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Energy and momentum.

  1. Nov 14, 2013 #1
    Hi, I got homework and I need someone to check it, please. :)
    http://shrani.si/f/1G/JS/1UO1SWSw/skakalnica.jpg [Broken]

    Ski jumper (m=70kg) goes down the hill with v0=9m/s. The hill has got R (radius of curvature = 8m). There is no friction (the hill is 1m above earth - like on picture).
    1. a) What is the velocity when the ski jumper just jumps?
    My answer: Wk(begining)+Wp(begining)=Wk(end)+Wp(end)=
    =(70*9^2)/2+70*10*1=(7*vfinal^2)/2+70*10*3,34 (h0+h)=
    vfinal=5,84m/s

    b) Find the highest point!
    hmax=v0y^2/(2g)=(5,84*cos45)^2/20=0,85+h0=4,2m

    c) What is the impulse of the hill to the jumper (vertical) when he is on the hill?
    ∫Fydt=G(end)-G(begining)
    ∫Fydt=0-m*v(end)y=0-70*sin45=-290Ns

    2. Let the hill move without friction. (Mhill=1000kg) When jumper hits the hill (v0=9m/s) it is at rest.

    a) Does the energy of the system (jumper+hill) preserve? Prove with Work of outer and inner forces.
    I assume it does, but don't know how to prove it.?!

    b) What is the velocity (to the observer on earth) when jumper leaves the hill? At what angle does it point?
    I am not sure here, whether I use the conservation of energy equation or mumentum equation?:
    m*g*h0+(m*v0^2)/2=mgh1+(m*v1^2)/2+(Mhill*v1^2)/2=
    =1,5m/s angle=45°


    or:
    m*v0=(m+Mhill)v1=0,589m/s.

    c) What is the velocity of the hill?
    Is it here possible that I use here the equation of mumentum to get this vx and that in (b) I use energy conservation to get v=1,5. And get angle (in b) like cosθ=v/v1...θ=67°

    Thanks for help. :)
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 14, 2013 #2

    haruspex

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    Seems a bit low. If you keep getting that answer, please post all your working.
    The trouble with using numerical intermediate results is accumulation of errors. It's best to keep everything purely symbolic (algebraic) until the final step of each answer. It also makes it easier to spot errors.
    You now have two unknowns: the velocity of the jumper (at leaving) and the velocity of the hill. So you will need two equations. Work energy is certainly one. You can use momentum if there are no forces external to the system in the direction of the momentum you calculate.
     
    Last edited by a moderator: Nov 19, 2013
  4. Nov 18, 2013 #3
    Hi, thank you for your quick reply!

    a) From energy equation that I used I get v=5,92m/s but not more. Also the same from v^2=v0^2-2gh...

    b) When I put this new result in I get 4,23m for max height.

    2. b) I have no clue how to solve this. So, the energy equation is right? Is this velocity that I get the velocity of the jumper and the hill to the right side (x axis)?
    Can you give me some hints please? :)
    Thank you.
     
  5. Nov 18, 2013 #4

    haruspex

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    Hmm.. I agree with your answer now. Don't know what I did wrong before. Sorry about that.
    Your energy equation assumes the hill and the jumper will have the same speed at take-off. That cannot be true.
    To keep it simple, let's allow a third unknown. You have the horizontal speed of the hill, the vertical speed of jumper at take-off and horizontal speed of jumper at take-off. That last can be either relative to the hill or relative to the ground - your choice - as long as you are consistent.
    In addition to the energy equation and horizontal momentum equation, you should be able to find an equation relating those three speeds to the slope of the hill at the take-off point.
    You understand why momentum of the skier+hill system is conserved horizontally but in no other direction, right?
     
  6. Nov 19, 2013 #5
    Ufff...just went trough maths and I really hope it is right. :D I have to hand this homework in tomorrow.

    So for 2.b)
    Momentum is conserved in x axis. So m1*v0+0=m1*vx1+m2*v2. From this equation I get v2=(m1*(v0-vx1))/m2.
    This vx1 is the x component of skiers velocity!
    Then I used energy equation (kinetic and potential energy of skier+hill in the beginning is equal to the kinetic and potential energy in the end). So:
    m1*v0^2/2 + 0 + m1*g*h0 + 0 = m1*v1^2/2 + m2*v2^2/2 + m1*g*h1 + 0 (This v1 is the whole velocity not only in x component).
    Then I put momentum v2 into this equation and get the equation to the power 4, hehe. Difficult to not do any mistake. :D And got v1^2=27,28 ==>v1=5,22m/s. When I put this into momentum v2 (again *cos45) I get v2=0,37m/s.
    Those results seems quite intuitive to me (v1 is a little bit less then when the hill is at rest and v2 is also small because of its mass).
    Please tell me this is right, hehe. ;D
    Now I only wonder how to prove (with work of forces) that energy is conserved?

    Thank you a lot! Have a nice day!
     
  7. Nov 19, 2013 #6

    haruspex

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    That all looks ok.
    Not sure how you got to there (not saying it's wrong). You had two equations and three unknowns: v1x, v1y, v2. What was your third equation?
    Yes, the answer certainly looks reasonable.
     
  8. Nov 19, 2013 #7
    I was thinking that if I get v1x I don't need v1y, because I know the angle? So v1=v1x/cos45?
    I spotted now one mistake (in previous reply I calculated v1=v1x*cos45, but it is devided by cos45). So now I get v1=4,18m/s and v2=0,42m/s.
     
  9. Nov 19, 2013 #8

    haruspex

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    No. Your v1x is relative to the ground. The relationship between horizontal and vertical motion of the skier that's determined by the hill slope will be for speeds relative to the hill.
    That said, it should only introduce a small error.
     
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