2011 #5

    PeterO

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    The larger mass was coming back!!!, so its velocity will be -52.604, and the other one will be going much faster.
     
  7. Nov 7, 2011 #6
    second try

    vf=vi + at
    t=11.006

    x = vit
    x = 578.947

    12 v(horizontal) = 3v(small) + 9v(large)
    12 (52.60) = 3v(small) + 9(52.60)
    v(small) = 52.6

    x = vt
    x = 578.9156

    x total = 1157.8626

    and wrong again.... :confused:
     
  8. Nov 7, 2011 #7
    m1v1 + m2v2 = 0
    9(-52.604) +3v2 = 0
    v2 = 157.812

    x = vt
    x = 157.812 (11.006)
    x = 1736.879m

    it was a typo for the first one but the result is the same
     
  9. Nov 7, 2011 #8

    PeterO

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    Sorry, didn't notice your eroneous zero.

    The momentum at the time of the explosion was not zero. the 12 kg mass was travelling at 52.504 m/s at the time!!
     
  10. Nov 7, 2011 #9
    m1v1 + m2v2 = mv
    9(-52.604) +3v2 = 12(52.604)
    v2 = 368.228

    x = vt
    x = 368.228(11.006)
    x = 4050.5m

    still wrong hahaha
     
  11. Nov 7, 2011 #10

    PeterO

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    Looks like you have worked out how far the 3 kg moved on after the explosion. It had already travelled quite a distance before the explosion so will presumably land further away that 4050.
     
  12. Nov 7, 2011 #11
    oooh yeaaa i forgot to add it to the original x -__-

    now how do i find the energy??

    .5mv^2 + mgh??
    .5(12)(52.604)^2 + 12(9.8)(578.947)
    84687.25J

    guess not
     
    Last edited: Nov 7, 2011
  13. Nov 7, 2011 #12

    PeterO

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    The explosion took place at a specific height - so potential doesn't come into it - pure kinetic.

    You know how fast the 12 kg was going before explosion 0.5 mv2 for that

    You know how fast each of the pieces were going after explosion. 0.5 mv2 a couple more times.

    Presumable the energy after is greater than the energy before → the energy from the explosion.
     
  14. Nov 7, 2011 #13
    you are the best man, thanks so much
     
  15. Nov 7, 2011 #14

    PeterO

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    Fine: Now my method.

    As you carefully calculated, this projectile takes 11 seconds to reach maximum height, covering a horizontal distance of 580m
    [these are rounded off figures for explanation purposes]

    In the explosion, the two parts [3/4 and 1/4 of the original] head off horizontally.

    the large part lands back where the original launch took place, so came back at the same speed as it reached the point.
    The velocity of that part changed by 2V [from V to -V] so the other mass, being 1/3 the mass will undergo a 6V change, reaching +7V.

    In the 11 seconds taken to get back to the ground, the 9kg travel the 580m [back to where it started]
    In the same time, the 3kg will travel 7 times as far [7 times as fast], so will finish a total of 8 times the original 580m → 4640m

    Energy wise.

    The 12 kg mass arrives at maximum height at speed V so has 1/2 x 12 x V2 of KE. [call that quantity X for the moment]

    3/4 of that mass travels at the same speed after explosion, so has 3X/4 units of energy.

    1/4 of that mass is travelling at 7 times that speed, so from a mass point of view will have 1/4 the enrgy, but from a speed point of view will have 49 times the energy, so a total of 49X/4.

    total energy is thus 13X

    Thus the explosion contributes 12X joules of energy

    X = 0.5 x 12 x 52.62

    You do the numbers - best done with the actual values with all their decimal places.

    NOTE: that if you can keep this, what I call global, view of the situation you can keep focussed on what is happening.

    I refer to it as living the problem to see the answer, and resorting to the calculator to get the second decimal place of the answer.

    Good Luck
     
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