Bouncing ball
I've just found this. I haven't got an official answer, but here goes:
A ball is dropped from a height of 1 m on to a surface. During the bounce, the ball loses half of it's energy. How long does the ball bounce for?
Assume everything else is ideal.
(I know Ragknot will have a field day with this.)
A ball is dropped from a height of 1 m on to a surface. During the bounce, the ball loses half of it's energy. How long does the ball bounce for?
Assume everything else is ideal.
(I know Ragknot will have a field day with this.)
Labels: funphysics
posted by Chris at
11:46 AM
15 Comments:
It depends what you mean by "everything else" being ideal.
If both the ball and the surface it is bouncing on are completely non-elastic, i.e. there is no deformation at the time of contact, and there is no air friction (as in a vacuum) then it would bounce indefinitely.
However, the fact that the ball loses half its energy each bounce inplies that either it is in a dense fluid/gas or that there is some deformation of either the ball or the surface.
I'm not sure how to deal with the former case, but in the latter situation the height of the bounce would eventually become less than the amount of deformation and it would then cease to bounce.
I'm sure someone out there could explain this in better scientific terms!
Hi Wiz. Everything else ideal - e.g. it's in a vacuum,, deformations are neglible - i.e. more of a maths problem than a physics problem. I partly covered that by saying the energy is lost during the bounce, and not during the fall.
Hint calculate the time - then deal with a "paradox".
it has to do simply with ball cynetic energy ... i'm not english so... let's try. the energy owned by the ball will get half every bounce. that means that the height reached by the ball is the half every bounce cuz there is a direct proportionality with cynetic energy and the "bouncing coefficient". that means tht the ball will never reached the amount 0 of its energy becaus it is not possible just by getting everytime the half. the ball will bounce forever but reach the half the hight every time it bounce.... woah... my english is awful... hope you understood
.... so what's that in seconds?
time ~= 0.9s
explaation to follow
Cam
forever
Hmmm. Quite a wide range of times there. Anyone want to make a bet on the result? ;)
Ok, I will make a guess, since it loses 1/2 of it's energy, is will bounce 1/2 the height, in 1/2 the time for the next bounce.
But as the time per bounce becomes infinitely small, so does the bounces. When the bounce becomes to small to measure, The ball will appear to stop, and you can stop counting the unmeasurable split seconds.
Hi Ragknot. You're right about half the height, but that makes the time become 1/sqrt(2) of the previous time. I thought you were going to give me a harder time than that.
I'm going to bed now.
... oh, and you mean infinitesimally small - that'll teach you not to laugh when people say circle instead of disc.
Bouncing ball
Given:
-ball at height of 1m
-ball loses half of energy each bounce
call Ho height before bounce and Hb after bounce
Energy of ball drop from a height is E= m*g*h
0.5(m*g*Ho)=m*g*Hb
m*g are constant
Hb=05*Ho
distance travelled by an accelerated object is
distance =integral (velocity)
velocity= integral (acceleration)
velocity= integral(a) =a*T+Vo (vo is the constant)
distance=integral(velocity)=0.5*a*T^2+Vo*T+x (x is constant for starting position)
d=0.5*a*T^2+Vo*T
a=g
Drop
Ho=0.5*g*To^2
To= sqrt(2*Ho/g)
Rebound
0.5*m*v^2=m*g*Hb
v=sqrt(m*g*Hb/0.5*m)=sqrt(2*g*Hb)
Hb=-0.5*g*Tb^2+ sqrt(2*g*Hb)*Tb
0.5*g*Tb^2=sqrt(2*g*Hb)*Tb
Tb= sqrt(2*g*Hb)/(0.5*g)= sqrt(2*g*Hb/(4*g^2))
Tb= sqrt(Hb/(2*g)) = sqrt(Ho/g)
Redrop
Hb= 0.5*Ho= 0.5*g*Tb^2
Tb= sqrt(Hb/(2*g))=sqrt(Ho/g)
Time of Rebound+Redrop=2*Tb
Tb/To= sqrt(Ho/g)/( sqrt(2*Ho/g))=1/(sqrt(2))
Thus making the series
Ttotal= To+2*To/(sqrt(2))+ 2*To/(sqrt(4))+ 2*To/(sqrt(8))+ 2*To/(sqrt(2^n))....
sqrt(2)*Ttotal= sqrt(2)*To+2*To+ 2*To/(sqrt(2))+ 2*To/(sqrt(4))+ 2*To/(sqrt(8))+ 2*To/(sqrt(2^(n)))....
(sqrt(2)-1)*Ttotal= sqrt(2)*To+2*To-To
(sqrt(2)-1)*Ttotal =(sqrt(2)+1)*To
Ttotal= To*(sqrt(2)+1)/ (sqrt(2)-1)
simplify by multiplying by 1=(sqrt(2)+1)/( sqrt(2)+1)
Ttotal= To*(sqrt(2)+1)* (sqrt(2)+1)/ ( (sqrt(2)-1) * (sqrt(2)+1))
Ttotal= To*(2+2* sqrt(2)+1)/ (2+sqrt(2)-sqrt(2)-1)
Ttotal= To*(3+2* sqrt(2))
From before we have To= sqrt(2*Ho/g)
To=sqrt(2*1/9.81)=0.4515s
Ttotal=0.45158*(3+2* sqrt(2))=0.45158*5.82842
Ttotal=2.63s
Hopefully no mistakes this time.
Answer
Ttotal=2.63s
Cam
Hi Cam. That's the time I was expecting.
So.... where's the paradox?
As far as a paradox goes:
one could say that there is an infinite number of terms for the height of the bouncing ball and conclude that the ball never stops bouncing, but the time required for bouncing is, in fact finite and calculable.
The paradox ceases to be if you replace infinite terms with a ridiculously large number of terms like a googolplex (10^googol=10^(10^100)). The ball stops bouncing a height that is unmeasurable (due to Heisenburg uncertainty principle) and the time is clearly finite.
Cam
Hi Cam. I'd not thought of Heisenberg' uncertainty principle, but I think that is a pretty good response. I'll try to apply it whenever a suitable opportunity arises.
I think Wiz had a fair point about the ball simply not having enough energy to actually bounce, but was more inclined to simply suggest that no real system could be modelled so simply as the question implied.
what planet are you from Cam?? Really? Good work.
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