PhoenixFire
  • PhoenixFire
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.
Physics
chestercat
  • chestercat
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anonymous
  • anonymous
velocity of separation = e (velocity of approach)
PhoenixFire
  • PhoenixFire
A little more explanation please?
anonymous
  • anonymous
|dw:1347425035783:dw|

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anonymous
  • anonymous
does it make sense?
anonymous
  • anonymous
|dw:1347424692991:dw|
PhoenixFire
  • PhoenixFire
why are you subtracting 0?
anonymous
  • anonymous
Velocity of approach = V' - zero...which is velocity of ground
anonymous
  • anonymous
sorry...Velocity of approach = V - zero...which is velocity of ground
anonymous
  • anonymous
ball is colliding with the ground..and use this velocity of separation = e (velocity of approach)
anonymous
  • anonymous
similarly velocity of separation = V'-0
PhoenixFire
  • PhoenixFire
okay. so e is the math constant.. what am i solving for?
anonymous
  • anonymous
so 1st calculate V, the velocity of ball from which it hits the ground 1st time..
anonymous
  • anonymous
no...e is the coefficient of restitution...and for ur problem e =0.85(given)
PhoenixFire
  • 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
  • anonymous
but see 4th line in ur solution..
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
The ball will bounce an infinite number of times, but the process will have a finite duration.
anonymous
  • anonymous
velocity of separation = e (velocity of approach) it does not mean that... Energy after collision =e( energy before collision)?
PhoenixFire
  • 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
  • anonymous
But the correct formula is for velocity not for energy velocity of separation = e( velocity of approach)
anonymous
  • anonymous
http://scienceworld.wolfram.com/physics/CoefficientofRestitution.html check this
PhoenixFire
  • PhoenixFire
So our professor gave us the wrong equation... awesome.
anonymous
  • anonymous
where are you from?
PhoenixFire
  • PhoenixFire
I'm studying in New Zealand.
anonymous
  • anonymous
ok
anonymous
  • anonymous
anyway...u can solve this problem by using correct formula
PhoenixFire
  • 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
  • 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
  • anonymous
|dw:1347427306605:dw|
anonymous
  • anonymous
t' = time of reaching at the peak after 1st collision
anonymous
  • anonymous
does it make sense?
PhoenixFire
  • PhoenixFire
So far yes.
anonymous
  • anonymous
|dw:1347427668995:dw|
anonymous
  • anonymous
good...:)
anonymous
  • anonymous
e, e^2, e^3...will keep decreasing on...
PhoenixFire
  • 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
  • anonymous
yes...now u can tell the answer :)
anonymous
  • anonymous
the ball will stop when v(n)=0...right?
PhoenixFire
  • 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
  • anonymous
and restitution <1..so tell me the answer?
anonymous
  • anonymous
the ball will stop when v(n)=0
PhoenixFire
  • PhoenixFire
v(n)=0 \[0=\mu_r^n \sqrt{2gh}\] Right?
anonymous
  • anonymous
but now the question is when v(n) will be zero?
anonymous
  • anonymous
yes...are u noticing something? e, e^2, e^3... with each v(n) for n=1,2,3,4....
anonymous
  • anonymous
and e=0.85 which is less than 1
anonymous
  • anonymous
now tell me... e^n will approach zero when n approaches....?
PhoenixFire
  • PhoenixFire
infinity
anonymous
  • anonymous
yes...perfectly right:)
anonymous
  • anonymous
so the ball will bounce infinite times...
PhoenixFire
  • PhoenixFire
well i got that answer like 3 hours ago... but i figured that there would be an actual answer.
anonymous
  • anonymous
ok
anonymous
  • anonymous
actually after each collision h keeps on decreasing...and theoretically h will approach zero after infinite bounces..
PhoenixFire
  • 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
  • anonymous
but practically ball can't go up to a size of atom after some bounces for large n
anonymous
  • 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
  • Vincent-Lyon.Fr
What you CAN work out is not number of bounces (infinite), but total time taken.
PhoenixFire
  • PhoenixFire
@Vincent-Lyon.Fr Out of curiosity how would you go about doing that?

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