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Finding velocity on flat part of ramp

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Greetings, I'm new here :).
    Straight to my problem. I have a ramp. A ball accelerates down at varying angles, and then it reaches the flat part and travels only in the x-direction. I need to find the velocity of the ball as it moves on this flat part, as it is no longer accelerating.


    2. Relevant equations
    suvat, trigonometry


    3. The attempt at a solution
    First, I did this: 9.81×sin[itex]\theta[/itex]. This gives me the acceleration of the ball down the ramp at whatever angle.
    (9.81×sin[itex]\theta[/itex])cos[itex]\theta[/itex] = x component of acceleration down ramp.

    s(length of ramp)=1/2at^2
    t= [itex]\sqrt{}2s/a[/itex]

    By the time ball reaches flat part, vx=axt?
    Here is an extremely simplistic model of the situation.
    Ramp_zps928a0a9c.jpg
     
  2. jcsd
  3. Sep 23, 2012 #2

    BruceW

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    Homework Helper

    Hi, Fieldguise, welcome to physicsforums :) Your working all looks good so far. I would agree that just before the ball reaches the flat part, the horizontal speed is given by the x component of acceleration down the ramp, multiplied by the time it has been moving. But when the ball actually gets to the flat part of the ramp, what will happen to its speed? will its vertical motion be turned into horizontal motion, or will the vertical motion simply be lost and the energy given off as sound, etc ?
     
  4. Sep 23, 2012 #3
    Yes, exactly. I don't how to determine how the y-component of the acceleration adds in to the final horizontal velocity. The ball is smooth steel against a plastic ramp. Friction is small so the vertical motion definitely doesn't just disappear.

    In some way, the vertical motion turns into horizontal motion. For example, when I angled the ramp steeply at about 75°, there was little acceleration in the x-direction, yet the ball still traveled further when it left the ramp than when the ramp was angled at 45°, at which point there was MORE acceleration in the x direction. So obviously the high vertical component of the acceleration at 75° added in to the horizontal motion, I just don't know how to calculate it exactly, as I need an actual theoretical value.
     
  5. Sep 23, 2012 #4
    Here's an experiment I'd suggest: Roll the ball for various inclinations of the ramp but keep the vertical height of the ramp fixed for all inclinations. What do you notice about the final velocity of the ball?
     
  6. Sep 23, 2012 #5
    Should the velocity be the same each time? I can't do further testing as this was lab work. Keeping the vertical height fixed means I would have to change the length of the ramp for each test angle, and that is too many independent variables.
     
  7. Sep 23, 2012 #6
    Yes! What does this suggest for the final velocity? Does it depend on length of ramp? inclination of ramp? vertical height?

    PS - Have you studied the concepts of work, energy and conservation of energy?
     
  8. Sep 24, 2012 #7
    Yes, I have studied those concepts.

    So.. if the final velocity is constant when the vertical height of the ramp is also constant, it simply depends on the gravitational potential?

    Then, by law of conservation of energy, the potential energy of the ball turns into kinetic energy. So I can use:

    gΔh = 1/2v^2 , and solve for v? and I can ignore mass because I'm not actually finding energy?
     
  9. Sep 24, 2012 #8
    Yes :smile:

    I'm not sure what exactly your question was, but if that's what you intended to find, then good!
     
  10. Sep 24, 2012 #9
    Awesome!! It isn't necessary to explain my whole procedure, but yea, I just needed to find the velocity on that part of the ramp.

    Honestly, thank you a lot! I wasn't completely sure how to solve it and you helped me use logic to find the solution.
     
  11. Sep 24, 2012 #10
    No problem!
     
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