Bouncing Balls: Newsletter Challenge (03/06/07)

Yes, you will be hit in the eye with the smaller ball.

Since the mass of the smaller ball is negligible compared to the larger ball and since the collision is elastic, the momentum of the larger ball will transfer to the top smaller ball and drive it upward to your perched location.

But the puzzle did not say that the top ball was the smaller ball of neglible weight! The weightless ball could be the lower ball.

NO a direct hit is 1000/1

Ddin's the first sentence say: "A very small ball is on the top of another ball"

Yes, but it didnennt say which was larger; a very small ball can be atop a smaller ball, don't you see?

It said which was heavier. Also, the first ball is 10-cm in diameter. Would this be classed as 'very small'?

Quote:

A very small ball is on the top of another ball with a radius of 5 cm; the mass of the smaller ball is negligible compared to the larger ball.

UnQuote:

yes, it does.

it did state the Lower Ball was the Lager ball and the drop height Was measured from the base of the larger ball

You better read the question again!

How do you know that the smaller ball will reach a height of 4.5 m or more?

You have to prove it!!

See #23?

Here the question identifies three conditions :

1. There are two balls.

2. One ball is small and another one is smaller.

3. Smaller ball weight is negligible to the larger ball.

Conditions specify that the small ball is on top of the smaller ball.

Now when they are dropped, inspite of the "smaller ball" (lighter) on the lower side, the "small Ball" (heavier) will touch the ground earlier due to its heavier weight than the "smaller ball" and will rebounce.

As both balls are moving in same line of motion and due to point # 3, the smaller ball (i.e. the ball which is lighter) will still be in downward motion and shall touch the ground later then the small ball in other conditions.

But however as in the mean time the small Ball (being the heaviest) will touch the ground earlier and will rebounce and change it motion 180 Deg opposite upwards against the gravity, and while trying to go upward will collide with the smaller ball (who was still in downwards motion due to its ligher weight) will transmit its total "motion energy" to the "smaller ball" in midway resulting changing the smaller Ball's direction from downward to upwards with the higher velocity (Which it has received from the upcoming and heavier "small Ball") than what it (the smaller ball) had while coming downward due to mass,stiffnes and damping cofficient of the balls.

Now the "small ball" from the position where its all energy have been dissipated will fall to the ground, but the "smaller ball" due to this "attained energy" status will move upward after changing direction 180 Deg at a higher speed in line with the same upward path of motion of small ball (heavier) and will hit the person who directly above their line of motion path.

It's painful to see how much you're neglecting, but that my be for a purpose. From my perspective there are three things you're missing:

1, it is specified for the spirit of the question that the 'weightless ball' is positioned vertically on top of the small ball for the duration of the until it's impact with the ground.

2, you assume that the smaller ball has a higher coefficient of drag, causing it to fall faster, but it is also tucked behind the small ball, lowering that. For those nascar fans out there, it's similar to drafting. I will admit, however, that they make no mention of the size, just that the ball of neglegent weight is smaller, so I will stick with the assumption that it will remain in contact with the upper edge of the ball.

3, the impact will not occur instentaneously. It was specified that the collision was elastic, but not immediate, which refers to the amount of energy before and after the interaction with the ground. In the time of the collision, the bottom of the small ball will suffer an almost immediate stop and a similarly immediate start when the interaction is over. The dynamics of the system become complex at that time, but it is my suspicition from empirical evidence that the smaller ball will attain twice the impact velocity in the opposite direction as a result of the impact with the ground.

Yes, I tend to agree. but what if there were three balls of desending weight?

If ideal, and each lower ball completes their bounces before the next up hits it, and each balls mass is 'negligible' compared with the one immediately below it:
the speeds should be 1, 3, 5 ,7... times the initial impact speed. Effects of finite ratios can be calculated based on #23, for example

Fyz

I make that 1, 3, 7, 15.....

Agreed - and thanks, careless of me.

Fyz

Quote from #4:

"Now when they are dropped, inspite of the "smaller ball" (lighter) on the lower side, the "small Ball" (heavier) will touch the ground earlier due to its heavier weight than the "smaller ball" and will rebounce."

higher mass (or weight) has no effect on the aceleration of gravity! g is a constant- two objects of similar density and geometry (density & geometry only figures in due to air resistance) will hit the ground at the Exact same time, no matter their mass/weight.

This is one of the first things taught in basic physics. Imho, any assumption of a "rebounce" is faulty as well (if the balls are dropped when touching, they will be touching upon the large balls impact with the ground)

as the mass of the small ball is not given (negligible) and is a critical component to how high it will go after the impacts, it cannot be said if you will be hit in the face or not! if small ball mass = effectively 0 (eactly how heavy is "negligible"?), its rising speed will be infinite, and you won't just get hit, it will pass right through your head!

