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Collision Physics 1 problem with distances and angles

  1. Jul 22, 2012 #1
    1. The problem statement, all variables and given/known data

    A 3000-kg Cessna airplane flying north at v1 = 70 m/s at an altitude of 1750 m over the jungles of Brazil collided with a 7000-kg cargo plane flying at an angle of θ = 34° north of west with speed v2 = 111 m/s. As measured from a point on the ground directly below the collision, the Cessna wreckage was found 1000 m away at an angle of 25° south of west, as shown in the figure. The cargo plane broke into two pieces. Rescuers located a 4000-kg piece 1800 m away from the same point at an angle of 22° east of north. Where should they look for the other piece of the cargo plane? Give a distance and a direction from the point directly below the collision. (Let to the east be the +x-direction and to the north be the +y-direction.)

    http://www.webassign.net/bauerphys1/7-p-064a-alt.gif
    http://www.webassign.net/bauerphys1/7-p-064b.gif




    2. Relevant equations

    P0=P.
    Px=7000kg(111 m/s) cos34
    Py=3000kg(70m/s)+7000kg(111 m/s)sin34

    P=mv

    Perhaps kinematics to find distance.


    3. The attempt at a solution

    px=7000(111)cos34=644,162.2 kg m/s
    py=7000(111)sin34+(70*3000)=644,492.9 kg m/s
    P=sqrt (px^2+py^2)=911216.8 kg m/s

    Now here's where I'm grasping for straws:

    Py/Px=1.00. arctan (1.00)==45

    Final angles: 22+25+90=137.

    Tan(137)=-.933.
    1.00x=.933=>x=.933

    .933*911216.8=849718.8.
    849718.8/ (mass of third piece=3000kg)=283.24 m/s

    Tried kinematics to get a final distance traveled, but to no avail.
    For the angle: 1.0-.93=.07. arctan.07=4 degrees. which was incorrect.

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 22, 2012 #2

    TSny

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    Hi jamesbiomed. Welcome.

    [EDIT: The problem wants you to take west as the negative x direcion.]

    :uhh: Don't think you grabbed even one straw there .

    You might explain how you set up your kinematics.

    So far you've avoided the most important principle that should be invoked in a collision problem.
     
    Last edited: Jul 22, 2012
  4. Jul 22, 2012 #3
    Conservation of momentum (p0=p). I totalled P0. But with distances and no times, there are how can you total final momentum?

    Kinematics I tried were vy^2=(0)-g(0-alt). This could give me the final velocity in the y-direction--That in turn could give me the time through vy=vy0-gt.

    If I knew the velocity in the x-direction, than the distance would be doable. If I initial velocity (v (terminal) in this case) then I could separate vy, get vx (which would be constant) and then get the distance that way.

    So the gap in my efforts is in going from the theory that momentum is conserved into appropriating the velocities, and finding these angles. Do you understand what I'm saying?
     
  5. Jul 22, 2012 #4

    TSny

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    OK, good. Conservation of momentum is central and you can get the time of free fall.

    You are given the horizontal distance traveled by two of the pieces. Can you use this info along with the time to get the horizontal speed of these pieces right after the collision?
     
  6. Jul 22, 2012 #5
    yes, actually I believe I can. Let me try that :)
     
  7. Jul 22, 2012 #6
    So I'm doing that, and getting all the initial velocities, which look reasonable. Then I'm multiplying those initial x velocities (vy0=0) by the mass and using the conservation of momentum to solve for the momentum for the final object.

    P0=Mcessna*vxcessna+ Mpiece*vxpiece+Ppeice2

    Then I can divide by the remaining mass to get its initial velocity, then use that with x=x0 + vx0*t to get the distance.

    Unfortunately, that distance value is coming out incorrect. One a practice version, I'm off by a factor of a few thousand. (I've double checked the math)

    I guess my question is, is my method correct so far? Do I need to take direction into account when multiplying their initial velocities?

    Thanks a lot for the help so far !
     
  8. Jul 22, 2012 #7

    TSny

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    Well, lets suppose you want to find the speed of the Cessna immediately after the collision (still 1750 m high). Use the horizontal distance the Cessna travels while falling and the time of fall. What do you get for the speed of the Cessna immediately after the collision?

    [Sorry, I meant to add that your method looks correct.]
     
    Last edited: Jul 22, 2012
  9. Jul 22, 2012 #8

    TSny

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    Work with the x and y components separately.

    The x-component of the total momentum must be conserved in the collision.

    Likewise for the y-component.

    Thus, you an find the x and y components of the momentum of the second piece after the collision.
     
  10. Jul 22, 2012 #9
    No problem. Sorry for the late reply also, I see you're answering almost immediately. I keep checking but it seems delayed to show up...anyway--

    For the Cessna:

    first I need time:

    v^2=v0^2-(2)9.81(0-1750)=> Vf=185.30
    185.30=0-9.81t=> (ignoring negative) t=18.89 s

    Cessna lands 1000 m away: 1000=0+vx0t. =1000/18.89=vx=> 52.94 m/s

    again, sorry for the delay. hope you're still on. I'll keep the page up this time and be more timely if so
     
  11. Jul 22, 2012 #10

    TSny

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    Good. That's what I got. Can you now get the x and y components of the momentum of the Cessna after the collision?
     
  12. Jul 22, 2012 #11
    Well, total momentum should be (52.94*3000kg)=158826.1 kg m/s

    Do you mean x and y as in the horizontal plane the planes were originally flying on? Because in that case, no, I don't know how to do that.

    The problem I'd have would be how to divide the initial momentum into x and y.
     
  13. Jul 22, 2012 #12

    TSny

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    Yes, the horizontal plane: x axis points east, y axis points north.

    You can get the x and y components of the momentum using the angle that specifies the direction of the momentum in the x-y plane.
     
  14. Jul 22, 2012 #13
    On the vertical plane, then momentum in the x-direction would be the total momentum, and in the y-direction would be zero.

    But I think you mean horizontal, and since the problem for the angles, it seems there must be a way.

    Since I know the original x and y momentum's can that help me?
     
  15. Jul 22, 2012 #14
    sorry, didn't see that. Let me read that first:)
     
  16. Jul 22, 2012 #15
    The direction of the total momentum? Or x and y's seperately?

    (sorry, this is not intuitive yet)

    I know 34 degrees is the direction of part of the inital momentum and then 90 degrees to the horizontal is the direction of the other part. Could that help?
     
  17. Jul 22, 2012 #16

    TSny

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    Yes. But I thought we were working with the momentum of the Cessna immediately after the collision. You found the magnitude of this momentum. Now you need the x and y components of this momentum.

    [You'll need to go to page 2 to see further comments.]
     
    Last edited: Jul 22, 2012
  18. Jul 22, 2012 #17
    would arctan of the momentum give me the angle?
     
  19. Jul 22, 2012 #18
    Would the arctan of the momentum give me the angle?
     
  20. Jul 22, 2012 #19

    TSny

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  21. Jul 22, 2012 #20
    about 45 degrees north West?
     
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