Dropping the Ball: Newsletter Challenge (May 2013)

Since we don't know the two masses (unless we look them up in the rule books), and since two independent masses make the problem too thorny; we make the simplifying assumptions that the tennis ball holds the soccer ball down from bouncing, and that all the energy goes into sending up the tennis ball. Then equate the before-and-after gravitational potential energies....

If final height H is measured from the floor to the bottom of the tennis ball,

(m₁ + m₂) × 1 = m₁ (H - 0.22).

Rearranging terms, H = [(m₁ + m₂)/m₁] + 0.22.

Define h as the initial height and H as the bounce height of the tennis ball. I think you are correct.

H = [h * (m2/m1 + 1)] + 0.22

If h = 1 meter, then

H = [1 * (m2/m1 + 1)] + 0.22

How much lead do you have to put into the tennis ball to make that work? Well, off to the gym, I guess.

I'd ask Physicist

Is this in a vacuum? If not, there will be energy lost but the bigger ball will clear the way through the air so the balls will remain in contact on both cases.

If there are no losses, the potential energy at the end should be the same as at the beginning. The balls cannot be modeled as linear springs but we have no way of computing the spring rates.

Based on the lack of information to calculate much else, I deduce that there is a special answer and thence that both return to their original position, the bottom of the tennis ball at 1.22m above the ground.

Good link but it says that the challenge cannot be answered, although I'm not sure that experiment was properly done. The answer I offered was based on it being a special case because there was not enough information to describe the behavior in that link.

About this high....

How high does the Tennis ball bounce on its own? How high does the soccer ball bounce on it's own? Is it that m= "a set number" and multiplied by 1 or 2, or are you just stating that they have different masses? EXAMPLE: m1= 10 x 1, m2 = 10 x 2, or m1= 10, m2 = 25.

What is the air pressure inside the soccer ball?

I can't determine any thing from this information. The best thing about this question in the youtube link. Thank you, USBPORT

Since both balls fall at the same speed, they remain in contact as they are falling. When the soccer ball contacts the ground, the force will be proportional to the sum of the masses of the two balls and the height that they fall. Pressure altitude and humidity must be considered as they will impact acceleration and air resistance.

The part of the soccer ball which contacts the ground will flatten out causing the gas pressure on the inside of the ball to increase. When the pressure increases, the diameter of the ball will also increase. But the expansion will not be equal in all directions. The upper surface of the soccer ball which is in contact with the tennis ball will first move toward the ground. The distance that the upper surface moves toward the ground is proportional to the mass of the soccer ball skin and the mass of the tennis ball. Meanwhile the same phenomena is also happening to the tennis ball. When the pressure of the soccer ball increases to the point that the upper surface begins to expand it will launch the tennis ball upward. The increased pressures inside the tennis ball will contribute to its stiffness and contribute to how much force it contributes toward its upward launch.

The time that it takes for the soccer ball to collapse and then expand (rebound) may be different than that of the tennis ball. If the time constants were the same the forces could be added directly, but they are probably shifted in time from each other. Therefore the amount of force that the tennis ball contributes to the launch can be bound between -100% and +100%.

Other factors that must be considered are: surface conditions of the balls for air resistance, modules of elasticity of ball skins, gas pressures inside the balls, firmness of the ground under the soccer ball. Temperature will impact the air pressures inside the balls and the elasticity of the ball envelopes.

Also, are the balls really falling at the same speed? If we were to ignore the effects of aerodynamics this would be true. The aerodynamics could cause the balls to separate or a slight wind could also cause the balls to separate.

This problem is highly nonlinear and impossible to solve with simple equations. Being an engineer, I propose a series of controlled tests.

Being an engineer, you should assume the balls undergo an elastic collision.

2h+2r = 222cm:

If perfectly elastic and onto a solid surface the rebound height would be the original height, 1m, since the smaller ball only falls 1m. However since it is also on top of a perfectly elastic soccer ball, the soccer ball will store and release the same amount of energy to the smaller ball. Therefore the smaller ball reaches it original height, 1m+2*11cm, and an additional 1m for a total of 222cm. This assumes temperature, air resistance, etc are negligible. The mass of the balls is irrelevant.