One of the (few?) things that is clear in the question is that very small ball is on the upper side. I think we can assume this to mean that the 5-cm radius ball is not the "very small ball"

If we assume that the very small ball is not merely above the larger ball, but resting on it, they will separate, because the support force will not vanish instantaneously once the lower ball is released, but will decay gradually*. The separation will be small - but should be sufficient to treat the collisions of the 5-cm radius ball with the ground separately from its collision with the upper ball.
*This is taking the view that perfectly elastic balls could not adhere, because that represents a source of energy loss

Finally, the speeds: the lower ball bounces from the ground with its original speed V. The upper ball is still falling at this same speed V. So the balls are approaching at relative speed 2V. After an elastic collision, they will separate with relative speed 2V. If the lower ball is essentially unaffected by the collision, it will be moving upwards at V. The upper ball is now moving upwards at speed 2V relative to the lower ball - total speed 3V. No unlimited velocity after the first bounce. Of course, in theory, ideal multiple bounces could eventually build up an infinite velocity - but we'd have to have zero air drag and the smaller ball so light that it was really close to 3V, and the height-dependence of gravity made its bounce time exceed three times that of the lower ball. However, given that the smaller ball hits your face each time, that seems unlikely to ever happen (even if it didn't rotation of the earth would cause misalignment before too long)

Fyz

BTW, if the balls were in contact at the time when the lower ball hits, the speed of the upper ball after this first bounce would only be V, because it would be pushed away from the lower ball partway through the bouncing process.

why would the support force not vanish instantaneously? what would cause it to decay gradually? As the gravitational acceleration on both balls is constant, they would acelerate, and fall at identical speeds- I infer this to mean that they would stay touching throughout the fall.

in the calculations using V (velocity), I think there is an oversight. There is no such law as "conservation of velocity"- but conservation of momentum in perfectly elastic collisions is proven. If the balls had identical mass, V(5cm)+V(verysmall) = 2V (masses cancel), but a momentum sum Must include mass : P(total)= M(5cm)*V(5cm) + M(vs)*V(vs)

I say very small ball (VSB) upward velocity would be "infinite", because "negligible" calculates as zero (impossible, yes, but so is infinite velocity- divide by zero I take to mean infinity, in mathematical reality it is just impossible/no soloution)

Even if the balls are touching upon ground impact, momentum would still transfer from the 5 cm ball to the very small ball because the compression shockwave propigating(sp?) through the 5 cm ball starting at its bottom, and ending at its top , where the small ball Then begins its "impact". This shockwave energy would also transfer to the very small ball. With the VSB's zero mass, having any momentum means it must have infinite velocity. If you take "negligible mass" to mean "very small mass", then the VSB upvard velocity can only be described as "very fast" - I do not understand where 4.6 meter rise comes from with not having the mass of the VSB.

Is my logic off somewhere? I can't understand what I'm missing if this problem can be solved numerically.

I think most of this is sorted. The residues: post #82 showed practically what happens of the balls are in contact when they hit the ground. The reason is that the first pass of the bouncing wave in the lower ball merely stops the top surface, and by the time the second pass of the wave has started it moving up again the upper ball will have departed. However, this should not happen unless the top ball is excessively light or the balls are inelastic at low impact. Both ping-pong balls and basket balls rely on the air filling to achieve elasticity, and the energy lost in the shells is a high proportion for small impacts; in addition, the ping-pong ball is rigid compared with the basket ball, which is exactly the wrong way around. The reason is that the separation of the balls is driven by the flexibility of the smaller ball, whereas the time that it takes for the lower ball to complete its bounce is determined by the flexibility of the larger one. (As regards the separation, more flexibility means a greater distance of movement for the same force, and therefore more separating energy => velocity imparted to/by the small ball).

Fyz

Thanks for noting that the two balls are in contact with each other in this perfect world when the larger more massive and lower ball hits the ground eliminating the discussion of the rebounding massive ball slamming into the smaller upper ball and transmitting some of its energy to the smaller ball.

If this secondary collision does not occur, what is the nergy distribution between the two balls - they hit at the same time- change direction instantaneously - and so should rebound instantaneously in a perectly elastic situation. I would expect that they bounce back to the same height neglecting friction, imperfect elasticity etc.

Once you address the pressure wave traveling through the larger ball, aren't you getting away from perfect elasticity?

Just my thoughts....

You are right that we can't combine pressure waves and localised surface deformation without residual losses to acoustic vibration in the ball. But we have to treat no loss as a limiting condition of some loss. I.e., we do the calculations for a particular type of situation, and then change the multipliers so that the loss tends to zero. But the resultant bounce is the same (=3x) regardless of whether the limiting condition is that all deformation is local to the contact point, or all deformation generates a coherent wave that meets at the opposite side of the ball to the contact point. Of course, no real ball bounces silently, so even balls made of lossless material would be lossy.

