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Particle Travels Between Two Walls

  1. Sep 5, 2009 #1
    Hi all, I am a long time reader but first time poster. I figure I would start posting my homework problems because I seem to have a bunch of them and I do not have the ability to see my professor on a daily basis. Thanks for any future help.

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

    A particle travels horizontally between two parallel walls separated by 18.4 m. It moves
    toward the opposing wall at a constant rate of 8.7 m/s. Also, it has an acceleration in the
    direction parallel to the walls of 2.7 m/s^2.

    Part 1 - What will be its speed when it hits the opposing wall?
    Part 2 - At what angle with the wall will the particle strike?

    2. Relevant equations

    I first tried the constant acceleration equations but then I re-read the question to see that the particle does not have a constant acceleration in the Y-direction.

    Vf = Vi + at

    3. The attempt at a solution

    After twenty minutes or so I tried this:

    v^2 - u^2 = 2aS

    where u = 8.7 and S = 18.4

    v^2 = u^2 + 2aS
    v = (u^2 + 2aS)^(1/2)
    v = (8.7^2 + (2)(2.7)(18.4))^(1/2)
    v = 13.23 m/s

    I found this equation in my notes I took during class but wasn't sure if it was the right one to use or not.

    I did not attempt part two.

    Thanks for any help,
    Ben
     
    Last edited: Sep 5, 2009
  2. jcsd
  3. Sep 5, 2009 #2

    Doc Al

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    Staff: Mentor

    Lets call the horizontal direction the x-axis. The way I understand the problem is that you have a particle that starts at one wall with a velocity in the x-direction. It experiences an acceleration in the y-direction. Hint: Treat the x and y motions separately.

    How long does it take to reach the opposite wall?
    What will be the y-component of its velocity at that time? The x-component?
     
  4. Sep 5, 2009 #3
    Thanks for the reply Doc Al.

    How long does it take to reach the opposite wall?

    (18.4 m) / (8.7 m/s) = 2.11 s

    What will be the y-component of its velocity at that time? The x-component?

    I am unsure about the components part but here is what I think:

    x-component = 18.4
    y-component = (2.11) x (2.7) = 5.697
     
  5. Sep 5, 2009 #4
    hmm personally i would find out the angle by seeing how far it's gone in the x and y direction between the hits, also remember though that it's a constant acceleration, not constant velocity,so i would use this equation

    x = ut + 0.5at²

    then it's geometry

    Edit:also where did you get that 2.7 from? 0.o, you've said earlier the acceleration was 2.3
     
    Last edited: Sep 5, 2009
  6. Sep 5, 2009 #5

    Doc Al

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    Staff: Mentor

    Good.

    Careful! Only the y-component is accelerated.
    Is the acceleration 2.7 or 2.3 m/s^2?
     
  7. Sep 5, 2009 #6

    Doc Al

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    Staff: Mentor

    What determines the angle is its velocity when it hits, not where it hits the wall.
     
  8. Sep 5, 2009 #7
    Whoops, right.

    In that case, the x-component would be 8.7 m/s.

    I typed the problem wrong, the actual acceleration is 2.7 m/s^2; I will fix the original post.

    Just to be clear I believe the y-component is (2.11) x (2.7) = 5.697
     
  9. Sep 5, 2009 #8

    Doc Al

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    Staff: Mentor

    Looks good.
     
  10. Sep 6, 2009 #9
    I solved both parts last night, thanks for all the help.
     
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