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.

- PhoenixFire

- chestercat

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- anonymous

velocity of separation = e (velocity of approach)

- PhoenixFire

A little more explanation please?

- anonymous

|dw:1347425035783:dw|

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## More answers

- anonymous

does it make sense?

- anonymous

|dw:1347424692991:dw|

- PhoenixFire

why are you subtracting 0?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

similarly velocity of separation = V'-0

- PhoenixFire

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

- anonymous

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

- anonymous

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

- 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.

- anonymous

but see 4th line in ur solution..

- Vincent-Lyon.Fr

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

- anonymous

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

- 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.

- anonymous

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

- anonymous

http://scienceworld.wolfram.com/physics/CoefficientofRestitution.html
check this

- PhoenixFire

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

- anonymous

where are you from?

- PhoenixFire

I'm studying in New Zealand.

- anonymous

ok

- anonymous

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

- 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

- 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

- anonymous

|dw:1347427306605:dw|

- anonymous

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

- anonymous

does it make sense?

- PhoenixFire

So far yes.

- anonymous

|dw:1347427668995:dw|

- anonymous

good...:)

- anonymous

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

- PhoenixFire

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

- anonymous

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

- anonymous

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

- 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

- anonymous

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

- anonymous

the ball will stop when v(n)=0

- PhoenixFire

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

- anonymous

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

- anonymous

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

- anonymous

and e=0.85 which is less than 1

- anonymous

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

- PhoenixFire

infinity

- anonymous

yes...perfectly right:)

- anonymous

so the ball will bounce infinite times...

- PhoenixFire

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

- anonymous

ok

- anonymous

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

- 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.

- anonymous

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

- 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.

- Vincent-Lyon.Fr

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

- PhoenixFire

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

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