Fyz

<<in the calculations using V (velocity), I think there is an oversight. There is no such law as "conservation of velocity"- but conservation of momentum in perfectly elastic collisions is proven. If the balls had identical mass, V(5cm)+V(verysmall) = 2V (masses cancel), but a momentum sum Must include mass : P(total)= M(5cm)*V(5cm) + M(vs)*V(vs)

I say very small ball (VSB) upward velocity would be "infinite", because "negligible" calculates as zero (impossible, yes, but so is infinite velocity- divide by zero I take to mean infinity, in mathematical reality it is just impossible/no soloution)>>

Even if the lighter ball has a negligible mass comparing to the other, it can not take from the heavier body more than the heavier body's speed. It's as if the lighter ball would collide with a wall or with the ground. In this case, yes, conservation of the impulse or momentum tanslates into conservation of the velocity and the lighter ball bounces back with the relative speed between the two bodies before the impact.

Continuing the idea, the lighter ball goes up with only 2v and nobody gets hit in the face.

Sorin.

Nul points - all covered before in this thread, but: try bouncing a ball from the ground early in the morning - does it come back at the same speed relative to the big ball that it went down? Is the speed the sum of the big ball's speed around the sun and the relative velocity of the balls? Do you dispute Nicolas Copernicus conclusions?

The larger ball only hits the ground first by virtue of being underneath the smaller ball. Both balls fall at the same rate (-9.81m/s^2). C'mon now, that is high school science. Weight or mass has nothing to do with the rate at which something falls. Its terminal velocity is determined by its mass, shape and size, but 50cm is hardly a long enough drop to reach terminal velocity unless these balls are made from Styrofoam.

This is being really pedantic, but you won't be hit in the face WHEN the balls bounce, you'll be hit AFTER the balls bounce.

A man after my own heart. Get things right, say EXACTLY what you mean.

Sieve?

I think we can assume that very small means a lot smaller than 10 cm (4") diameter.

downward journey:

½mV² = mgh
h = ½
so V = SQRT(g)

Now: if you watch a Newton's cradle for a while you will notice that the results of collisions are identical whether or not the balls are in contact or not when they jointly collide.

So the result of the "bounce" is as if the larger ball has completed its re-bound before the smaller ball hits it.

So upward journey: V_initial = 3*SQRT(g)

mgh = ½ m*9*g
So h = 4.5 m
But the smaller ball starts its upward journey from 10 cm off the ground. So yes: it hits you in the face.

Hi Randall, yes, I think you've got it perfectly right!

My first reaction was that your V_initial = 3*SQRT(g) is wrong and should have been a factor 2, not 3. Then I remembered the sums made for the gravity assist "swing-by" of a spacecraft near a planet, which gives you the equivalent of the factor 3, not 2!

Regards, Jorrie

Randall may have the math correct for his energy calculation, but in collisions it is momentum (= mv) that is conserved, not energy ( = ¹/₂ mv² ) ...(the balls deform during the collision, and that takes some elastic energy, which is where the coefficient of restitution comes into play) .

but again, momentum is always conserved, and all of the big balls upward Mv goes into the top, smaller (negligible mass) ball, sending it soaring.....as guest #1 correctly noted.

Physics is Phun.....

The standard Newtons cradle is a very special case, because with equal size balls there is only one solution that conserves both momentum and energy. This is not the case for different size balls, where the speed distribution can change at each bounce - so we really do need to look at the detail. If the balls are close enough that only the first wave has reached the surface when the smaller ball hits, then the smaller ball will only go back to its original height on the first bounce - but the separation in time that this involves means that everything should be put right by the third collision between the balls. If there is time between the collisions for the wave to go up-down-and-up-again, the first bounce of the lower ball will be complete, and the upper ball will go bounce 3H. For any electrical engineers out there, a joint bounce is somewhat akin to reflections in mismatched transmission lines

Fyz

OK: I had to sleep on this, I must have played with too many balls as a kid (oh alright & as an adult), because I still have a very strong gut feel that the result is the same whether or not the small ball sticks with the larger one on the first "fall" or not.
There's a lot of speculation (and inference) in here, but, as I say it agrees with my gut feel.
1.) I know what the person who posed the question had in mind. That's why he placed the observer 4.5 m above the ground. So my interpretation of "elastic" is the same as his. I also believe that for very high coefficients of restitution the following argument is reasonable.
2.) When a collision occurs between two elastic bodies the interaction at the interface is very closely related to "simple harmonic motion". This means that irrespective of the amount of energy transferred/stored (during impact) there is a time constant associated with any two bodies. Also because the force increases/reduces at exactly the right rate (depending on displacement): there is no ringing.
3.) If the time constant for the "Big" ball with a hard immovable object (granite floor) is T_B and that for the "Little" ball is T_L : then the time constant for the collision will be T_(B,L) = ƒ(T_B , T_L) It is my hypothesis that this function always produces a result which is greater than or equal to the larger of the two individual time constants. [Clearly if the balls are "Identical" then ƒ(T_I , T_I) = T_I ]
4.) Hence: if the balls remain in contact on the downward journey then the "upper collision" always takes longer than the "lower collision".

I can see that point three above is very speculative and contentious: it just seems to coincide with my experience.

