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Classical Mechanics Kleppner Problem

  1. Jan 16, 2012 #1
    1. The problem statement, all variables and given/known data

    An Instrument carrying a projectile accidentally explodes at he top of its trajectory.The horizontal distance b/w the launch point and the point of explosion is L. The projectile breaks into 2 pieces which fly horizontally apart. The larger piece has three time the mass of the smaller one and returns to earth at the launching station. How far away does the larger piece land??

    2. Relevant equations



    3. The attempt at a solution
    My answer is 4L/3. Just want to verify if this is correct...
     
  2. jcsd
  3. Jan 16, 2012 #2
    it comes out as L+ L/3.
     
  4. Jan 16, 2012 #3
    Can anyone solve this please and verify my answer!!!!!!!!!!!
     
  5. Jan 16, 2012 #4

    Delphi51

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    Homework Helper

    Suppose there was no explosion. Then the whole mass would go another L, total of 2L before hitting the ground. The explosion increases the speed of the small piece, so it will go further than the 2L. So 4L/3 is too small.

    You do have the right approach, doing a conservation of momentum at the turnover point where the explosion takes place. I agree with your initial speed of L/t. Maybe show me your momentum calc for the one mv before the collision and two after the collision, with 2 of the 3 speeds known. That must be where your difficulty is.
     
  6. Jan 17, 2012 #5

    gneill

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    Staff: Mentor

    It may be profitable to think in terms of conservation of momentum for the center of mass of the system. Regardless of of what happens to the projectile, even if it explodes, the free-fall motion of its center of mass will remain the same! Where does that put the center of mass when the parts land? Where does that put the parts?
     
  7. Jan 17, 2012 #6
    An Instrument carrying a projectile accidentally explodes at he top of its trajectory.The horizontal distance b/w the launch point and the point of explosion is L. The projectile breaks into 2 pieces which fly horizontally apart. The larger piece has three time the mass of the smaller one and the smaller one returns to earth at the launching station. How far away does the larger piece land??
     
  8. Jan 17, 2012 #7

    gneill

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    Staff: Mentor

    It lands just as far as the first time the question was asked :smile:
     
  9. Jan 17, 2012 #8
    My apologies. I had made an error in typing. It is the smaller piece that returns back to the launching station.
    Thinking in terms of the conservation of momentum of center of mass. My answer seems correct. The centre of mass must be at the distance of L as it falls freely.

    But shouldnt it be at a distance of 2L????
     
  10. Jan 17, 2012 #9

    gneill

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    The center of mass will fallow the same trajectory as the projectile would have followed if it had not exploded.
     
  11. Jan 17, 2012 #10
    The smaller piece comes back top the launching station. Hence the
    V(s)/V(l)= 3/1

    IF V(s)xt= L then it implies that V(l)xt =L/3.
     
  12. Jan 17, 2012 #11
    wait let me show you the momentum conservation
    that I had applied
     
  13. Jan 17, 2012 #12
    V - velocity of unbroken piece.
    V(s)- velocity of smaller piece
    V(l)- Velocity of larger peice.
    t(f)- time of ascent or descent
    all velocities in x axis.

    before explosion momentum = MV
    after explosion momentum = M/4 x[V-V(s)]+ 3M/4[V(l)+V]

    equating and removing M and solving we get
    3V(l)= V(s)

    V(s)xt(f)= L

    therfore V(l)xt = L/3
     
  14. Jan 17, 2012 #13

    Delphi51

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    I used m as the mass of the small piece, 3m for large piece, total 4m.
    momentum before = momentum after
    MV = MV + MV
    4m*L/t = -m*L/t + 3m*v --> v = 5/3*L/t
    Very tricky; better check to ensure that this results in the landed center of mass being 2L as Gneil saw so cleverly!
     
  15. Jan 17, 2012 #14
    Yes, it does. the answer comes as 8L/3 and this gives the centr of mass cordintes as (2L,0) The trick is if we take assume velocities (like I did) it is important to define wrt to ground. The relation that i derived for the velocities is correct but it is wrt to each other.
    Hence

    V(s)xt is not equal to L rather [V(s)-V]x t = L

    the larger piece travels father than the smaller piece aided by the horizontal velocity making the velocity wrt ground of the larger piece as V + V(l).


    An excellent problem sheerly because of its simplicity and I must thank you guys for this excellent discussion. I feel that I have learnt a lot more than this problem aimed to teach.
     
  16. Jan 17, 2012 #15

    Delphi51

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    My sentiments, too! I must try to remember the center of mass approach.
     
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