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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.
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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0velocity of separation = e (velocity of approach)

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0A little more explanation please?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347425035783:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347424692991:dw

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0why are you subtracting 0?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Velocity of approach = V'  zero...which is velocity of ground

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry...Velocity of approach = V  zero...which is velocity of ground

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ball is colliding with the ground..and use this velocity of separation = e (velocity of approach)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0similarly velocity of separation = V'0

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0okay. so e is the math constant.. what am i solving for?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 1st calculate V, the velocity of ball from which it hits the ground 1st time..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no...e is the coefficient of restitution...and for ur problem e =0.85(given)

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Potential 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
 3 years ago
Best ResponseYou've already chosen the best response.0but see 4th line in ur solution..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The ball will bounce an infinite number of times, but the process will have a finite duration.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0velocity of separation = e (velocity of approach) it does not mean that... Energy after collision =e( energy before collision)?

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0That'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
 3 years ago
Best ResponseYou've already chosen the best response.0But the correct formula is for velocity not for energy velocity of separation = e( velocity of approach)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://scienceworld.wolfram.com/physics/CoefficientofRestitution.html check this

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0So our professor gave us the wrong equation... awesome.

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0I'm studying in New Zealand.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anyway...u can solve this problem by using correct formula

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes...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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347427306605:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0t' = time of reaching at the peak after 1st collision

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347427668995:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0e, e^2, e^3...will keep decreasing on...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes...now u can tell the answer :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the ball will stop when v(n)=0...right?

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0well 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
 3 years ago
Best ResponseYou've already chosen the best response.0and restitution <1..so tell me the answer?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the ball will stop when v(n)=0

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but now the question is when v(n) will be zero?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes...are u noticing something? e, e^2, e^3... with each v(n) for n=1,2,3,4....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and e=0.85 which is less than 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now tell me... e^n will approach zero when n approaches....?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes...perfectly right:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the ball will bounce infinite times...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually after each collision h keeps on decreasing...and theoretically h will approach zero after infinite bounces..

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 3 years ago
Best ResponseYou've already chosen the best response.0but practically ball can't go up to a size of atom after some bounces for large n

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0as 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What you CAN work out is not number of bounces (infinite), but total time taken.

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0@VincentLyon.Fr Out of curiosity how would you go about doing that?
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