I note that some one has tried the experiment with a ping pong ball and a basket ball and produced a contradictory result. I claim that the collision between a ping pong ball and a basket ball is not elastic.

I am not sure if I've read all the posts since last night so I apologise if I am covering ground that someone else already has.

There are some good points there. The "no ringing" approximation is not necessary for perfectly elastic collisions. However, you are probably right that it is what the questioner had in mind. And I agree that, if we make this approximation, the smaller ball will indeed always bounce at the higher velocity (tending towards 3x original as the mass of the top ball reduces).

Can it be a realistic approximation? Certainly, it is easy to see that, for a 'simplified' situations such as a uniform rod bouncing end-on, where is a wave that propagates three times through the rod to develop a bounce. It seems unlikely to me that the rounded part of the sphere that impacts the ground is sufficiently decoupled from the remainder of the sphere to modify this behaviour, but this is just gut feel, and I'm not equipped either to simulate or to measure the effects. Gas-filled spheres will certainly have wave-like properties, and this corresponds to the basket-ball and ping-pong ball that have been tried. (These won't be the most elastic examples you could imagine - but the basic behaviour will be similar)
It is of course possible to create a ball that is quite close to being free of ringing - a hard solid ball with a more compliant (but elastic) outer - but I doubt the question setter had anything that unusual or complex in mind.

In the end, I don't think it matters a lot, so long as we use heavy solid balls, because I expect the balls will naturally separate during the drop - assuming the lower one was supporting the upper one in the first instance.

Regarding the ping-pong ball and the basket ball - definitely not perfectly elastic, but I have no doubt that a highly elastic gas-filled ball would show identical behaviour.

If anyone has a couple of ball-bearings of different sizes to hand, they could perhaps check the true situation by lightly adhering them together before dropping them.

Hi Randall

Perhaps I should have followed your example and taken a good nights sleep before replying. I forgot the effect of the "free" surface of the top of the lower ball on the surface velocity - I now realize (thinking more about the cylinder) that when the wave reaches the top surface, the change in velocity will be twice the change in the travelling wave. So the upward wave will first stop the ball, and the reflected downward wave will start the upwards motion.

Given the applicability of your ball-play - presumably the balls you have been involved with were solid-cored, mostly of similar size and bouncing off each other?

I believe that the reason that the basket ball and ping-pong balls did not work like this must be that the waves in the lower ball that are reflected from the impact with the ground do not all reach the top at the same time. I don't know that this absolutely means that the collisions cannot be elastic, but the conditions for elasticity would be hard to meet.

Finally to return to the issue of elastic collisions - I don't see how you could have an elastic collision with an inifinite mass (ground), unless the ground was either infinitely stiff or infinitely dense, because any deflection of the ground must initiate a propagating wave that will not return. I'm sure that different size spheres could be satisfactory if the relative sound velocities match the relative sizes - but not whether there are other viable solutions.

Fyz

Not enough information. Need coefficient of restitution for both balls and also the ground.

The question stated "If all collisions are elastic".

The real answer to the question truly depends on facts omitted from the question.

ARE WE IN A VACUUM?

Answer yes, the smaller ball.

IS THERE AN ATMOSPHERE?

Answer no. (Energy lost to atmosphere is not returned.)

Yes, you will be hit in the face by the larger ball. You are lying on a roof lower than then the balls are dropped. You must be looking upward seeing the larger ball.

This must be some new system of measurement where 50 cm (0.50 m) is greater than 4.5 m

So now we have roofs that are below the ground. Ah well, I suppose that is one way to pretend to answer the qestion without doing the groundwork.

Sorry Guys, I'm in error. I don't use the metric system on a daily basis.

I've actually seen this done with a tennis ball and a basketball. The smaller ball takes off like a rocket. To put a bit of physics to it, the kinetic energy from the larger ball is transferred (assume a total theoretical transfer) to the negligible mass smaller ball. Smaller mass ball now has greater velocity, able to shoot up and reach the height of your face.

Dave Meador

The smaller ball could in fact be hollow and then absorb the bounce of the larger ball. Also the smaller ball has a smalller wind resistance due to being above the larger which is obtaining all the wind resistance and therefore will hit at the same time.

No. I'm smart enough to move out of the way. :D

_Lloyd_

The real question is that when the larger ball bounces from .5m, will it have generated enough force on the smaller ball to propel it upwards 4.5m (Nine times the height of the original drop). The elasticity of both balls must be taken into account, based upon what material they are made of, and if they are the same material. I.E. Steel has a much lower elasticity than rubber, therefore would not bounce as high. Not enough information given in the question.

See post #39.

Laying on the roof 4.5 meters above the ground means that your face is a bit less than 4.4 meters from the top of the smaller ball (4.5m - (2 x 5 cm) - smaller ball diameter). The following discussion assumes that losses due to friction with air are negligible.

Balls falling

Fall height = 0.5 m means the balls' downward speed is as follows:

0.5m V²=mgh=mg(0.5m) or V=g^0.5 downward

The larger ball rebounds with perfectly elastic results and so its upward speed is also g^0.5.

