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Conservation of Energy / Projectile motion Problem

  1. Nov 4, 2011 #1
    Conservation of Energy- A Skier on a Snowball

    1. The problem statement, all variables and given/known data

    A skier of mass "m" starts at the top of a very large frictionless hemispherical snowball of radius "r" with a very small initial speed and skis straight down the side. Treat the skier as a point partical.

    a) At what angle with the vertical does he lose contact with the surface?

    b) After flying through the air, the skier hits the ground level with the center of the circular arc. Determine the horizontal distance from the center of the circle to the point of impact on the ground.

    2. Relevant equations

    For part a

    Vi=0 Initial Velocity
    Y1=r Initial Y
    Y2=rcos[itex]\alpha[/itex] Final Y
    a(rad)=(v^2)/r Radial Acceleration
    n[itex]\rightarrow[/itex]0 The skier loses contact when n goes to zero


    For part b

    X = xi + ViT + (a(T^2))/2

    3. The attempt at a solution
    Part a
    I think that I have the correct angle. From the sum of the forces in the y direction, Vf^2=rgcos[itex]\alpha[/itex]

    Pluging that into conservation of energy...

    Part b is where I fall apart.
    I know that at the point that the skier loses contact with the snowball, he can be treated as a projectile. And that if I can find the time it takes for the object to move from y1 to y2, I can use it to solve for the horiaontal distance. But I am confused with the initial x and y positions, and the initial velocities in the x and y direction.

    Attached Files:

    Last edited: Nov 5, 2011
  2. jcsd
  3. Nov 5, 2011 #2


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    Re: Conservation of Energy- A Skier on a Snowball

    Solution of part a is correct.

    When in air, the skier is a projectile, starting at the height and with initial velocity it had, when lost contact with the slope. So you have two equations, one for the horizontal and one for the vertical motion. What are they? Find the components of the initial velocity of the projectile. You know the magnitude of v from the centripetal force equal to the radial component of mg. The velocity is tangent to the circle, so what angle does it enclose with the horizontal?
    You also find the initial position from the angle and the radius of the circle, using your very good drawing.

  4. Nov 5, 2011 #3
    Thank you very much for responding ehild...
    Two equations? Are you talking about the constant acceleration equation I provided? One in terms of X, and the other in terms of Y? and Is the angle you're refering to, 41.8 making...

    Xi = r[itex]\sqrt{5}[/itex]/3

    Yi = 2r/3

    Xf = d

    yf = 0

    Vxi = [itex]\sqrt{5}[/itex]/3*[itex]\sqrt{2gr/3}[/itex]

    Vyi = -(2/3)*[itex]\sqrt{2gr/3}[/itex]

    Vxf = [itex]\sqrt{5}[/itex]/3*[itex]\sqrt{2gr/3}[/itex]

    Vyf = Vyf

    Acceleration(x) = 0 Acceleration(y) = -g
  5. Nov 5, 2011 #4


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    Vyf=Vyi-gt, but the others are all right, go ahead.

  6. Nov 5, 2011 #5
    Ok... I've done it four times now. Each time getting a different answer. lol Does d = R(1.66) sound reasonable to you?
  7. Nov 6, 2011 #6


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    No. If the skier would move along a straight line it would reach the ground at distance R/sin(alpha)=1.342 R, so you have to get a shorter distance.

    It is really ugly, I always get different answers, too.:yuck: Try to do the calculations symbolically and plug-in the values at the end.

    It might help that eliminating t, you get the equation for y in terms of d=x-xi:

    yi+(viy/vix)d-g/2 d2/vix2=y.

    y=0. Solve for d and add xi. My last result is about 1.12 R.

    Last edited: Nov 6, 2011
  8. Nov 6, 2011 #7
    Wow thank you!! With all the trouble I'm having in solving T, I would have never thought of getting rid of it. I am a little confused in the calculation though. Now I'm getting d = 1.44. When you found that R/sin[itex]\alpha[/itex] = 1.342 R, you used the angle 48.2 (the angle from the vertical to the initial point)... but I have been using [itex]\beta[/itex] = 41.8 from the very begining. Is that where I went wrong?
    Last edited: Nov 6, 2011
  9. Nov 6, 2011 #8


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    I can not follow you with the angles. You draw a very good sketch in the first post, now I redraw it, with the initial velocity vector shown. Initial mean the instant when the skier leaves the slope. The angles shown are all equal to alpha, whose cosine is 2/3. Xmax is the distance from the centre of the circle if the skier moved along a straight line with its initial velocity.

    I suggested to calculate d= x(final)-x(initial). You get it from the quadratic equation. Finally, add x(initial) =rsin(alpha) =r√5/3 to it. I made a typo in the last post, the final distance was 1.12 R instead of 1.2 R.


    Attached Files:

    Last edited: Nov 6, 2011
  10. Nov 6, 2011 #9
    AWWW.... I have it now. I was adding R to the final result instead of R[itex]\sqrt{5}[/itex]/3. Thank you for all you help and patience.
  11. Nov 6, 2011 #10


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    You are welcome. Good job.

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