I think that the upward velocity of the tennis ball is just shy of three times the downward velocity, based on the 3 elastic reactions alone. Then considering the very bottom elastic reaction is greater than the very top reaction due to the combined mass of the stack, it maybe more. The upward velocity of the tennis ball will determine its maximum height, somewhere around 3 meters.

Building on the idea of Randall but tweaking some details of the formulation:

neglecting air resistance

1 is the ratio of the mass of the soccer ball to its own mass. All equations are simplified by dividing by the mass of the soccer ball, m2. The energy equation has been simplified by multiplying by 2.

This solution seems to me to be the best so far, but it seems to assume that the collision between the ground and the soccer ball occurs instantaneously an infinitesimal instant before the collision between the tennis ball and the soccer ball. Some practical reasons argue that this may be okay, but a solution that explicitly addresses the simultaneous collisions would be better.

I've never actually played with a Newton's cradle, but, I've seen several videos of them. The collisions, however complex, always result in the same overall effect as would occur if all collisions were simple individual ones.

I know that this is only observational rather than analytical evidence: but I can't get my head around a way to prove it analytically.

ideal the two balls don't loose contact (falling the same time with the same speed; they are one body) while falling, than the maximum height is the same as the falling height.

if they loose contact than the maximum height differs with the speed of the falling balls the moment they contact again: maximum height of the tennis ball then is (m2/m1+1)²*1m +2r(SoccerBall)+2r(TennisBall) at the upper boundary of the tennis ball.

Assuming that the question contains sufficient data for arriving at a solution, let us see how far we can proceed. Given quantitative data is limited to the height h (=1m) and sball radius r (=11cm). Reasonable assumptions could be made for masses m1 and m2 if needed.

Actual distance of fall is h (=1m) for both sball and tball, plus a small amount due to local flattening of the spherical surfaces for the duration of contact when bounce occurs. Let us neglect this extra height, at least for a start, and take the fall distance as 1m for both balls.

Each ball has its own datum level for zero potential energy, from which they are raised by 1m, so that we have PE1 = m1.g.h, and PE2 = m2.g.h.

We are told that the collisions are elastic, which implies that no mechanical energy is lost/dissipated. After making some appropriate* assumptions we can state that the combined energy of the two balls will remain as PE1+PE2, with the the distribution between potential and kinetic energy varying with time. There is also a phase of stored elastic energy during the impact.

At this point a little digression may be useful. If there is only one ball being dropped, a fully elastic impact will by definition send it back to its original height, and it will continue to bounce indefinitely (with PE+KE = constant) until someone stops it or some dissipative energy leakage occurs.

For the present challenge we assume a one-dimensional straight line path for the mass centres of the two balls. This rules out glancing impact, friction effects and rotational energy terms. When the two balls hit the ground, the bottom of the sball gets progressively flattened and released over a definite time interval as the ball decelerates and reverses direction, while the air/gas inside gets adiabatically compressed. The elastic membrane also stretches a bit and stores a small part of the energy momentarily. All this gets returned to the ball as KE when contact is lost.

What happens between the tball and sball? In principle the situation is similar, but the extent of flattening and duration of contact would be much less than at the ground. The impact is against a decelerating body, so the transient mechanics of motion would be a bit complex, but the tball will take off before the sball leaves the ground, and bounce back to its original height. [this is where I differ from jerryglen in post #13]. Then when it is falling again it meets the rising sball for a second mid-air encounter and bounces back once more while the sball velocity gets diminished in accordance with the conservation laws for momentum and mechanical energy.

My conjecture is that there will be a continuing sequence of bounces in random positions (perhaps predictable if sufficient data is assumed), but the tball will always go back to its original height of 1.22m from the ground. Someone who is less mathematically challenged than me could investigate whether a series of little mid-air bounces will bring the pair back to their original starting position after the first ground impact. However that may be, the balls will keep bouncing ad infinitum and ad nauseam.

This is of course an approximation, because the surface deflection during impact has been neglected, but I'll leave it to the CR4 pundits to judge whether the original challenge has been adequately addressed in my qualitative treatment (provided anyone has the patience to wade through this post).

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* For the benefit of the ubiquitous nit-picker community in CR4, I should perhaps try to spell out some of the implied assumptions in greater detail.