The smaller ball is travelling downward with speed V=g^0.5. Its perfectly elastic rebound from the now upwardly moving larger ball, moving with upward speed V=g^0.5, produces an upward speed for the smaller ball of 3g^0.5. The relative approach speed between the two balls at the instant of rebound is 2g^0.5 (one upward speed of g^0.5 and one downward speed of g^0.5). Therefore, the resultant smaller ball upward speed relative to the larger ball is 2g^0.5 due to the perfectly elastic rebound. However, the upward speed of the smaller ball relative to the observer and to the ground (both stationary) is the speed of the larger ball plus the relative speed between the two balls.

Smaller ball moving upward

0.5 mV²=mgh=0.5 m (3g^0.5)² =0.5 m(9g)

Solve for h

h=4.5meters

Thus, the smaller ball would travel upward about 4.5 meters but it starts its upward motion 10 cm above the ground and it must have some diameter.

The observer would be hit in the face provided the balls were perfectly aligned so the smaller ball bounced exactly vertically and provided the observer remained stationary. However, the smaller ball would be moving at greatly reduced speed after 4.4 meters of upward travel; one must expect that the observer likely would flinch and avoid being hit. OR the smaller ball now moving at about 1.4 m/s and having negligible mass would not be noticed by the observer when it hit them in the face.

I don't understand how this can be correct -- the conclusion that the smaller ball's resulting velocity is a fixed factor of three after a collision with the larger ball (in this & other inputs) seems to violate the conservation of momentum equation that requires initial & final mass-velocity multiples to be equal:

M1v1i + M2v2i = M1v1f + M2v2f

A number of responses seem to determine the second ball's final velocity independent of any relative mass ratios -- no matter what the relative masses, the smaller ball ALWAYS has the same resulting velocity after a collision?

3 is the limiting case for mass of the upper ball tending to zero. See #83 for fuller solution to the bounce, and #23 for air resistance

If the big ball is filled with explosive, then maybe you will be hit by something. Otherwise, without some sort of stored energy nothing is going to bounce higher than it was dropped from, else its good old perpetual motion.

Whoops...missread this ridiculous question.

(I'd suggested the ball may be filled with explosive)

Who uses cm? You'd have to move ridiculously fast to view the balls from 4.5m when they were dropped from 5m (why mix units?) or do you just let them hit the back of your head on the way down.

Mr A. Pragmatist

You're right that no real pragmatist would try this experiment as posed. Although it allows you to get someone else to drop the balls, you'd certainly prefer to watch from the side (preferably with a cine camera to persuade those who think that transferring the energy to the smaller ball is sime kind of perpetual motion). But regarding units, cm. are good pragmatic units for this sort of problem - right sort of size, and closely related to a simple coherent structure (but it helps to know that 50-cm is 0.5-m, not 5-m). Any volunteers to do the air-loss calculations in feet, inches, lbs etc.?

Provided the mass of the small ball is suitably chosen (not too small, or air resistance will stop it) it could indeed hit you in the face following the first bounce - at least theoretically. Somewhat lighter balls could hit your face afer the second bounce. This could also be the result if the balls and ground have practial coefficients of restitution (e.g. being of steel and granite respectively)

The single bounce would require that "elastic" is taken to mean no losses, the ground is also elastic and flat, and the balls are centred above each other, and both balls are spherically symmetrical, and the balls are dropped without imparting excessive spin, there isn't too much of a breeze - oh - and you are looking directly down onto the balls and don't move you head out of the way. The two-bounce solution requires even better flatness and symmetry, but could be achieved with realistic coefficients of restitution.

The first approximation ignores the action of air and takes the mass of the small ball to be truly negligible, the upward velocity of the small ball after the first bounce will be 3x that of the large ball, so the small ball would be rising to a peak height of (50 x 9 + 10) cm = 4.6 metres. So, in the absence of air, the small ball would hit you in the face after the first bounce. After the second bounce, the velocity of the small ball would be 5x, so it could reach 12.6-metres if unimpeded...

So the remaining question is whether air resistance would allow this.

As a check, I made an initial choice of mass that would allow half the height lost in the first bounce to be due to it having a mass that is significant compared with the larger ball. I.e. the mass of the smaller ball is chosen so that it would rise to 4.55-metres (in the absence of air resistance). This gives a required velocity of 2.984 x the velocity of the larger ball immediately prior to bouncing. The relative mass of the smaller ball is given* by M=(3+2.V/V-V^2)/(1+V)^2, where M is the ratio of the balls' masses, and V is the ratio of the speed of the smaller ball (after first bounce) to the (immediately) pre-bounce speeds. This gives a relative mass of 0.415%, or a smaller radius of 8-mm if the balls are of similar construction.
*Calculated by conserving energy and momentum...

Air loss will be (approximately) the larger of what we would calculate using laminar and CD approximations.

