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Explosion and momentum

  1. Dec 30, 2016 #1
    1. The problem statement, all variables and given/known data
    A rocket is fired from from the ground at initial velocity of ##v_0## and at an angle ##\theta##. At its highest height it splits into 2 parts of equal masses. The first part is fired straight up and at velocity ##v_0/2##. Find the angle and intensity of the second part.
    2. Relevant equations
    3. The attempt at a solution

    At its maximum height the velocity has only the x component which is ##v_0cos\theta##. The momentum is ##p_0=mv_0cos\theta##. Since it has only the x component and the part which divides has the y component the second part must have an equal y component. Mening the ##v_{2y}=v_{1y}=v_0/2##. That would make the change in the y component of the momentum ##0##. Since that starting condition has total momentum in x direction the ##v_2## part must have the x component equal to the ##v_0cos\theta##. The intensity would then be ##\sqrt{v_{2y}^2+v_{2x}^2}##. Does this seem right couse when i use it in the problem given by the books where this intial velocity is ##1000## the angle ##60## i get ##v_y=500\sqrt{2}## and the book gives ##v_y=1118##?
     
    Last edited: Dec 30, 2016
  2. jcsd
  3. Dec 30, 2016 #2
    When the problem says that the first part is fired straight up, I think it means straight up relative to someone on the ground - not straight up relative to the moving projectile.
     
  4. Dec 30, 2016 #3

    Merlin3189

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    Gold Member

    Your method sounds correct in general, though I get a different answer from you when using that approach. I think the book is correct.
    Perhaps you could show your working in detail.
    I think you may be going wrong in the horizontal component, what you label v2y
     
  5. Dec 30, 2016 #4
    At the max heigth the velocity has only the x-component which is ##v_0cos\theta=500m/s##. Since it then splits into two parts they both must have 0 net y velocity and 500m/s in the x-component. ##v_{1y}=500m/s## and ##v_{1x}=0##. From there it follows that the vertical component of the second velocity must be ##500m/s## in order for the y-components to cancel out.
    The x component must be equal to ##m*500m/s##. So ##\frac{m}{2}v_{2x}=m500##. So it should be ##1000m/s## Now i get it right. I forgot the fact that the mass splits so i thought that i have full mass of velocity 2. Good now.
     
  6. Dec 30, 2016 #5

    Merlin3189

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    Gold Member

    Exactly where I went wrong myself at first attempt!
     
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