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Momentum of falling objects on impact; p = mv not working?

  1. Jun 26, 2007 #1

    exi

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    Got the first part; the second part of the question asks what angle of inclination the bale will strike, given between -180° and 180°. Drawn out, this seems to resemble a right triangle with a 65 m height, 200.4188 m length (initial X velocity = 55 m/s, time of 3.6422 s), and the inverse tangent of that is 17.970° - yet neither that nor ~162° seem to work. Why not?

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

    A plane, flying due east at 55 m/s, drops a bale of hay from an altitude of 65m.

    Acceleration due to gravity: 9.81 m/s².

    If the bale of hay weighs 173 N, what's the momentum of the bale the moment it strikes the ground? Answer in kg∙m/s.

    3. The attempt at a solution

    Since momentum is just the product of mass and velocity, what I tried to do is consider y-axis movement only; that is,

    [tex]V_0 = 0 \frac{m}{s}[/tex]

    [tex]a = 9.8 \frac{m}{s^2}[/tex]

    [tex]\Delta y = 65 m , \mbox so:[/tex]

    [tex]v^2 = v_0^2 + 2a\Delta x[/tex]

    Solving that, velocity comes out to 35.711 m/s. Since the bale's weight is 173 N, then 173 / 9.8 = 17.653 kg. Multiplying that weight by the above velocity would seem to me that it should produce the correct answer, but it does not; why?
     
    Last edited: Jun 26, 2007
  2. jcsd
  3. Jun 26, 2007 #2
    because momentum is a vector and you've only considered the y component. you have to consider both and then find the resultant vector
     
  4. Jun 26, 2007 #3

    exi

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    My mistake; got it now.

    Thanks for the reminder.
     
  5. Jun 26, 2007 #4

    exi

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    Edited original post to try and figure out what I'm screwing up now.
     
  6. Jun 27, 2007 #5

    Astronuc

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    Staff: Mentor

    Did you get the answer?

    momentum and velocity are vectors.

    In Cartesian coordinates, [tex]v\,=\,\sqrt{v_x^2\,+\,v_y^2\,+\,v_z^2}[/tex], where v is the speed or magnitude of the velocity, and then momentum is just the product of the mass (scalar) and velocity (vector).

    In the OP, just use vx and vy.
     
    Last edited: Jun 27, 2007
  7. Jun 27, 2007 #6

    exi

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    Yes, I did - it's the angle of impact that's got me now.
     
  8. Jun 27, 2007 #7
    [tex]tan(\theta)=\frac{opp}{adj}[/tex]
     
  9. Jun 27, 2007 #8

    Dick

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    Science Advisor
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    True. But apply this to the final velocity vector itself. The path of the bale (approximate hypotenuse) of the triangle you are looking at is actually curved.
     
  10. Jun 27, 2007 #9

    Astronuc

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    Staff: Mentor

    One has vx and vy

    So, what does one do with vy / vx with components measured at impact.
     
  11. Jun 27, 2007 #10

    exi

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    Not sure what you're getting at. Using a straight line for the approximate trajectory of the bale would seem to suffice, and since I had what I thought to be both vertical and horizontal displacement, it would seem to me that taking the tangent of those figures would've done it - but it doesn't.
     
  12. Jun 27, 2007 #11

    Doc Al

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    Staff: Mentor

    Treating the trajectory as a straight line would a lousy and unnecessary approximation. Forget about displacement. Focus on the velocity at impact. (Reread Astronuc's hints.)
     
  13. Jun 27, 2007 #12

    exi

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    If only I were graded on my ability to needlessly complicate things.

    [tex]65.5761cos\theta = 55[/tex]

    [tex]\theta = cos^{-1}\frac{55}{65.5761} = 32.994^{\circ}[/tex]

    Much appreciated.
     
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