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Finding distance traveled by rebound when measured for surface

  1. May 16, 2017 #1
    1. The problem statement, all variables and given/known data
    Ball falling from 1m above a slanted surface 5 degrees from horizontal. KE conserved
    Find distance travelled along slanted surface on its second rebound



    2. Relevant equations
    S=(1/2)(at^2)
    (V^2)=(u^2)-2as

    3. The attempt at a solution
    V before hitting =sqrt(2*9.81)
    =4.42
    4.42/9.81=0.4515
    0.4515*2=0.9030
    0.9030*4.42*cos(85 degrees)
    =0.0355
    0.0355/cos(85)=3.98
     
  2. jcsd
  3. May 16, 2017 #2

    BvU

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    Hello Itilrandir, :welcome:

    I can follow you up to ##v=4.42## m/s (always include units). Could you explain what you are doing after that ? And make a sketch of the situation and post it ?

    I do notice there is no question in your post ?

    (all in the spirit of PF culture, you mostly don't get a yes/no answer and we don't do the exercise for you. But we help where we can !)
     
  4. May 16, 2017 #3
    Sorry for not being very clear with myself.
    The question basically asks for us to find distance between the first and second impact on a infinitely long slanted surface tilted at 5 degrees.
    Pic attached is what I attempted to do and get the final ans of around 4(3.98)
     

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  5. May 16, 2017 #4

    BvU

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    I do understand the question.

    Let me guess (this is NOT PF culture -- instead, you should explain, as I asked :rolleyes: ) :
    You use ##v = v_0 - gt = 0 \Rightarrow t = {v_0\over g} ## to find the time to reach the highest point after the bounce, 0.903 m.

    However, ##\vec v_0## is not in the same direction as ##g## is acting, so you lose me at that point.
    Furthermore, I also don't understand (understatement) the 85 degrees in the next line...

    In your sketch I see 1 m above the surface as a constant -- which it may well be. Can you prove that it is a constant ?
     
  6. May 16, 2017 #5
    It was just a random guess/ attempt at solving it cos the question was timed when i was doing it.:sorry: I was hoping that someone has a better explaination for the question.
     
  7. May 16, 2017 #6

    BvU

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    After the first bounce, ##v=v_0=4.42## m/s, I agree.
    Can only be if it is at 10##^\circ## wrt the vertical (##E=\vec p^2/(2m)## is conserved).

    So you have ##v_x = v_0\cos(80^\circ),\ v_y = v_0\sin(80^\circ),\ ## for a parabolic trajectory.
    That intersects the surface ##y = y_0 - x\sin 5^\circ## at the second bounce point.
    Set up the equations and solve the quadratic equation.
     
  8. May 16, 2017 #7
    Do you have the correct answer? I want to double check after doing it. No time right now.
     
  9. May 16, 2017 #8

    haruspex

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    sin 5°?
     
  10. May 17, 2017 #9

    BvU

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    PF culture is YOU do the work, WE try to help :smile: . If that necessitates working it all out, then I am always willing to do the work. Not in this case.
     
  11. May 17, 2017 #10

    haruspex

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    I was querying the trig function, not the angle.
     
  12. May 17, 2017 #11

    BvU

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    With good reason o:) . Should have been ##\tan## ! Thanks!
    upload_2017-5-17_11-24-21.png
    Now we both wait for Itil to have some time to work it out ....
     
  13. May 18, 2017 #12
    I got the answer its apparent 0.68something meters
     
  14. May 18, 2017 #13

    ehild

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    Show please, how did you get that result.
     
  15. May 18, 2017 #14
    Sorry I was busy.
    The method to solve it is to resolve the components of gravity and velocity and turn it into a projectile motion
     
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