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Impulse on ball involving angles

  1. Dec 2, 2008 #1
    1. A ball of mass 0.40kg is moving to the left at a velocity v0=-20 m/s. It is kicked and given a velocity, at 45 degrees upwards to the right, of vf30[/sub] m/s.

    1) Find the impulse of the force

    2) Find the average force

    The collision time is 0.010s



    2. I = (delta) P

    I = F (delta)t



    3.

    1) before p = 0.4x-20= -8
    after 30x0.4=12

    Change in p = 20Ns Where do the angles some in?

    2) F=(delta)P/t=2000

    Which is obviously wrong. They didn't include the angles for fun, so there something wrong in my working. Do I have to choose a line to work out the momentum across? ie, linear momentum. A bit confused. Any help appreciated.
     
    Last edited: Dec 2, 2008
  2. jcsd
  3. Dec 2, 2008 #2

    Doc Al

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    Momentum and velocity are vectors and must be treated as such.

    [tex]\Delta \vec{v} = \vec{v}_f - \vec{v}_i[/tex]

    Hint: Break the vectors into components, then subtract them. That's where the angles come in.
     
  4. Dec 2, 2008 #3
    OK, so, 30cos45=21.2

    21.2-(-20)=41.2

    correct?
     
  5. Dec 2, 2008 #4

    Doc Al

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    So far, so good. You've correctly found the change in the horizontal component of the velocity. What about the vertical component? (Then find the magnitude of the total change in velocity.)
     
  6. Dec 2, 2008 #5
    well, sin 45 and cos45 are the same, so the vertical change is just sin45=21.2

    So the total change in velocity is the total of the two, 21.2+41.2=62.4

    So no to find the change in momentum I times 62.4 by the mass to get 24.96 ? which is the impulse?
     
  7. Dec 2, 2008 #6

    Doc Al

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    OK, the change in the vertical component of velocity is 21.2 - 0 = 21.2.
    No, you can't just add horizontal and vertical components of a vector. How do you find the magnitude of a vector from its components?
     
  8. Dec 2, 2008 #7
    Pythagorus. So [tex]\sqrt{41.2^2+21.2^2}=46.3[/tex]

    so I = 46.3x0.4 = 18.5
     
  9. Dec 2, 2008 #8

    Doc Al

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    Good. (Impulse will have the same units as momentum, thus kg-m/s.)
     
  10. Dec 2, 2008 #9
    And my working out tghe average force will be correct with this new value? ie,

    F=18.5/0.01=1850N.

    Thanks for you help btw. This has got to be handed in in 20 minutes! Worth 25 marks too. Silly really for how easy it is, once you've got started, i'm just too hungover to think properly and needed a nudge to get going. Not even sure there are 25 stages in the question to mark.
     
    Last edited: Dec 2, 2008
  11. Dec 2, 2008 #10

    Doc Al

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    Staff: Mentor

    Looks good.
     
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