## 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. one year ago one year ago

1. akash123 Group Title

velocity of separation = e (velocity of approach)

2. PhoenixFire Group Title

3. akash123 Group Title

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4. akash123 Group Title

does it make sense?

5. akash123 Group Title

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6. PhoenixFire Group Title

why are you subtracting 0?

7. akash123 Group Title

Velocity of approach = V' - zero...which is velocity of ground

8. akash123 Group Title

sorry...Velocity of approach = V - zero...which is velocity of ground

9. akash123 Group Title

ball is colliding with the ground..and use this velocity of separation = e (velocity of approach)

10. akash123 Group Title

similarly velocity of separation = V'-0

11. PhoenixFire Group Title

okay. so e is the math constant.. what am i solving for?

12. akash123 Group Title

so 1st calculate V, the velocity of ball from which it hits the ground 1st time..

13. akash123 Group Title

no...e is the coefficient of restitution...and for ur problem e =0.85(given)

14. PhoenixFire Group Title

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 Group Title

but see 4th line in ur solution..

16. Vincent-Lyon.Fr Group Title

The ball will bounce an infinite number of times, but the process will have a finite duration.

17. akash123 Group Title

velocity of separation = e (velocity of approach) it does not mean that... Energy after collision =e( energy before collision)?

18. PhoenixFire Group Title

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 Group Title

But the correct formula is for velocity not for energy velocity of separation = e( velocity of approach)

20. akash123 Group Title
21. PhoenixFire Group Title

So our professor gave us the wrong equation... awesome.

22. akash123 Group Title

where are you from?

23. PhoenixFire Group Title

I'm studying in New Zealand.

24. akash123 Group Title

ok

25. akash123 Group Title

anyway...u can solve this problem by using correct formula

26. PhoenixFire Group Title

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 Group Title

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 Group Title

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29. akash123 Group Title

t' = time of reaching at the peak after 1st collision

30. akash123 Group Title

does it make sense?

31. PhoenixFire Group Title

So far yes.

32. akash123 Group Title

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33. akash123 Group Title

good...:)

34. akash123 Group Title

e, e^2, e^3...will keep decreasing on...

35. PhoenixFire Group Title

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 Group Title

yes...now u can tell the answer :)

37. akash123 Group Title

the ball will stop when v(n)=0...right?

38. PhoenixFire Group Title

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 Group Title

and restitution <1..so tell me the answer?

40. akash123 Group Title

the ball will stop when v(n)=0

41. PhoenixFire Group Title

v(n)=0 $0=\mu_r^n \sqrt{2gh}$ Right?

42. akash123 Group Title

but now the question is when v(n) will be zero?

43. akash123 Group Title

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

44. akash123 Group Title

and e=0.85 which is less than 1

45. akash123 Group Title

now tell me... e^n will approach zero when n approaches....?

46. PhoenixFire Group Title

infinity

47. akash123 Group Title

yes...perfectly right:)

48. akash123 Group Title

so the ball will bounce infinite times...

49. PhoenixFire Group Title

well i got that answer like 3 hours ago... but i figured that there would be an actual answer.

50. akash123 Group Title

ok

51. akash123 Group Title

actually after each collision h keeps on decreasing...and theoretically h will approach zero after infinite bounces..

52. PhoenixFire Group Title

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 Group Title

but practically ball can't go up to a size of atom after some bounces for large n

54. akash123 Group Title

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 Group Title

What you CAN work out is not number of bounces (infinite), but total time taken.

56. PhoenixFire Group Title

@Vincent-Lyon.Fr Out of curiosity how would you go about doing that?