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Impulse and average force

  1. Oct 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A ball of mass m moving with velocity [tex]v_{i}[/tex] strikes a vertical wall. The angle between the ball's initial velocity vector and the wall is [tex]\theta_{i}[/tex] as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between the ball and the wall is [tex]\Delta[/tex]t, and this collision is completely elastic. Friction is negligible, so the ball does not start spinning. In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis.


    2. Relevant equations

    What is the magnitude F of the average force exerted on the ball by the wall?

    How do you even start this question....? i dont know what its talking about. Any comments would be great
  2. jcsd
  3. Oct 13, 2007 #2


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    What is the definition of impulse?

    or rather...

    the average force = mass*average acceleration

    so what is the average acceleration...
    Last edited: Oct 13, 2007
  4. Oct 13, 2007 #3
    ok well impulse is the change in linear momentum so, F[tex]\Delta[/tex]t = m[tex]\Delta[/tex][tex]\stackrel{\rightarrow}{v}[/tex] but in this case it isnt linear....and im not sure what you mean by average acceleration. but the thing is i was told by my teacher, the answer does not depend on the variable [tex]\Delta[/tex][tex]\stackrel{\rightarrow}{p}[/tex] or m[tex]v_{i}[/tex]
  5. Oct 13, 2007 #4


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    Hmmm... I'm not sure how the answer can't depend on mvi...

    Momentum is linear... think of the momentum along the x-axis (perpendicular to the wall), and the momentum along the y-axis (parallel to the wall).

    [tex]F_x*(\Delta t) = mv_{xfinal} - mv_{xinitial}[/tex]

    [tex]F_y*(\Delta t) = mv_{yfinal} - mv_{yinitial}[/tex], but Fy is just 0 according to the question... so

    [tex]0 = mv_{yfinal} - mv_{yinitial}[/tex]

    You are also given that the collision is perfectly elastic... therefore the final kinetic energy = initial kinetic energy.

    so try to use these 3 equations (impulse in the x-direciton, impulse in the y-direction, conservation of kinetic energy)

    to solve for Fx.
  6. Oct 13, 2007 #5
    How come you need the three equations? why cant you just rearrange the Fx to get you

    [tex]F_{x}[/tex]= [tex]\frac{mv_{xfinal} - mv_{xinitial}}{(\Delta t)}[/tex]

    note: my mistake, the answer does depend on [tex]mv_{xi}[/tex] but it does not depend on [tex]mv_{xf}[/tex]

    so in light of this information, i guess i have to ask you how these 3 equations actually relate to one another.
    Last edited: Oct 13, 2007
  7. Oct 13, 2007 #6


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    To get rid of the extra variables... if you just use that equation, you have vf, vinitial, thetafinal, thetainitial...

    you should be able to write the final answer just in terms of vinitial and thetainitial...
  8. Oct 13, 2007 #7
    so is the final answer just F = m[tex]v_{ix}[/tex]/t
  9. Oct 13, 2007 #8


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    not quite... how did you get that?
  10. Oct 13, 2007 #9
    i dont know.....i dont know how to set the three equations equals to one another...
  11. Oct 13, 2007 #10


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    we'll start with this one:

    [tex]F_x = \frac{mv_{xf} - mv_{xi}}{(\Delta t)}[/tex]

    write vxf in terms vf and thetaf... write vxi in terms of vi and thetai... vf is the magnitude of the final velocity... vi is the magnitude of the initial velocity. vxf = -vfsin(thetaf) (because it is going towards the left)

    [tex]F_x = \frac{-mv_{f}sin(\theta_{f}) - mv_{i}sin(\theta_{i})}{(\Delta t)}[/tex] (1)

    KEfinal = KEinitial

    (1/2)mvf^2 = (1/2)mvi^2


    vf = vi.

    next equation:

    [tex]0 = mv_{yf} - mv_{yi}[/tex]


    [tex]0 = v_{yf} - v_{yi}[/tex]

    [tex]v_{yf} = v_{yi}[/tex]

    [tex]v_fcos(\theta_f) = v_icos(\theta_i)[/tex]

    using vf = vi from the kinetic energy conservation equation, we get:

    [tex]v_icos(\theta_i) = v_icos(\theta_i)[/tex]

    [tex]cos(\theta_f) = cos(\theta_i)[/tex]

    hence [tex]\theta_f = \theta_i[/tex]

    So the point of all this was to show that vf = vi and [tex]\theta_f = \theta_i[/tex]

    what do you get when you plug these 2 into (1)
  12. Oct 13, 2007 #11
    So....final answer is F = [tex]\frac{-2msin\theta_{i}v_{i}}{(\Delta t)}[/tex] ?
    Last edited: Oct 13, 2007
  13. Oct 13, 2007 #12


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    Yes, but since they only want the magnitude you'd leave off the minus in the final answer.
  14. Oct 13, 2007 #13
    thanks you thank you thank you
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