## PhoenixFire 3 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. akash123

velocity of separation = e (velocity of approach)

2. PhoenixFire

3. akash123

|dw:1347425035783:dw|

4. akash123

does it make sense?

5. akash123

|dw:1347424692991:dw|

6. PhoenixFire

why are you subtracting 0?

7. akash123

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

8. akash123

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

9. akash123

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

10. akash123

similarly velocity of separation = V'-0

11. PhoenixFire

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

12. akash123

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

13. akash123

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. akash123

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. akash123

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. akash123

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

20. akash123
21. PhoenixFire

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

22. akash123

where are you from?

23. PhoenixFire

I'm studying in New Zealand.

24. akash123

ok

25. akash123

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. akash123

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

|dw:1347427306605:dw|

29. akash123

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

30. akash123

does it make sense?

31. PhoenixFire

So far yes.

32. akash123

|dw:1347427668995:dw|

33. akash123

good...:)

34. akash123

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. akash123

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

37. akash123

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. akash123

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

40. akash123

the ball will stop when v(n)=0

41. PhoenixFire

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

42. akash123

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

43. akash123

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

44. akash123

and e=0.85 which is less than 1

45. akash123

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

46. PhoenixFire

infinity

47. akash123

yes...perfectly right:)

48. akash123

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. akash123

ok

51. akash123

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. akash123

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

54. akash123

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?

Find more explanations on OpenStudy