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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
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

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akash123Best ResponseYou've already chosen the best response.2
velocity of separation = e (velocity of approach)
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
A little more explanation please?
 one year ago

akash123Best ResponseYou've already chosen the best response.2
dw:1347425035783:dw
 one year ago

akash123Best ResponseYou've already chosen the best response.2
dw:1347424692991:dw
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
why are you subtracting 0?
 one year ago

akash123Best ResponseYou've already chosen the best response.2
Velocity of approach = V'  zero...which is velocity of ground
 one year ago

akash123Best ResponseYou've already chosen the best response.2
sorry...Velocity of approach = V  zero...which is velocity of ground
 one year ago

akash123Best ResponseYou've already chosen the best response.2
ball is colliding with the ground..and use this velocity of separation = e (velocity of approach)
 one year ago

akash123Best ResponseYou've already chosen the best response.2
similarly velocity of separation = V'0
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
okay. so e is the math constant.. what am i solving for?
 one year ago

akash123Best ResponseYou've already chosen the best response.2
so 1st calculate V, the velocity of ball from which it hits the ground 1st time..
 one year ago

akash123Best ResponseYou've already chosen the best response.2
no...e is the coefficient of restitution...and for ur problem e =0.85(given)
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
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.
 one year ago

akash123Best ResponseYou've already chosen the best response.2
but see 4th line in ur solution..
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
The ball will bounce an infinite number of times, but the process will have a finite duration.
 one year ago

akash123Best ResponseYou've already chosen the best response.2
velocity of separation = e (velocity of approach) it does not mean that... Energy after collision =e( energy before collision)?
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
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.
 one year ago

akash123Best ResponseYou've already chosen the best response.2
But the correct formula is for velocity not for energy velocity of separation = e( velocity of approach)
 one year ago

akash123Best ResponseYou've already chosen the best response.2
http://scienceworld.wolfram.com/physics/CoefficientofRestitution.html check this
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
So our professor gave us the wrong equation... awesome.
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
I'm studying in New Zealand.
 one year ago

akash123Best ResponseYou've already chosen the best response.2
anyway...u can solve this problem by using correct formula
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
Wait, what do I do now that I've got the velocity before and after collision? vel approach =14.01 vel separation = 11.91
 one year ago

akash123Best ResponseYou've already chosen the best response.2
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
 one year ago

akash123Best ResponseYou've already chosen the best response.2
dw:1347427306605:dw
 one year ago

akash123Best ResponseYou've already chosen the best response.2
t' = time of reaching at the peak after 1st collision
 one year ago

akash123Best ResponseYou've already chosen the best response.2
dw:1347427668995:dw
 one year ago

akash123Best ResponseYou've already chosen the best response.2
e, e^2, e^3...will keep decreasing on...
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
general formula \[v_n=\mu_r^n \sqrt{2gh}\]\[t_n=\mu_r^n \sqrt{2h \over g}\] I need to find n.
 one year ago

akash123Best ResponseYou've already chosen the best response.2
yes...now u can tell the answer :)
 one year ago

akash123Best ResponseYou've already chosen the best response.2
the ball will stop when v(n)=0...right?
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
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
 one year ago

akash123Best ResponseYou've already chosen the best response.2
and restitution <1..so tell me the answer?
 one year ago

akash123Best ResponseYou've already chosen the best response.2
the ball will stop when v(n)=0
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
v(n)=0 \[0=\mu_r^n \sqrt{2gh}\] Right?
 one year ago

akash123Best ResponseYou've already chosen the best response.2
but now the question is when v(n) will be zero?
 one year ago

akash123Best ResponseYou've already chosen the best response.2
yes...are u noticing something? e, e^2, e^3... with each v(n) for n=1,2,3,4....
 one year ago

akash123Best ResponseYou've already chosen the best response.2
and e=0.85 which is less than 1
 one year ago

akash123Best ResponseYou've already chosen the best response.2
now tell me... e^n will approach zero when n approaches....?
 one year ago

akash123Best ResponseYou've already chosen the best response.2
yes...perfectly right:)
 one year ago

akash123Best ResponseYou've already chosen the best response.2
so the ball will bounce infinite times...
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
well i got that answer like 3 hours ago... but i figured that there would be an actual answer.
 one year ago

akash123Best ResponseYou've already chosen the best response.2
actually after each collision h keeps on decreasing...and theoretically h will approach zero after infinite bounces..
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
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.
 one year ago

akash123Best ResponseYou've already chosen the best response.2
but practically ball can't go up to a size of atom after some bounces for large n
 one year ago

akash123Best ResponseYou've already chosen the best response.2
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.
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
What you CAN work out is not number of bounces (infinite), but total time taken.
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
@VincentLyon.Fr Out of curiosity how would you go about doing that?
 one year ago
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