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Need help (not answer just help) on a problem

  1. Nov 1, 2005 #1
    Alright, I am asking for help on how to start this problem. I did all the other homework problems, but this ones is confusing me. Maybe I am just stupid or something. Anyways, any help would be appreciated, I don't know where to start and have tried a lot of things (spent about an hour just on this problem).

    A shell is shot with an initial velocity Vo of 20 m/s, at an angle of 60 degrees with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrainis level and air drag is negligible?

    It deals with Newton's Second Law for a System of Particles and the answer is 53m. Please help. Thank you for your time.
  2. jcsd
  3. Nov 1, 2005 #2
    This question can be done quite easliy if you apply conservation of momentum for when the shell splits into 2 pieces, you assume that both pieces are of equal mass, that means that one of the pieces will have 0 momentum while the other has all the momentum equal to the shell prior to exploding. Thus you can calculate the horizontal component of velocity of the exploded shell and work out the time it takes for it to reach the ground, thus the horizontal distance travelled would just be the horizontal component of velocity multiplied by the time taken for it to fall to the ground. Then add this to the horizontal distance travelled by the shell the before exploding and you got 53m.
  4. Nov 1, 2005 #3
    as you mentioned in the question the shell split in to 2 pieces , in order to caculate what the question wants you , you should equalize to equation 1)all the kenetic energy that shell has shot (which means E=1/2 MV^2)
    2)second equation of the time when shell explode into 2 equal peices (which is (1/2*1/2MVx^2)+(1/2MGH)where His the elevation of trajectory
    by equlizing these 2 equation you will get to the quantiy of Vx
    after that it is like a shell which is shot in horizental way with the initial speed of Vx at the elavation of H
    so you can easily get the distance where the second half of the shell will land
    Last edited: Nov 1, 2005
  5. Nov 1, 2005 #4
    Answer tot your problem

    20 m/s vertical speed is 20m/s x sin(60) = 17,36 m/s
    white a = 9,82 m/s2 the time until the top is reached is
    17,36 / 9,82 = 1,7638 s

    s = v x t
    avarage speed = 20 : 2 = 10 m/s
    s = 10 m/s x 1,7638 = 17,638 m to the top

    De total distance of the complete particle wil be 2 x 17,638 m

    Because de total mass x everage speed = sum of mass x average speed of each particele
    If one particle fals down the speed = zero so
    we can concude that
    de mass is 1/2 de speed must be twice as mutch
    conclusion de distance is alse twice as much

    17,638 m for the top en 2 x 17,638 m after the top
    s = 52,91 m

    Sie olso Giancoli Center of mass and Translational Motion
  6. Nov 1, 2005 #5
    shell exploding at the maximum height

    I feel that since one fragment falls vertically its reversemementum is eaual to the forward momentum of the second fragment. coresponding velocity is equal to the horizontal component of the velocity of the intial projectile. Asuming that the fragment falls vertically on gravity

  7. Nov 1, 2005 #6
    Thank you all so much for the help. I really appreciate you all for bearing with me. I guess I just have a hard time with trajectory problems or something. I'll learn eventually; practice makes perfect!
  8. Nov 5, 2005 #7
    Solution to the problem

    Consider the motion of the particle on the topmost point.The vertical component of the particle's velocity vanishes ,only the horizontal component remains and that is equal to 10m/s(since the horizontal component remains the same throughout the motion).Now when the particle breaks down into 2 pieces one of it has 0 velocity.Calculate the velocity of the other piece by the equation of the centre of mass.
    The velocity of the second particle comes out to be 20m/s.Now shift the origin to the topmost point.Now you know the horizontal component of the particle,just calculate its range and add it to half the range of the motion if the particle would not have broken down.The answer cones out to be 53 meters.
  9. Feb 1, 2006 #8

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