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PhoenixFire Group Title

Does anyone know how to calculate how many times a ball will bounce if dropped from some height? Information: Restitution = 0.85 Mass = 0.7 kg Drop Height = 10 m Gravity = 9.81 m/s Ground is Horizontal. No air resistance.

  • 2 years ago
  • 2 years ago

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  1. akash123 Group Title
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    velocity of separation = e (velocity of approach)

    • 2 years ago
  2. PhoenixFire Group Title
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    A little more explanation please?

    • 2 years ago
  3. akash123 Group Title
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    |dw:1347425035783:dw|

    • 2 years ago
  4. akash123 Group Title
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    does it make sense?

    • 2 years ago
  5. akash123 Group Title
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    |dw:1347424692991:dw|

    • 2 years ago
  6. PhoenixFire Group Title
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    why are you subtracting 0?

    • 2 years ago
  7. akash123 Group Title
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    Velocity of approach = V' - zero...which is velocity of ground

    • 2 years ago
  8. akash123 Group Title
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    sorry...Velocity of approach = V - zero...which is velocity of ground

    • 2 years ago
  9. akash123 Group Title
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    ball is colliding with the ground..and use this velocity of separation = e (velocity of approach)

    • 2 years ago
  10. akash123 Group Title
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    similarly velocity of separation = V'-0

    • 2 years ago
  11. PhoenixFire Group Title
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    okay. so e is the math constant.. what am i solving for?

    • 2 years ago
  12. akash123 Group Title
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    so 1st calculate V, the velocity of ball from which it hits the ground 1st time..

    • 2 years ago
  13. akash123 Group Title
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    no...e is the coefficient of restitution...and for ur problem e =0.85(given)

    • 2 years ago
  14. PhoenixFire Group Title
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    Potential Energy completely becomes Kinetic Energy just before collision \[E_{system}=0+mgh=(0.7)(9.81)(10)=68.67\]\[E_{kinetic}=E_{system}\]\[v_{before}=\sqrt{2E_{kinetic} \over m}=14.01[ms^{-1}]\]\[\mu_r = {E_{kin. after} \over E_{kin.before}}\]\[E_{kin. after}=\mu E_{kin. before}=(0.85)(68.67)=58.67[J]\]\[v_{after}=\sqrt{2E_{kinetic} \over m}=12.91[ms^{-1}]\] I have both the velocities. Before and after the bounce.

    • 2 years ago
  15. akash123 Group Title
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    but see 4th line in ur solution..

    • 2 years ago
  16. Vincent-Lyon.Fr Group Title
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    The ball will bounce an infinite number of times, but the process will have a finite duration.

    • 2 years ago
  17. akash123 Group Title
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    velocity of separation = e (velocity of approach) it does not mean that... Energy after collision =e( energy before collision)?

    • 2 years ago
  18. PhoenixFire Group Title
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    That's one of the formula we were given.. that restitution is energy after divided by energy before. so i just used that to calculate energy after collision and then the velocity.

    • 2 years ago
  19. akash123 Group Title
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    But the correct formula is for velocity not for energy velocity of separation = e( velocity of approach)

    • 2 years ago
  20. akash123 Group Title
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    http://scienceworld.wolfram.com/physics/CoefficientofRestitution.html check this

    • 2 years ago
  21. PhoenixFire Group Title
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    So our professor gave us the wrong equation... awesome.

    • 2 years ago
  22. akash123 Group Title
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    where are you from?

    • 2 years ago
  23. PhoenixFire Group Title
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    I'm studying in New Zealand.

    • 2 years ago
  24. akash123 Group Title
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    ok

    • 2 years ago
  25. akash123 Group Title
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    anyway...u can solve this problem by using correct formula

    • 2 years ago
  26. PhoenixFire Group Title
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    Wait, what do I do now that I've got the velocity before and after collision? vel approach =14.01 vel separation = 11.91

    • 2 years ago
  27. akash123 Group Title
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    yes...If u have done correctly... Better use kinematics relation..and use symbols dont plug in the value now..just use h,g and all..then u'll get better understanding of this problem

    • 2 years ago
  28. akash123 Group Title
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    |dw:1347427306605:dw|

    • 2 years ago
  29. akash123 Group Title
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    t' = time of reaching at the peak after 1st collision

    • 2 years ago
  30. akash123 Group Title
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    does it make sense?

    • 2 years ago
  31. PhoenixFire Group Title
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    So far yes.

    • 2 years ago
  32. akash123 Group Title
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    |dw:1347427668995:dw|

    • 2 years ago
  33. akash123 Group Title
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    good...:)

    • 2 years ago
  34. akash123 Group Title
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    e, e^2, e^3...will keep decreasing on...

    • 2 years ago
  35. PhoenixFire Group Title
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    general formula \[v_n=\mu_r^n \sqrt{2gh}\]\[t_n=\mu_r^n \sqrt{2h \over g}\] I need to find n.

    • 2 years ago
  36. akash123 Group Title
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    yes...now u can tell the answer :)

    • 2 years ago
  37. akash123 Group Title
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    the ball will stop when v(n)=0...right?

    • 2 years ago
  38. PhoenixFire Group Title
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    well not really. because I want to find when it stops bouncing right? so v=0 or t=0. ends up just being n=0.. unless i'm failing at algebra

    • 2 years ago
  39. akash123 Group Title
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    and restitution <1..so tell me the answer?

    • 2 years ago
  40. akash123 Group Title
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    the ball will stop when v(n)=0

    • 2 years ago
  41. PhoenixFire Group Title
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    v(n)=0 \[0=\mu_r^n \sqrt{2gh}\] Right?

    • 2 years ago
  42. akash123 Group Title
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    but now the question is when v(n) will be zero?

    • 2 years ago
  43. akash123 Group Title
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    yes...are u noticing something? e, e^2, e^3... with each v(n) for n=1,2,3,4....

    • 2 years ago
  44. akash123 Group Title
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    and e=0.85 which is less than 1

    • 2 years ago
  45. akash123 Group Title
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    now tell me... e^n will approach zero when n approaches....?

    • 2 years ago
  46. PhoenixFire Group Title
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    infinity

    • 2 years ago
  47. akash123 Group Title
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    yes...perfectly right:)

    • 2 years ago
  48. akash123 Group Title
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    so the ball will bounce infinite times...

    • 2 years ago
  49. PhoenixFire Group Title
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    well i got that answer like 3 hours ago... but i figured that there would be an actual answer.

    • 2 years ago
  50. akash123 Group Title
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    ok

    • 2 years ago
  51. akash123 Group Title
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    actually after each collision h keeps on decreasing...and theoretically h will approach zero after infinite bounces..

    • 2 years ago
  52. PhoenixFire Group Title
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    Well, thanks for the help! Showed me one of the equations was wrong which let me fix the first bit of the question. I guess I'm happy with infinite bounces for now. But I'll have a talk with my professor next week and see what he says.

    • 2 years ago
  53. akash123 Group Title
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    but practically ball can't go up to a size of atom after some bounces for large n

    • 2 years ago
  54. akash123 Group Title
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    as according to ur problem..u r starting with h=10m...later on ...h will reach in nanometer scale...and it's not practical that ball will bounce up to some nm height.

    • 2 years ago
  55. Vincent-Lyon.Fr Group Title
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    What you CAN work out is not number of bounces (infinite), but total time taken.

    • 2 years ago
  56. PhoenixFire Group Title
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    @Vincent-Lyon.Fr Out of curiosity how would you go about doing that?

    • 2 years ago
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