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PhoenixFire

  • 2 years ago

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.

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  1. akash123
    • 2 years ago
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    velocity of separation = e (velocity of approach)

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

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

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

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

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

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

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

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

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

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

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

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

  14. PhoenixFire
    • 2 years ago
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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.

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

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

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

  18. PhoenixFire
    • 2 years ago
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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.

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

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

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

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

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

  24. akash123
    • 2 years ago
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    ok

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

  26. PhoenixFire
    • 2 years ago
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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

  27. akash123
    • 2 years ago
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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

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

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

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

  31. PhoenixFire
    • 2 years ago
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    So far yes.

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

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

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

  35. PhoenixFire
    • 2 years ago
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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.

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

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

  38. PhoenixFire
    • 2 years ago
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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

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

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

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

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

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

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

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

  46. PhoenixFire
    • 2 years ago
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    infinity

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

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

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

  50. akash123
    • 2 years ago
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    ok

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

  52. PhoenixFire
    • 2 years ago
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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.

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

  54. akash123
    • 2 years ago
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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.

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

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

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