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Potential energy exercise

  1. Apr 5, 2014 #1
    Hi everyone , I'm really stuck with an exercise I can't solve any question and it is really important ! I need help :

    Two identical billiard balls start at the same height and at the same time and roll along different tracks,as shown in figure attached

    1- Which ball has the highest speed at the end ?
    2- Which one get to end first ? (clearly same questions)
     

    Attached Files:

  2. jcsd
  3. Apr 5, 2014 #2

    mfb

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    Where are you stuck?
    Please show what you did and thought of so far.

    What do you know about potential energy, kinetic energy, energy conservation and so on?

    Those are different questions.
     
  4. Apr 6, 2014 #3
    I know that Ug(y)=mgy+constante and ΔUg= -Wg with Wg = -mgh

    I also know that ΔK= Wg so ΔUg= -ΔK and therefore : ΔE=ΔU+ΔK=0 which explains the conservation of energy for conservative forces !

    I wantek to use ΔK= -ΔU and I know that 1/2mv²=mgh , but then what is h ? and I don"t even know how I can solve the second question
     
  5. Apr 6, 2014 #4

    mfb

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    Formulas are pointless without a description of their variables.

    Anyway, the main point is: if both balls drop by the same height (in total, so after the ramp), what can you say about the kinetic energy they get? How is that related to their speed?

    What about the point in between, where they have different heights?
     
  6. Apr 6, 2014 #5
    After the ramp , and as they drop from the same height , they should have the same velocity and therefore the same kinetic energy .

    The point in B where there isn't a straiht path represent a hard thing for me to analyze and understand its impact on velocity or speed
     
  7. Apr 6, 2014 #6

    haruspex

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    Unless my eyes deceive me, the net drop in height from start to end is less in the second picture.
    That makes it impossible to answer the second question.
    Is it perhaps the case that the posted diagram has tilted the lower picture, and that in the original the gradient is zero at the end?
     
  8. Apr 7, 2014 #7
    no they are the same
     
  9. Apr 7, 2014 #8

    haruspex

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    OK, so it's just the way the diagrams have been uploaded, with the lower one being tilted up to the right.

    Compare the horizontal components of velocity, at various horizontal distances, between the two arrangements.
     
  10. Apr 7, 2014 #9
    How can I compare it , I'm really stuck ! I won't be asking for help if I knew the answer
     
  11. Apr 7, 2014 #10

    haruspex

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    I'm just suggesting something for you to think about. For reference, let's label points as follows:
    P = start in top picture, P' = start in lower picture.
    Q = bottom of first slope in top picture, Q' etc.
    R' = start of second downslope in lower picture; R = corresponding point in top picture.
    S' = end of second downslope in lower picture; S = corresponding point in top picture, in terms of horizontal distance.
    T' = start of upslope in lower picture, U' = end of upslope in lower picture; T, U being corresponding points in top picture.

    At Q and Q', the speeds are the same, right? And this will continue until R, R'.
    At a point between R' and S', how does the horizontal speed compare with at the corresponding point (in terms of horizontal distance) between R and S? Is it the same, less, or more? What does that tell you about the time taken to cover R' to S', compared with R to S?
    Do the same analysis for ST versus S'T' and TU versus T'U'.
     
  12. Apr 8, 2014 #11
    well, the horizontal speed in RS is bigger than R'S' as Kf-Ko=mgh !
     
  13. Apr 8, 2014 #12

    haruspex

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    I think you have it backwards - I defined R'S' etc. as referring to the lower diagram, the one with the extra dip in it.

    Energy works ok for the ST versus S'T' sections, but RS versus R'S' is a bit trickier. The speed is greater in R'S' than in RS, but is the horizontal component greater? To answer that, it's easier to think about horizontal momentum. When travelling down the R'S' slope, what forces act? What does that tell you about change in horizontal momentum?
     
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