Physical Science Question

Discussion in '9/11' started by Katzenjammer, May 24, 2016.

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  1. Katzenjammer

    Katzenjammer New Member

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    Greetings from the new kid on the block

    I have a question: ( I've already done some work on this at the Physics 101 level )
    Case of a physical object traveling at 240 meters/sec and having a mass of 100,000 kg
    meeting a stationary object of 2,000 kg ( inelastic collision )

    My concern here is for what happens to the original object that was traveling at 240 m/s
    that is what consequences to this collision would there be ( obviously loss of speed, however, how much,
    and over what span of time )

    Any help for a marginal student? Thanks in advance.
     
  2. contrails

    contrails Active Member

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    You don't have enough information here to calculate anything about the consequences to this collision. Here is a link that might help you out though.

    http://www.real-world-physics-problems.com/inelastic-collision.html
     
  3. Katzenjammer

    Katzenjammer New Member

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    On the alleged insufficient data angle,

    There is sufficient data to know that there would be some tonnage of force applied to an airliner striking a bird at say 300 mph,
    So what would be the consequences of striking something considerably more massive than a bird?
    My estimate is that the airliner would experience a >100 g jolt and that would be rather serious, however I wanted to run it by
    some other people to double check my work. are there any serious academics on the interwebz?
     
  4. perdidochas

    perdidochas Well-Known Member

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    Not enough information to determine span of time, at least based on the physics I know (Physics 101 and 102).

    Inelastic collision--based on perfect conditions--no friction, no other loss of energy, etc.
    m1 mass of first object, 100,000kg
    v1 velocity of first object, 240 m/s
    m2 mass of second object
    v2 velocity of second object
    v, final velocity combined

    v=(m1v1+m2v2)/(m1+m2)= (100,000kgx240 m/s + 2000kgx0m/s)/(100,000kg +2000kg)=(24,000,000kgxm/s + 0)/(102,000kg)=(24,000,000kgxm/s)/(102,000kg)=235.29 m/s would be the final velocity for both objects moving together.
     
  5. rahl

    rahl Banned

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    why does this feel like it's going to be a truther thread?
     
    DarkDaimon likes this.
  6. perdidochas

    perdidochas Well-Known Member

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    Assuming ideal conditions (no friction, etc.), you can calculate the new velocity of the system--which will be about 235 m/s.
     
  7. perdidochas

    perdidochas Well-Known Member

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    There are too many unknown variables to calculate--composition of objects, etc. Also, the main damage to a plane by a projectile is the result of the parts of the plane that are destroyed.
     
  8. Katzenjammer

    Katzenjammer New Member

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    Given the example of an aircraft, the wing mounted jet engines would experience >100 X their mass as stress upon the mounts
    and guaranteed for certain rip away in <35 milliseconds, and the engines would rotate / tumble as they disconnected.
    anybody have any alternate theory?
     
  9. contrails

    contrails Active Member

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    Wouldn't those assumptions make the collision elastic? The OP specifically asked about an inelastic collision.
     
  10. contrails

    contrails Active Member

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    The amount of force experienced in any collision is proportional to the mass of the other object. While I don't doubt that the bird would see > 100g in such a situation, obliterating it immediately, the aircraft would barely even notice it. US Airways Flight 1549 was going around 200 mph when it struck several birds before ditching in the Hudson River back in 2009. And while it cause both engines to fail, there was very little force ( < 9g) applied to the airframe.

    http://www.exosphere3d.com/pubwww/pages/project_gallery/cactus_1549_hudson_river.html
     
  11. perdidochas

    perdidochas Well-Known Member

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    No, that was inelastic. Think about a bullet hitting a ball, without friction. That's inelastic. In an elastic collision we have to assume that each part goes a different direction, and it gets much more complicated.

    Here are the equations I used.
    http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html#c1
     
  12. contrails

    contrails Active Member

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    My bad. I though by keeping kinetic energy constant you were making it elastic as opposed to an extremely inelastic collision. Still, without more data, calculating things like g-forces or acceleration time is not possible.
     
  13. Hoosier8

    Hoosier8 Well-Known Member Past Donor

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    There is a computer simulation of an aircraft striking the tower at MIT I believe. Basically an airliner would shred like paper in a shredder. Steal also deforms due to the forces applied.
     
  14. Katzenjammer

    Katzenjammer New Member

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    So you compare striking a few birds to having to encounter a minimum of 3 tons of mass ...... right?
     
  15. Katzenjammer

    Katzenjammer New Member

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    Elastic or inelastic collision, we are talking about having to displace no less that 3 tons of mass
    how is that done in a seriously short time without catastrophic effects upon the aircraft?
     
  16. contrails

    contrails Active Member

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    What happens when you apply equal forces to opposite sides of a solid object?
     
  17. Katzenjammer

    Katzenjammer New Member

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    The solid object compresses, and depending upon the composition of the "solid object"
    the is movement or not, also dependent on the force applied.

    however, relative to the force applied, the laws of physics will remain in force.
    that is there will be equal and opposite re-action to said forces.

    are we in sync here?
     
  18. Katzenjammer

    Katzenjammer New Member

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    BTW: there is NO computer simulation of any of the events of 9/11/2001 where the source data has been published
    and without any sort of foundation, the cartoons produced by these "simulations" are totally useless.
     
  19. contrails

    contrails Active Member

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    Solids are generally defined by the property of incompressibility, meaning they don't compress when force is applied to them. Yes, if you apply enough force, the solid eventually yields, but for standard aircraft aluminum that occurs at around 25 tons per square inch, not 3 tons over the whole aircraft.
     
  20. Hoosier8

    Hoosier8 Well-Known Member Past Donor

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    They are based on physics so I guess you deny that they can simulate anything.
     
  21. Alien Traveler

    Alien Traveler New Member

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    Actually, solid means not gaseous, liquid or plasma. Rubber is solid. Children clay is solid.
     
  22. Hoosier8

    Hoosier8 Well-Known Member Past Donor

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    6160 aluminum used in aircraft has a maximum tensile strength no more than 120 MPa (18,000 psi), and maximum yield strength no more than 55 MPa (8,000 psi). That does not include deformation of sheets once the structure is compromised. Doing a simple mass calculation based in weight and speed, the aircraft would have imparted about 68.5 thousand tons upon impact.
     
  23. contrails

    contrails Active Member

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    I was using numbers for 2024-T3 aluminum which has a maximum tensile strength around 60 MPa and a maximum yield strength around 40 MPa. Then there is 6061-T6 which gets around 290 MPa ultimate and 240 MPa yield. Either way you go, 3 tons of pressure from displacing air is nothing.
     
  24. Alien Traveler

    Alien Traveler New Member

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    However you can easily dent a wing of an aircraft with a hammer. Even a child can.
     
  25. Katzenjammer

    Katzenjammer New Member

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    You can guess all you want but proper simulations publish the source data so that the simulation can be verified as being proper,
    Who is to say that the parameters that were set up in producing the famous simulation cartoons were correct in regards to the
    strength of the steel in the tower wall and the failure modes displayed by the aluminum aircraft?

    Real physics takes into account the energy required to perform a given action
    like that of having to move tons of mass in order to make penetration into the tower.
     

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