## PhoenixFire 4 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.

1. anonymous

velocity of separation = e (velocity of approach)

2. PhoenixFire

3. anonymous

|dw:1347425035783:dw|

4. anonymous

does it make sense?

5. anonymous

|dw:1347424692991:dw|

6. PhoenixFire

why are you subtracting 0?

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

similarly velocity of separation = V'-0

11. PhoenixFire

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

12. anonymous

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

13. anonymous

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

14. PhoenixFire

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. anonymous

but see 4th line in ur solution..

16. Vincent-Lyon.Fr

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

17. anonymous

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

18. PhoenixFire

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. anonymous

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

20. anonymous
21. PhoenixFire

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

22. anonymous

where are you from?

23. PhoenixFire

I'm studying in New Zealand.

24. anonymous

ok

25. anonymous

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

26. PhoenixFire

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. anonymous

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. anonymous

|dw:1347427306605:dw|

29. anonymous

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

30. anonymous

does it make sense?

31. PhoenixFire

So far yes.

32. anonymous

|dw:1347427668995:dw|

33. anonymous

good...:)

34. anonymous

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

35. PhoenixFire

general formula $v_n=\mu_r^n \sqrt{2gh}$$t_n=\mu_r^n \sqrt{2h \over g}$ I need to find n.

36. anonymous

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

37. anonymous

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

38. PhoenixFire

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. anonymous

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

40. anonymous

the ball will stop when v(n)=0

41. PhoenixFire

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

42. anonymous

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

43. anonymous

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

44. anonymous

and e=0.85 which is less than 1

45. anonymous

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

46. PhoenixFire

infinity

47. anonymous

yes...perfectly right:)

48. anonymous

so the ball will bounce infinite times...

49. PhoenixFire

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

50. anonymous

ok

51. anonymous

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

52. PhoenixFire

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. anonymous

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

54. anonymous

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

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

56. PhoenixFire

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