The only external 'force' on the balls is due to a constant value of 'g' (local variations, and effects of earth rotation or interventions by Coriolis, Einstein, et al being ignored).

The 'ground' is massive, flat and hard compared to the bouncing objects.

Both sball and tball are hollow spherical shells made of some impermeable elastic membrane of constant wall thickness, filled with some ideal gas at sufficient pressure to achieve the necessary bounce quality.

The surrounding medium (assumed to be air) is still, has constant temperature and humidity, and is in thermal equilibrium with the sball and tball. Aerodynamic drag on the falling objects is considered negligible (or else a vacuum medium could be assumed).

[Anyone who is more seriously inclined is welcome to extend this list to cover electrostatic and electromagnetic conditions, sound decibel levels, quantum effects, or whatever.]

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This challenge question seems to be a reincarnation of the one on bouncing balls which was posed in March 2007, with the essential difference that there the balls were solid. That changes the entire nature of the problem. Compression of gas in a hollow elastic sphere is quite different from shock waves propagating from the point of impact at the speed of sound through a solid sphere. The theory behind Newton's cradle may not be so simple!

I cannot find any fault in Randall's analytical treatment in Post #17. My qualitative approach does not consider the ratio of masses, but there is a tacit assumption of m1 << m2. My arguments will break down if m1 is close to or exceeds m2! I've just taken the challenge question literally. A general solution is for the theoreticians, not an ornery engineer like me.

=TeeSquare=

and bounce back to its original height

You really should go and try it.

Thanks for the feedback. I take it you (and others) would have ascertained that in reality the tball bounces (much?) higher than I had predicted. I'm not presently able to try the experiment myself (No balls! Arranging the set-up would be a major logistics exercise for me now.)

But I was prepared for this kind of response. My tentative 'conjecture' was based entirely on the preceding assumptions and explanations. Something was obviously incomplete or wrong there. I'm all too aware of the pitfalls of theoretical predictions, as much of my learning, insights and experience as a design engineer has come from being proved wrong in practice!

That aside, what could be the flaw in my previous arguments? A closer look at the problem suggests that in addition to the air compression and increase in membrane tension which I assumed would temporarily absorb all the kinetic energy, there would also be a shock wave travelling around the sball from bottom to top. This may arrive as a concentrated pulse which can give a substantial jolt to the tball.

I suppose there would be some treatise or monograph dealing with vibrations of spherical shells which deals with the mathematics of such situations, but it would lie far beyond my competence. I wouldn't know whether a 'soft' or 'flexible' membrane would behave in the same way as a rigid shell.

Anyway, the fact is that there is at least one more significant term in the energy balance during the impact period. How these are distributed as a function of time, and what proportion of the total gets transferred to the tball and when, would be more like a research problem. I guess it would involve advanced mathematical techniques, and several assumptions regarding elastic constants, etc. I would say that lies beyond the scope of this challenge.

My verdict for now is that there is no 'reasonable' solution which can be derived from the basic mechanical principles of the conservation laws and elastic impact. Maybe the official answer will throw up a surprise. =TeeSquare=

I share your difficulty at arriving at an intuitive way of analysing what I can only describe as a compound bounce. However, observation has shown that any compound bounce can be treated as a sequence of simple bounces.

Treating this as ball 2's bounce followed immediately by collision of the upward-moving ball 2 with the downward moving ball 1, we have (assuming 1 m is the distance of the CM of ball 2 from the ground*):

Ball 2's post-bounce upward speed = its speed at ground impact = sqrt(2gh) = sqrt(2*9.81*0.89) = 4.179 m/s.
Ball 1, with a downward velocity of 4.179 m/s, immediately collides with ball 2.
In an elastic head-on collision, the post-collision opening speed of the bodies equals the pre-collision closing speed. The opening speed vo, which equals 2*4.179 = 8.358 m/s, is divided between the two bodies as follows:
vo1 = vo*m2/(m1+m2)
vo2 = -vo*m1/(m1+m2)
(Since vo has no direction, the two body vos must have differing signs. Since ball 1 is moving upward we make it positive.)
The actual speeds v are obtained by adding the speed of the CM of the two-body system, which due to momentum conservation equals its downward speed, 4.179 m/s.
So the upward speed of ball 1, v1 = 4.179 + 8.358*m2/(m1+m2) m/s.
Since ball 1 starts from a height (now we have to assume it's the height of the bottom, since no radius was given) of 0.22 m, it reaches a height of 0.22 + v1^2/(2g) m.

