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Momentum and collisions for a fired projectile

  1. Apr 23, 2010 #1
    1. The problem statement, all variables and given/known data
    2. A 20.0 kg projectile is fired at an angle of 60⁰ above the horizontal with a speed of 80.0 ms-1. At the highest point of its trajectory, the projectile explodes into two fragments of equal mass, one of which falls vertically with zero initial speed.
    Ignoring air resistance, a) how far from the point of firing does the other fragment strike, if the terrain is level? (5 marks)

    hey i just need someone to confirm my answer of 1334 metres. part two of the question asked for the energy released during the explosion and i have no idea how to find this.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 23, 2010 #2
    Are you sure you're using the right formulas? I get 848m as my final answer. First I calculated the vertical component of the velocity. From this I calculated time to reach the top. Then using that time I multiplied by the horizontal constant component to get horizontal distance to that first point. Then, since the projectile splits in half, the second fragment has twice the horizontal speed. Thus it goes twice as far. So I multiply my initial first distance by 3. Hopefully someone else can confirm.

    As for part two, you would just compare the energy before and after the collision using 1/2mv^2.
     
  4. Apr 24, 2010 #3
    hey for the horizontal distance before the explosion did you get 282.5m. why is the speed after the explosion just 2 times the initial speed??? i used conservation of momentum to find the final speed and got 80*sqrt(3). please reply i need the right answer soonish... thanks.
     
  5. Apr 24, 2010 #4
    Yes I got 282.5m before the explosion. Let m1 be the mass before the explosion and v1 be its velocity. Let m2, m3, v2, and v3 be the masses of the parts after the explosion. So, using conservation of momentum, m1v1 = m2v2 + m3v3. Since v2 is zero, we get m1v1 = m3v3. Since m3 is one half v1, we get m1v1 = m1v3/2. From this, we get v3 = 2v1. It's exactly double the initial horizontal velocity. Since it falls the same length, time is the same, and from this we get distance. Not quite sure how you got 80*sqrt(3).
     
  6. Apr 25, 2010 #5
    hey, the initial velocity(just before the explosion is) 80cos60 so 2 times that is 80. (the sqrt(3) was a mistake sorry). i was doing exactly what you said only i didnt realise that it ended up being 2* the initial... Anyway long story short i now get 850.5m do you think that is right?? (i got t=7.1s is that what you got?.) thanks for all your help too.
     
  7. Apr 25, 2010 #6
    sorry, one more thing is the energy released = 16000J?
     
  8. Apr 25, 2010 #7
    850.5m sounds about right but it looks like you might have rounded off a bit early. 16000J sounds good to me.
     
  9. Apr 25, 2010 #8
    thanks ill check back over my work and see where the error is.
     
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