Laminar calculations give the retarding force (Stokes law) as 6.pi.r.eta.v = 6*3.14*8E-3*1.73E-5*v= 2.6E-6*v Newtons. For a bounce height of about 4.5-metres, the velocity is about 9.4m/sec, so the energy loss during the bounce will be about 2/3*9.4*4.5*2.6E-6=7.36E-5 Joules. If the specific gravity of the ball is about 9, the mass will be about 0.0021kg, which would correspond to a reduction in bounce height of about 0.39-mm.

For turbulent calculations, we take the drag coefficient (CD) as 0.1 (polished sphere - from Wikipedia), and the energy loss will be about 1/2*height*CD*air-mass-in-column*initial_KE/sphere-mass = 1.27-cm

So, the air loss is predominantly turbulent, but sufficiently small for the smaller ball to bounce above face level on the first bounce.

Fyz

Hi Fyz, you closed with: "So, the air loss is predominantly turbulent, but sufficiently small for the smaller ball to bounce above face level on the first bounce."

I suppose this must the difference between engineers and physicists - the latter analyzes it to death and the former simplifies it to death!

I also like the analysis part, but in the end, there are just too many unknowns and we better perform a test...

So how would we achieve the small ball to hit the large ball precisely "center" so that it will bounce straight and truly vertical?

Regards, Jorrie

No, I think we should be similar - we do just the minimum analysis to be certain of the result. So I'm not quite certain what you are saying: do you think you could be confident of the result without checking the air losses, or are you saying that an engineer does not care whether the analysis is correct unless it's for real? I have some sympathy with the second viewpoint, but tend to take a somewhat extreme approach - either I don't start at all, or I work it as if it is a real problem (otherwise it doesn't do much to keep the rust at bay).

Back to levels of analysis - my first shot was limit in vacuo (10-cm to spare on first bounce). Second check is whether air resistance prevents it. This is possibly where there is a difference - unlike an engineer, I had no idea whether we would be above or below onset of turbulence, so I had to do both sets of sums. In the end there is significant margin if we use steel balls - but it require two bounces to reach your face if the density of the balls is less than 2.28.

Regarding the test - I could design and build a release mechanism to do the job - a concentric pair of non-magnetics tubes of inside diameters 5-cm and 8-mm (half the ball diameters) to locate the balls, and an electromagnet to retain and release them. I'm pretty certain that would give adequate alignment and stability for a single-bounce solution. And for a single bounce, standard engineering flatness limits would be adequate. Just shield form draughts... On second thoughts, I'm confident enough now that I don't think I shall bother - but is the mass of the second ball actually "negligible relative to the first"?

Just to play devil's advocate ... should we assume that value of gravity is normal as that is not stated in the question. Also if the mass is negligible would not the effect of gravity on that mass also be negligible?

If you look at the third equation in #34 you'll see that the 'g' (acceleration due to gravity) cancels out.

The 'effect' of gravity can be expressed as a force or an acceleration. Yes, the force becomes negligible as mass isreduced, but acceleration remains the same.

No equations in post #34 - which post did you mean? BTW, the question says elastic, but also specifies that you can occupy the same space - shouldn't you take account of air resistance? Mind you, that has no effect on the height so long as the flow is turbulent - the height will only reduce once the balls are moving slow enough for the flow to become laminar

Fyz

Sorry, I meant #37 which was a reply to #34. I see four main areas being addressed:

Release: - Is it realistic to say that the two balls are released simultaneously? Can we release them accurately enough to hit the target?

Downwards travel: Would the balls separate while travelling downwards even if they were released simultaneously?

Collision: Is there one collision (balls have stayed in contact) or two (balls have separated)? If only one, then how much momentum is transfered to the lighter ball before contact is lost?

Upward travel: How much height is lost due to air resistance? How much accuracy is lost due to air movement?

My position is that the release and/or downward travel is very likely to result in enough separation for the balls to collide from opposite directions. I base this on observation of similar experiments.

Regarding accuracy, the upper ball would have to land within ~0.3mm of top dead centre to hit a 100mm diameter disc 4.5m above. This is unlikely for any individual trial, but becomes more likely if we increase the number of trials (much cheaper than building apparatus). Cross wind would also complicate aiming.

Regarding air resistance, most of the discussion has been centred around how much drag the smaller ball would experience on it's upward journey, but little attention has been given to the drag experienced on the downward path of the larger ball. Also, most equations for air resistance apply to steady state conditions. It would be much more complicated to model the air resistance of a small ball travelling the first 0.5m through the draft of a larger ball in the opposite direction of travel to the larger ball. (Don't get me wrong: I don't mean to heap contempt on people efforts to calculate these effects).

I believe that a perfectly elastic ball could travel high enough to strike an observer perched 4.5m above the ground.

The only value I've seen mentioned for C was 0.98 in your #54. This reduces travel of lighter ball from 4.5 to 4.15 making contact impossible. Any experiment would have to take this into account by reducing height of target or increasing initial drop height.