*If it's the bottom of ball 2 that's 1 m from the ground, replace 0.89 by 1, 4.179 by 4.429 and 8.358 by 8.858.

If we call the initial impact speed vi, the speed ratio v1/vi = 1+2m2/(m1+m2) and the height ratio (not including the starting height) is the square of that.
I think I must disagree with Randall's (#17) graph. I agree with the height ratio of 9 when m1=0. But the shape of the graph suggests the height ratio will approach 0 as m1/m2 increases without limit. Less than 1 seems a physical impossibility. And for m1=m2, I get a height ratio of 4, not 1. When the two equal masses collide, ball 1 still gains upward speed and ball 2 loses some. A ratio of 1 suggests that m2 is assumed to be at rest on the ground, or perhaps that the composite system speed was somehow not accounted for.

When the two balls have the same mass, effectively what happens is this:-

1. The bottom ball hits the ground and rebounds.
2. The two ball collide and reverse directions but keep the same speeds.
3. The bottom ball hits the ground and rebounds, and, follows/tracks the top ball up to the starting point.

Again, as I've said in the past, analysing the "compound bounce" is not very intuitive, but, observation shows that the overall effect is the same as a sequence of simple bounces.

_{(Look at the picture at the top of the thread: the height h = 1m refers to the height of the bottom of the soccer ball at the start of the experiment)}

Thanks, Randall. Your clear explanation made it obvious I fired before aiming. Energy wasn't conserved, etc. My error was assuming that the ground bounce changes the sign of the total system momentum whereas it (obviously) only changes the sign of ball 1's momentum. In the m1=m2 case, this means the system momentum becomes zero after the bounce and the validity of your graph is evident. Although in this case there is no 2nd collision after ball 1's 2nd bounce, a program I wrote shows that for a very small mass ratio m1/m2 (in the approximate range 0 to 0.3), at least four bounce-collision pairs are possible.
On the question of a bounce-then-collide analysis vs. one that integrates both events, I think the integrated approach would be preferable in the situation of balls with high compliance, i.e. our situation, since the balls are in contact with each other for most or all of the bounce-collide phase. But it seems complex and difficult.
There's an interesting (free!) paper (http://www.physics.usyd.edu.au/~cross/PUBLICATIONS/38.%202BallBounce.pdf) that compares bounce-then-collide with an integrated model involving a spring between the ground and m1, and another between m1 and m2. Each spring is attached to either but not both of the bodies adjacent to it so that compression can occur but not extension. The state progression in time must be solved by numerical integration.
Choosing the spring constants k1 and k2 is critical. The paper notes that energy division between the two masses is a function not only of their mass ratio m1/m2 but of the spring stiffness ratio k1/k2. This can result in ball 2's bounce height deviating considerably from the value yielded by bounce-then-collide. With elastic balls, the values of k can be nonlinear due to sphericity and denting. It appears that a lot of measurement work is needed to make the integrated model useful.

Thanks kirchwey,

I'm hoping to try the "springs experiment" in phun when I get the time: so I'll hold off reading that article 'til I've done it then compare results.

Could anyone who has looked at the AJP article give a quick synopsis of the analysis used in the paper, just so that the rest of us can decide whether or not to plunk down the cash to look at it in detail for ourselves?

I am no mathematical wiz, nor a brilliant deducer of numbers,(only got through high school algebra with b's) but here goes. My mind works with true values. I'm a welder/fabricator who works with his hands. Put these two balls in front of me and away I go. The general consensus in my brain considers the following; 1(The #'s don't lie. m1,m2 etc. are all inconsequencial, al be it, 0's to me because of the lack of informative computational values.) The tennis ball is a solid mass, with a weight factor, whereas the soccer ball is hollow, only determinant by its external thickness of the material (ie; shell) and whether or not it is inflated. Hollow mass is pre-determined by outside sources like whether or not the air within the ball is precisely the same as the outside ambient pressure and temperature. Internal/external temperatures and pressures directly effect the elasticity of any sphere. Given both balls equal values of pressure and temperature, more adjustments need to be considered. The soccer ball will, under normal physics, loose directional force once it impacts the ground. Energy is lost to the ball, and absorbed in the ground. Once lost, its ability to 'spring back' to its origin is a mute point. Energy from the falling tennis ball will be transferred to the ground as well, through its outer shell, because of deformation, being there is no solid mass within the soccer ball to transfer. And one more note. r=11cm? is that the radius of the soccer ball alone, or the combination of both. This directly effects any and all formulas that need to be considered. Thanks for listening. Ray

