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AP Problem - Elastic Collision at an Angle

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data

    SEE BELOW: 4th Reply has a Diagram

    2. Relevant equations

    Conservation of Momentum: m1(v1i) +m2(v2i) = m1(v1f) + m2(v2f)
    Vf^2=Vi^2 + 2a(delta y)
    Conservation of Kinetic Energy (Elastic Collision): .5m1(v1i^2)+.5m2(v2i) = .5m1(v1f^2)+.5m2(v2f^2)

    3. The attempt at a solution

    See 4th post for newest question:
    Last edited: Dec 8, 2007
  2. jcsd
  3. Dec 8, 2007 #2

    Doc Al

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    Staff: Mentor

    Since the surface is frictionless, only the component of velocity normal to the surface is affected by the collision.
  4. Dec 8, 2007 #3
    What? Sorry I don't understand what you're saying....does that mean the angle doesn't change?

    (By the way thank you for being so amazingly helpful!!!! :) )
  5. Dec 8, 2007 #4

    Doc Al

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    Staff: Mentor

    Yes, the angle the velocity makes with the surface will not change.
  6. Dec 8, 2007 #5
    This is the problem and diagram:

    http://img255.imageshack.us/img255/1966/55320227sg8.png [Broken]

    I have done a-c, and these are my answers that I got. I know they're correct because I checked them with the solutions:

    L= 4[tex]\sqrt{}2[/tex] (h)
    Delta Y = Delta x = L/[tex]\sqrt{}2[/tex]

    For d I used the kinematic equation
    Vf^2=Vi^2 + 2a(delta y)
    I got delta y=4h
    and vf = [tex]\sqrt{}8gh[/tex]

    but the ap board says the answer is (conservation of energy):

    mgh + mgL/[tex]\sqrt{}2[/tex]=.5mv^2

    v = [tex]\sqrt{}10gh[/tex]

    But why is my answer for part d wrong?

    note: [tex]\sqrt{}2[/tex]
    (this symbol is the square root symbol)
    Last edited by a moderator: May 3, 2017
  7. Dec 8, 2007 #6
  8. Dec 8, 2007 #7


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    Science Advisor
    Homework Helper

    You didn't add the Vi^2 (=2gh) to the 2a(delta y) (=8gh).
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