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An elastic ball bouncing up a slope, where does it bounce first?

  1. Feb 13, 2012 #1


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    1. The problem statement, all variables and given/known data
    An elastic ball is projected with speed V from a point O of a smooth plane inclined at an angle theta to the horizontal. The initial direction of projection is up the plane, makes an angle theta + phi with horizontal and lies in a vertical plane containing a line of greatest slope of the plane. (i)Calculate the distance between O and the point of the first bounce.
    (ii)If the ball ceases to bounce at the instant it returns to O show that tan(theta)tan(phi)=1-e

    2. Relevant equations
    The way I tried this for both parts was to make the x axis follow the line of the slope NOT the horizontal which means splitting gravity into components, but this became really confusing.

    3. The attempt at a solution
    I chose the x axis to be the line of slope and z to be at right angles
    to this as the other axis. Using F=ma and F=mg I found a=g and the
    integrated to find the position vector r(t) = g(t^2/2)+vo, where vo is
    the starting velocity in component form, i.e. vo = kVsin(phi) +
    iVcos(phi), where k and i are unit vectors. Then because gravity has
    to be split up I used vector g = -igsin(theta)-kgcos(theta). Plugging
    both vector g and vo into the r(t) equation and splitting again into
    components, I get z(t) = Vtsin(phi)-(t^2/2)gcos(theta) and
    x(t)=Vtcos(phi)-g(t^2/2)sin(theta). When the ball hits the slope z(t)
    = 0 so using my equation for z(t) and rearranging to get t, I find
    t=2Vsin(phi)/gcos(theta). Then I put this t into x(t) and that should
    be the distance up the slope. To find the distance from O along the
    horizontal I used trig, but it comes out with this horrible looking
    equation and I'm pretty sure my error is in the gravity stuff because
    the change in axis confuses me massively! Sorry this is really
    rambling, I hope it makes sense.
  2. jcsd
  3. Feb 13, 2012 #2


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    Homework Helper

    how about staying in horizontal/vertical coordinate system - the motion of the ball will be parabolic, the place where it first bounces can be detrimned by finding where a line (the sloping surface) and the parabola intersect.
  4. Feb 13, 2012 #3


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    Thanks this helped a lot!
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