Sorry people. One mistake. Tennis balls are a hollow sphere as well, just of a different quality shell. It still would make an impact on the soccer ball at point of contact with the earth, and therefore absorb some of the energy released by the soccer ball after deflection of its shell. In order for the tennis ball to bounce beyond its origin, it must in fact absorb all of the energy from the soccer ball.

The question tells you to "Assume the balls undergo an elastic collision.": so you can ignore the composition (hollow solid etc.) of the balls, and, don't assume that any energy is "lost" when the soccer ball hits the ground.

Go and try the experiment with: a tennis ball and a soccer ball; a tennis ball and a basket ball; a pingpong ball and a golf ball, or, two different sized ball bearings. Watch some videos of a Newton's cradle. And, start to work out some basic hypotheses.

_{I don't know why someone marked your posts as off topic, it wasn't me.}

The problem is unsolvable without specific modulus of elasticity information of the interacting objects. I believe empirical testing to be extremely difficult as ideal vertical alignment is critical. Any off axis results would be meaningless, and wide ranging.

Bill

Time for a beer, then?

Congratulations CR4 on another wrong answer. Just like Februarys 'A Balance of Power' where we had antigravity, now we have energy creation.

This logic is flawed in a number of ways. First, the author assumes that the soccer ball will have the same velocity immediately after it hits the ground as just before it hits the ground. Second, must take into account the conservation of momentum. He also assumes that the masses are meaningless. And this will somehow make the tennis ball bounce over 9 meter high?

So let's run some numbers (I'm using these masses as an example). Tennis ball: m1=0.1kg Soccer ball: m2=1kg

Before bounce. Soccer ball Pe(soccer) = m2gh = 1*9.81*1 Pe(soccer) = 9.81J

Tennis ball Pe(tennis) = m1gh = 0.1*9.81* Pe(tennis) = 0.981J

Total energy is 10.791J

For either of these objects, they have only been lifted 1m and only fall a height of 1m.

Since the author assumes the soccer ball velocity is the same after collision, its final height must be the same as when it started; 1m.

And so

Pe(soccer) = m2gh = 1*9.81*1 Pe(soccer) = 9.91J

The tennis ball Pe is then

Pe(tennis) = m1gh = 0.1*9.81*9.22 Pe(tennis) = 9.045J

So the total Pe of the system after the bounce is then: Pe(soccer) + Pe(tennis) = 9.91J + 9.045J

Pe(total) = 18.955J

The energy of the system after the bounce is greater than before the bounce which is clearly impossible.

The authors' equations don't take into account the mass of the soccer ball. Think of it this way. What if the upper object (tennis ball) was 1000kg and the lower object (soccer ball) was 0.1kg. According to the authors formula this 1000kg object would get shot up 9.22m after falling 1m and bouncing against a 0.1kg object.

The only way the tennis ball can go higher than its initial starting point is have gained energy. And the only place to gain energy is from the soccer ball. And that being the case, the soccer cannot bounce up to its original starting position. Its velocity will be less after the bounce and it will bounce up to a height of less than 1m. This lost energy then goes into the tennis ball to increase its height. The final height of the tennis ball is related to the ratio of the two masses.

The authors' answer assumes the tennis ball has negligible mass compared with the soccer ball and in that case the final height would be 9.22m.

The ratio of a tennis ball to a soccer ball is approximately 0.133. Soccer ball = 0.420kg and the tennis ball = 0.056kg. See post #17 to find the height of the tennis ball.

Post #2 provided the correct answer on 5/1 with no fluff.
For these problems is is expected that we assume perfectly elastic collisions, no losses from friction or drag, and obviously "conservation of energy".

