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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.
velocity of separation = e (velocity of approach)
A little more explanation please?
why are you subtracting 0?
Velocity of approach = V' - zero...which is velocity of ground
sorry...Velocity of approach = V - zero...which is velocity of ground
ball is colliding with the ground..and use this velocity of separation = e (velocity of approach)
similarly velocity of separation = V'-0
okay. so e is the math constant.. what am i solving for?
so 1st calculate V, the velocity of ball from which it hits the ground 1st time..
no...e is the coefficient of restitution...and for ur problem e =0.85(given)
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.
but see 4th line in ur solution..
The ball will bounce an infinite number of times, but the process will have a finite duration.
velocity of separation = e (velocity of approach) it does not mean that... Energy after collision =e( energy before collision)?
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.
But the correct formula is for velocity not for energy velocity of separation = e( velocity of approach)
http://scienceworld.wolfram.com/physics/CoefficientofRestitution.html check this
So our professor gave us the wrong equation... awesome.
I'm studying in New Zealand.
anyway...u can solve this problem by using correct formula
Wait, what do I do now that I've got the velocity before and after collision? vel approach =14.01 vel separation = 11.91
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
t' = time of reaching at the peak after 1st collision
e, e^2, e^3...will keep decreasing on...
general formula \[v_n=\mu_r^n \sqrt{2gh}\]\[t_n=\mu_r^n \sqrt{2h \over g}\] I need to find n.
yes...now u can tell the answer :)
the ball will stop when v(n)=0...right?
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
and restitution <1..so tell me the answer?
the ball will stop when v(n)=0
v(n)=0 \[0=\mu_r^n \sqrt{2gh}\] Right?
but now the question is when v(n) will be zero?
yes...are u noticing something? e, e^2, e^3... with each v(n) for n=1,2,3,4....
and e=0.85 which is less than 1
now tell me... e^n will approach zero when n approaches....?
yes...perfectly right:)
so the ball will bounce infinite times...
well i got that answer like 3 hours ago... but i figured that there would be an actual answer.
actually after each collision h keeps on decreasing...and theoretically h will approach zero after infinite bounces..
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.
but practically ball can't go up to a size of atom after some bounces for large n
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.
What you CAN work out is not number of bounces (infinite), but total time taken.
@Vincent-Lyon.Fr Out of curiosity how would you go about doing that?