Hi Davo

Simultaneous release - now where does it say that? However, I think that the easiest jig-free way to perform this experiment is with the smaller ball resting on the larger one, and steadied to the central position. This give approximately simultaneous release. (Actually, the small ball will be slowed initially, and the large ball accelerated - another source of loss that I have not included) This leads to

Downwards travel - yes the balls will continue to separate, as air effects have to be small even for the smaller ball and the higher bounce. But I don't know whether they will separate enough for the
Collision - to be treated as separate events (time for an additional two transits of the sound wave across the lower ball after the top surface has stopped)

Upward travel: I think I covered loss due to air resistance - this places a lower bound on the size of the ball - 2-mm at specific-gravity=9. For the same height loss due to air, the diameter of the upper ball needs to be inversely proportional to the specific gravity - but heavier upper balls have smaller rebound velocities, which makes the lower limit for SG somewhere around 2. It would be straightforward to calculate the maximum wind speed vs ball size, but, based on the ball's bounce velocity (#9m/sec), and the height loss corresponding to that speed, we can be reasonably confident that the upward path would be little disturbed by winds of less than 5 m/sec (~10mph). The relative position of the balls prior to bounce is much more critical, so I would be looking for a virtually wind-free day (<~1mph?).

Regarding Accuracy: I think you meant radius, not diameter? - I estimated about 0.4-mm offset for a 100mm radius disc. That is also close to the size of my face.

Air resistance - I neglected the effect during the initial drop because its effect is approximately a factor R1/R2*0.5m/4.5m smaller - i.e at least a factor 27 for situations where the smaller ball rises high enough to hit the observer. Maybe I should have been even more pedantic and included it.

C - Of course, the question meant we didn't have to deal with this. But we could postulate an arrangement that gave better results than the 0.98 I observed, as it was for a ball and a flat surface, and didn't include correction for air loss (about 12-meters each way [down and up], and 2-cm diameter ball bearings). Identical steel balls are probably > 99%. The loss is not immediate absorption during impact, but due to the energy in the acoustic wave in the surface not being returned. If we can make the 'ground' more rigid and/or denser than the lower ball, and/or of the appropriate construction (a hemi-ellipsoid??), and the upper ball hollow or more compliant than the lower one (so that the acoustic transit times are equal), we could perhaps manage a high enough bounce.

I said in post #55 that this was becoming too complex for me...

Fyz

The larger ball needs to gain enough energy during it's 50 cm fall to acheive all of the following:

Rebound and propel itself upwards into the smaller

over come the smaller's momentum

re-direct the smaller

propel the smaller in the opposite direction

plus overcome the gravity now

reach a height of 4.5 meters

Seems impossible for that much energy to be created during a 50 cm drop...

The smaller would have to hit the face first...IF either did make it that high

We are dealing with numbers here not with "seems impossible". The colissions are elastic, so all the conditions you are talking about will be met, no matter what the heigth is. With one exception: the big ball doesn't need to reach 4.5 m. The small one will.

I say the two balls stay in contact and bounce 50 cm, thus not hitting me in the face. But lacking complete confidence in my answer, I would be wearing safety glasses, just in case.

-Erica

If the mass of the second ball is truly "negligible" you are probably right - dust sticks pretty well...

I should hope anyone performing this type of experiment would wear the minimum Personal Protective Equipment, which would include safety glasses, fall restraint harness (a 4.5 meter fall can hurt...), hardhat (in case the balls do go higher than you thought), gloves (the roof surface may be abrasive), hearing protection (knocking balls), and clean underwear (in case there's an accident).

Wearing safety glasses would eliminate the possibility of either ball striking you in the face, as you are "...looking directly to the falling balls."

The answer in a highly regulated world would be "no"...

I say that since the bottom ball sees nearly all of the drag, the smaller ball wants to fall faster in the semi-vacuum directly above and so begins to roll down the larger ball until it sees the same wind drag as the large ball. The ball will not stay directly centered on top because of this. It's hemispherical position at impact will determine the direction of travel upon rebound.

But the balls are only falling 0.5m. Do you know how far that is? It's about as high as your knee. I can't see the smaller ball rolling very far off centre.

When you say 'hemispherical position' do you mean that it's centre will be at the same height as the heavier ball? If this is true there would be no collision.

I think #1 is correct if all impacts are elastic the speed of the small ball bouncing back would be inversely proportional to their weight so it should approach infinity it's mass were zero.

Don

You've made me think again. The first bounce clearly gives nearly three-times the lower-ball's bounce speed, but what about subsequent bounces? What started me thinking again was the velocity you proposed - I guess you mean the height, not the velocity, as potential energy considerations would limit the height of the small ball to 50-cm x (mass ratio+1) - so the limit to the velocity would only be the square-root of this. Now, the question is whether the very-small-ball would ever approach this theoretical maximum?