When
m2 = 0.448kg soccer ball at h=1m
m1 = 0.056kg tennis ball at h=1.22m
m2/m1 = 8

conservation of energy equation reduces to

H = [(m1+m2)/m1)] + 0.22

same as

H = [1 + m2/m1] + 0.22

H = 9.22 meters

also see posts #6, #10, #28
I dispute the accuracy of #17 (though I also frequently use speadsheet solutions ) because when the masses are equal (m2/m1 = 1) one ball should remain on the ground and the other should bounce to 2 (+0.22) m, NOT 1 (+0.22) m as shown in the chart.

These formulas will not give the correct answer.

H = [(m1+m2)/m1)] + 0.22

same as

H = [1 + m2/m1] + 0.22

Let's say m2 = 100kg and m1 = 1kg

H = {1 + 100/1) + 0.22 So H = 101.22m

The higher the mass ratio, the higher the second object will go. This implies there is no limit on how high the second object can go.

The only way these formulas work is to assume that m2 hits the ground and stops. And all of its energy is transferred to m1 sending it upwards. This is clearly not what happens.

The soccer ball (or m2) bounces back upwards and so retains some or most of its energy.

The only way these formulas can work is if the soccer ball were to hit the ground and stop. And then all of its energy is transferred into the tennis ball sending it upward. If the soccer ball were to bounce up to 1m, then no energy is transferred to the tennis ball. It all stays with the soccer ball. The tennis ball doesn't go any higher. If the soccer ball hits the ground and stops, somehow all of its energy is transferred to the tennis ball sending it up to 9.22m.

The highest the second object can go is 9.22m. The answer alluded to this fact. The speed of the tennis ball is 3v. An object dropped from 1m will have a v of 4.429m/s. And 3v is 13.287m/s. A tennis ball moving at 13.287m/s will go 9.22m high (9m + 0.22m).

The true answer lies between these two extremes. The soccer balls' upward velocity cannot be v; this implies it will return to 1m high and would not have transferred any energy. So the soccer ball after collision must be v-k. And if it is v-k then the tennis ball must be less than 3v. Less than 3v would send it up to less than 9.22m.

The maximum height must always be less than 9.22m

#17 shows the maximum height of 9m plus remembering to add 0.22m for the diameter of the soccer ball.

I consider the "correct" answer given as the official solution to be numerically correct, though lacking a bit in explanation, and requiring a slight tweak in the way it is expressed.

The problem deals with a limit, based on two infinitesimals, which (thankfully) appear to be separable. First, referring to the balls by their respective mass, the mass m2 is considered to be infinitesimally smaller than the mass of ground, so its collision with the ground completely reverses its speed. That is how the relative speed between m1 and m2 becomes 2v, where v is the speed of either ball (relative to the ground) just prior to impact. Then, in the limiting case for highest bounce, mass m1 is taken as infinitesimally smaller than m2 which means, in the collision between m1 and m2, that a) the relative velocity between m1 and m2 is completely reversed as well, and b) m1 has no effect on the relative velocity between the ground and m2, which remains v. Adding the relative velocity between m1 and m2 to the relative velocity between m2 and the ground, the velocity of the infinitesimally small m1 (relative to the ground) is 3v, which is the limit of possible speed, from which we can calculate the limit of possible height.

Strictly speaking, because m1 will have an actual, finite mass, the limit of possible height is not the same as the maximum height to which the ball can bounce; it might be better to say that the tennis ball bounce cannot exceed 9.22 m measured from the ground to the bottom of the ball.

The correct answer didn't even answer the question. The question was 'Determine the maximum height that the tennis ball bounces'. The ratio of an official size tennis ball mass to an official soccer ball mass is approximately 0.13. Using the graph from #17, the maximum height is about 7m.

I agree if the mass of m₁ is insignificant to the mass of m₂, then the absolute maximum height of m₁ is 9.22m (9m + 0.22m).

But the equations in the answer don't even have either m₁ or m₂ in them.

½m₁(3v)² = m₁g(H - 2r) m₁ is divided out from both sides.

Then,

H - 2r = (3v)² /2g

Then,

H = 2r + 9v²/2g = 2r + 9(2gh)/2g

And

H = 2r + 9h = 0.22 + 9(1) = 9.22m

Where is the mass of m₁ and m₂? They are not even in the final equation.

This is not the height of the tennis ball. This is the height of m1 if m₂ >>>>m₁.