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Skiing down a hill.

  1. Nov 5, 2007 #1
    1. The problem statement, all variables and given/known data

    A 70-kg woman skies down a slope and then encounters the top of a huge snowball of radius 5 m while she is moving at 3 m/s. She continues over the surface of the "ball" until she ultimately parts company from the "ball" at an angle (alpha) which is measured by the radial line to her from the vertical. Consider the snow and ice of the "ball" to be frictionless. Employing both energy considerations and Newton's 2nd law, determine the height above the ground and the angle at which she leaves the "ball".

    2. Relevant equations


    3. The attempt at a solution

    I don't really know where to begin. Please help!
  2. jcsd
  3. Nov 5, 2007 #2


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    Staff: Mentor

    This is a variation of a mass sliding down a circular/spherical surface and determining when the mass leaves the surface, which is when the contact force = zero. Here however, rather than starting at rest, the skier has an initial velocity v0 = 3 m/s.

    Refer to this


    In addition to the change in gravitational potential energy, there is an initial kinetic energy.

    Note also that the energy terms all involve the same mass, so ultimately the mass divides out of the energy equation.
    Last edited: Nov 5, 2007
  4. Nov 7, 2007 #3
    I started by saying (1/2)mv0^2+mga = (1/2)mv^2+mgz
    then solved for v^2.
    Then I found in the book that R(the contact force)=mg(z/a)-m(v^2/a)
    I plugged in the v^2 that I got, and eventually came out to: R=mg/a((3z-2a)-v0^2/g)
    I set that to zero and found z=3.64=8.64 after adding 5 for radius.
    To find the angle: z/a=cos(theta)
    is this correct?
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