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Homework Help: Simple particle on slope confusion

  1. Jun 5, 2015 #1
    1. The problem statement, all variables and given/known data
    a particle on a slope with angle theta with no friction. v(0) = 0, x(0) = 0, with coordinates i down the slope and j normal to it.

    I am confused about why with the constant velocity forumla I get a different answer to my attempted method.. I cant see whats wrong..

    I need to find the velocity at "l"
    x is the top of the slope of a particle on a smooth surface, with no friction,
    v(0) = 0, x(0) = 0

    along i direction I am starting with:

    mgsin(theta) == ma

    gsin(theta) == a

    using constant acceleration formula

    v^2 = v0^2 + 2a0(x-x0)
    v^2 = 0 + 2gsin(theta)(x-0)
    v = sqrt( 2gl*sin(theta) )

    My original attempt below is wrong, but I cant see why. I want to know what it doesn't work the same.

    so from
    gsin(theta) == a

    Integrating wrt t

    gsin(theta)t == v + c

    Integrating wrt t again

    1/2*gsin(theta)t^2 == x + ct + d

    with v(0) == 0 and x(0) == 0

    0 = c and d = 0

    so if I now plugged in 'l' to the equation for position

    1/2*gsin(theta)t^2 == l

    and solve for t I get,

    t = sqrt[ (2l)/(gsin(theta)) ]

    so this is the time at which position == l ?

    If I then plug this time into the equation for velocity,

    gsin(theta)t == v

    gsin(theta)*sqrt[ (2l)/(gsin(theta)) ] = v

    not the same as with the constant velocity forumla..
    why ? what is wrong with this method

    Thanks for any help
  2. jcsd
  3. Jun 5, 2015 #2


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  4. Jun 5, 2015 #3
    If I plug in numbers I get a different answer for both??
  5. Jun 5, 2015 #4


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    Example ... ?
  6. Jun 5, 2015 #5
    Looks like I did something wrong when I checked it. I was so sure that my formula must be wrong (as it was a different way to the book and done by me) I didn't check twice.

    You are right, and with some simple rearranging it comes out the same... thank you for your help.
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