The Bouncing Ball
This puzzle was suggested by the following one by Sam Loyd:
The leaning tower of Pisa is exactly 179 feet high. An elastic ball is dropped from the tower and on each rebound rises exactly 1/10 of the height from which it fell. How far will it travel before it comes to rest?
Actually, this question is a bit simpler than Sam Loyd's. Also the Wham-O Super Ball is made of butadaiene rubber and is rather special for the height of its bounce. If you are a math genius perhaps you can tell me how high it bounces as a fraction of the distance it falls.
It's a bit of a curiosity that a bouncing ball actually does come to rest. That is because its motion is described by a series which though infinite, is convergent. The mathematical ball bounces an infinite number of times. Since no actual ball can do that, a real ball probably stops a little sooner.
Quote: puzzlenutIt's a bit of a curiosity that a bouncing ball actually does come to rest. That is because its motion is described by a series which though infinite, is convergent. The mathematical ball bounces an infinite number of times. Since no actual ball can do that, a real ball probably stops a little sooner.
You can do the math even with infinite number of bounces.
The time between two bounces depends on the kinetic energy left in the ball (i.e. how high it will bounce), and that quantity will get smaller each bounce.
The ball will bounce less high each time, but also in a more rapid fashion.
Actually the time between two individual bounces should scale (in a frictionless atmosphere that is) as the sqrt(energy) (calculated from the physics of free fall). If the ball loses a constant fraction q of its energy due to friction each time it bounces, the n'th time interval between bounces should then be proportional to sqrt(q^n) = sqrt(q)^n. The sum of sqrt(q)^n (that is the total time until the n'th bounce happens) is a geometric series, which does converge (to the time after which the ball is at rest).
That said, back to your original question. Time between two individual bounces with a maximum reverting height of h is T = sqrt(8h/g) = sqrt(8*E/mg^2).
So the total time between the first bounce and the time of rest is T0 = sum_n sqrt(8*E0*q^n/mg^2) = sqrt(8*h0/g) / (1 - sqrt(q)).
If we take h0 = 3 feet we overestimate the stated time by the first rising time (which is not included in the text, since we count the time from the first maximum height), which is sqrt(2h0/g), so 16s = sqrt(8*h0/g) / (1 - sqrt(q)) - sqrt(2h0/g), solving to q = 89.76%.
Increasing h0 to 10 feet with the same q yields then 29.21 seconds.
The logic is initially the ball (ignoring friction) accelerates at g (from rest) and hits the ground at a certain velocity V and after a time T. At all stages, before the bounce, the Energy is the same.
It then reverses such that it reaches a height of 1/10. The energy at this stage is therefore 1/10 of that when it started, so the height is 17.9 feet. Therefore by conservation of energy, the velocity at the ground was SQRT(1/10)*V ). Similarly you can work out the time since the velocity drops by the constant g, so is proportional to the starting speed, i.e. SQRT(1/10) * T * 2 (remembering to multiply by 2 as it goes up and then comes down).
Thus time gradually reduces, by a factor of SQRT(1/10). Total Time = T + 2 T (SQRT(1/10)+1/10+....). = 2T(1+x+x2...)-T = 2T/(1-x) - T
T = SQRT(D/g*2) = 3.344...
Total = 6.44s
For it to be 16 seconds, the ball must keep about 43% of its energy after bouncing.
Quote: MangoJ
Increasing h0 to 10 feet with the same q yields then 29.21 seconds.
I think you have correctly answered both questions.
A discussion of the bouncing ball from both the point of view of the distance traveled and the time spent bouncing may be found at http://www.sosmath.com/calculus/geoser/bounce/bounce.html
Curiosity? Not really.Quote: puzzlenutIt's a bit of a curiosity that a bouncing ball actually does come to rest. That is because its motion is described by a series which though infinite, is convergent. The mathematical ball bounces an infinite number of times. Since no actual ball can do that, a real ball probably stops a little sooner.
A "mathematical" ball generally deals with speed, gravity and distance.
A real ball also has to deal with friction. THAT'S why it eventually stops.
Quote: DJTeddyBear
A real ball also has to deal with friction. THAT'S why it eventually stops.
The mathematical ball stops too because its motion is defined by a convergent infinite series. A convergent series has a finite value even though it has an infinite number of terms. I suggest that rather than puzzling about mathematical questions such as how you can't take the square root of a negative number on your calculator you buy a Wham-O Super Ball and bounce it on a hard surface.
Quote: paisielloIs it friction? or is it the fact that the collision with the ground is, like all collisions in the real world, not 100% elastic? It is a sort of internal friction.
An elastic collision is defined as a collision without any energy dissipation. Hence a non-elastic collision is a collision with some energy dissipation.
Some people call energy dissipation simply "friction", as it usually also reduces linear momentum.
The reason the ball stops are two-fold: First the amount of energy dissipation, which depends on the elastic properties of the ball and the surface it bounces off. Second the geometric series in the bounces (which comes from the kinematics) which guarantees a full stop even if the dissipated energy at each bounce decrease, the rate of bounces increases.
Actually there is a third mechanism: the turbulences created while traveling through the air. And a fourth mechanism: the radiation of accoustic waves at each bounce. These do contribute to the energy dissipation.
Quote: puzzlenutThe Wham-O Company sells a rubber ball that will bounce for 16 seconds when dropped from a height of 3 feet. How long will it bounce when dropped from a height of 10 feet?
This puzzle was suggested by the following one by Sam Loyd:
The leaning tower of Pisa is exactly 179 feet high. An elastic ball is dropped from the tower and on each rebound rises exactly 1/10 of the height from which it fell. How far will it travel before it comes to rest?
Actually, this question is a bit simpler than Sam Loyd's. Also the Wham-O Super Ball is made of butadaiene rubber and is rather special for the height of its bounce. If you are a math genius perhaps you can tell me how high it bounces as a fraction of the distance it falls.
It's a bit of a curiosity that a bouncing ball actually does come to rest. That is because its motion is described by a series which though infinite, is convergent. The mathematical ball bounces an infinite number of times. Since no actual ball can do that, a real ball probably stops a little sooner.
Part 1: I get 29.21 seconds also (h2/h1)^0.5*T
Part 2:
Quote: puzzlenut
The leaning tower of Pisa is exactly 179 feet high. An elastic ball is dropped from the tower and on each rebound rises exactly 1/10 of the height from which it fell. How far will it travel before it comes to rest?
I remember this from ninth grade math.
If there is ANY distance, however minute it is, then one tenth of that distance, however minute it is, will always mean the ball keeps bouncing because there is some distance left. It never reaches zero by that definition which of course is very philosophical but rather absurd from a point of view of physics.
