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Circular elastic collision

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data
    a 6 kg, originally held at rest in position 1 as shown in the diagram below, slides down a frictionless quadrant of a verticle circle of radius 15 m. It then collides elastically with another 16 kg mass at rest at position 2. They bounce off each other and move on the horizontal frictionless surface and up the frictionless quadrant of a verticle circle on each side. Calculate the verticle height "H" for each ball at which they would momentarily come to rest after elastic collision.

    Diagram: http://s1176.beta.photobucket.com/user/LolaGoesLala/media/jkmlkm.jpg.html


    3. The attempt at a solution

    So i was thinking of starting with the kinetic energy formula btu then i realised i need to include the gravitational potential energy but i need the height for it... so can anyone help me out on starting with this problem?
     
  2. jcsd
  3. Nov 11, 2012 #2

    tiny-tim

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    Hi Lolagoeslala! :smile:
    to start this problem, you need to find the speed of the first mass just before the collision, and then use conservation of energy and conservation of momentum to find the speeds just after the collision :wink:
     
  4. Nov 11, 2012 #3
    so to start this problem...
    i would use the equation...

    1/2m1v1 + mgh = 1/2m1v2
    This represents the mass at rest state with the height which equals to the velocity at no height

    correct? but what would the height be then?
     
  5. Nov 11, 2012 #4

    tiny-tim

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    what is this ?? :redface:

    you can't add a momentum to an energy :frown:
     
  6. Nov 11, 2012 #5
    OOPS sorry im so sorry...
    the equation is all wrong...
    its supposed to be like this:

    1/2m1v1^2 = mgh + 1/2m1v1`^2

    this is for the conservation of energy just before it hits the collision for m1
     
  7. Nov 11, 2012 #6

    tiny-tim

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    ok, at least that makes sense now!

    but v1' is 0
     
  8. Nov 11, 2012 #7
    Wait isn;t the v1 zero.. that is the intially and that is the rest position....
    so wouldn;t it be

    mgh = 1/2m1v1`^2 and we are trying to find the speed with which it hits the second ball?
     
  9. Nov 11, 2012 #8

    tiny-tim

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    yes, that's correct …

    that gives you the initial speed (just before the collision) of the first ball :smile:

    (btw, i would have used "u" rather than "v", for reasons which may shortly appear obvious :wink:)

    now use conservation of energy and conservation of momentum for the elastic collision​
     
  10. Nov 11, 2012 #9
    Wait ...

    so i would use the equation
    m1gh = 1/2m1v1`^2

    now whats the height??? is that the radius?
     
  11. Nov 11, 2012 #10

    tiny-tim

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    yes …
    … it's a quadrant, so it goes from 3 oclock to 6 oclock … that's a height of one radius :wink:
     
  12. Nov 11, 2012 #11
    so...

    m1gh = 1/2m1v1`^2
    6kgx9.8m/s^2x15m = 1/2(6kg)v1`^2
    882 J / 3 kg = v1`^2
    √294 J = v1`
    17.14 m/s = v1`

    Is that correct? :D
     
  13. Nov 11, 2012 #12

    tiny-tim

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    (try using the X2 and X2 buttons just above the Reply box :wink:

    yes :smile:

    (but you could have deleted m1 from the equation completely, and so not used 6 kg at all :wink:)
     
  14. Nov 11, 2012 #13
    Oh ok.. so now would i used the conservation of momentum

    so like

    m1v1 + m2v2 = m1v1` + m2v2`
    m1v1 = m1v1` + m2v2`
    but then i have two unknown variables v1` and v2`
     
  15. Nov 11, 2012 #14

    tiny-tim

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    yes, so you also need to use conservation of energy (for the collision)
     
  16. Nov 11, 2012 #15
    umm but wouldn't the energy be lost thenn ...
    so they wouldn't be equal to one another would they...
    lets see...

    1/2m1v1`^2 = 1/2m1v1`^2 + 1/2m2v1`^2

    i still have two missing variables.. :cry:
     
  17. Nov 11, 2012 #16

    tiny-tim

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    but the question says it's elastic (ie, no energy is lost) …
    yes, you now have two equations and two unknowns …

    so solve it​
     
  18. Nov 11, 2012 #17
    v1`=(m1-m2/m1+m2)v1
    V1`=(6 kg-16kg/22kg)17.14 m/s [E]
    v1`= 7.79 m/s [W]

    v2` = (2m1/m1+m2)v1
    v2`= (12kg/22kg)17.14m/s[E]
    v2`=9.34m/s[E]
     
  19. Nov 11, 2012 #18
    YOU THERE?!?!?:redface:
     
  20. Nov 12, 2012 #19

    tiny-tim

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    (just got up :zzz:)
    yes, those equations are correct :smile:

    but how did you get them? :confused:
     
  21. Nov 12, 2012 #20
    school lesson we had to derive them so i had them.. now what should i do ?? :O?
     
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