Neglecting all imperfections, you would think that the speed of a near-zero-mass negligible-mass ball would approximate 3x, 5x, 7x ... the original peak speed after each bounce. However, the situation is the limit of a finite mass ball, and the smaller ball will have a bounce time that is slightly shorter than an exact multiple of the bounce time of the larger ball - so the balls will be moving in the same direction when they meet for the second bounce of the small ball, so it will be at exactly the original velocity thereafter. The third collision will be a double bounce, with the second ball back at 3x velocity. I'd need to do a lot more sums to see whether the timings ever allow the bounce velocity to exceed 3x (no chance).

.. Add a bit of air-loss, and there is variable dispersion between the speeds of the balls. This makes it possible that eventually the timings would result in the smaller ball bouncing quite high - but I can't be certain that the loss wouldn't prevent this happening. That is quite some modeling problem, I'm afraid, and quite different to what I thought in the first instance (see post #23). I now realise that the mechanism for obtaining sequentially increasing bounces of the smaller ball would be adequate loss in bounces on the ground (This would shorten the larger ball's period, so that the larger ball is always on the way up when the smaller ball returns).

Fyz

Hi Fyz and Jorrie,

I was thinking about a simple way to demonstrate the effect. It occurred to me that you could use appropriately sized steel balls with small holes drilled through the center and a thin wire threaded through the holes. You could then fix the wire at the top and bottom and use it the keep the balls aligned throughout the test.

The guide wire would add to the overall friction but if you used sufficiently massive steel balls the effect could be minimized. Admittedly it isn't perfect but it would keep the balls aligned sufficiently to demonstrate the effect and the friction component would be nothing when compared to the imperfect elastic properties of the steel.

Hi masu, I like your steel balls with guide wire better than post #51's (Randall, I think) golf ball and ping-pong ball in a tube - the ping-pong ball will fall slower than the golf ball, I think, hence always meeting the golf ball on its way up!

A small steel ball in the wake of a larger ball should attempt to fall faster than the larger one, I think...

Regards, Jorrie

If the upper ball is resting on the top of the larger ball initially, any compliance will delay the start of the fall of the smaller ball. So it's hard to know what the position will be at impact. However, if they are in contact, I would guess (used advisedly) they would remain in contact until the first wave from the impact with the ground reaches the top of the large ball. The result would be that the top ball only bounces to 0.6-m. As it starts out first, it would bounce off the large ball shortly before it hits the ground for the second time, so the second bounce would give the desired 4.6(ish) metres. This is getting too complex for me...

Fyz

Ouch. The reason I hacked the theory was to be certain of the result if we followed the drift of the question. However, real steel balls of different sizes are unlikely to be that loss-free in practice.

Returning to experimentation: you could construct a Newton's cradle but with different size balls to show the order of the effect, but that wouldn't necessarily give you the information on the second-order effects that might actually determine success

If you really want to perform the experiment, it should be reasonably straightforward to turn a brass fixture to restrain the balls and to house a soft iron rod above the balls; winding a coil round the outside should complete the rotation-free release mechanism. If the rod is by a fishing-line cradle from one side, and the brass fixture from the other, the mechanism could be made self-releasing, so it moves out of the way within the available 1/3 of a second and without creating too much disturbance.

Much will depend on the properties of the balls and of the flat surface. While equal size steel balls give coefficients of restitution that are very close to unity, I don't believe that this is necessarily the case for balls of unequal sizes - I think that the smaller ball needs to be more compliant than the larger one, or the energy imparted to the large ball will not be reflected in time to be useful. On the other hand, my recollection of dropping ball bearings down a stair well onto a solid surface (over forty years ago) was that the bounce height corresponded to a COR of about 0.98. This was a lunchtime activity at a standards research institute; as I remember, the floor had been specially conditioned for some specific measurements, so we can't regard that as typical.

Fyz

I'm not sure which law of physics predicts this happening Don. I view this question as being two similar collisions both of which conserve momentum. The first is a collision between the heavier ball and the Earth. The second collision is between the heavier ball and the lighter ball.

Immediately before the first collision the heavier ball is travelling towards the Earth with velocity V1 (downwards). The frame of reference for measuring this velocity is conveniently taken to be the Earth. After colliding with the Earth in a perfectly elastic collision the heavier ball bounces back up with a velocity -V1 (same magnitude, but upwards). If your line of reasoning was correct, then the heavier ball, of negligible mass in comparison with the Earth, would bounce away from the Earth with greater velocity than that with which it arrived. Everyday experience tells us that this is not so.

The second collision can be considered with the heavier ball as the frame of reference. The lighter ball will be traveling downwards with velocity V1 relative to Earth, but twice that speed relative to the heavier ball which is now travelling upwards. Since it's mass is negligible it will bounce away from the heavier ball in the same way that the heavier ball just bounced away from the Earth. That is, the lighter ball will have same magnitude velocity that it started with but opposite direction, all relative to the heavier ball. It's velocity relative to the earth will be it's velocity relative to the heavier ball, plus the velocity of the heavier ball relative to the earth. Very wordy, sorry.

So the velocity of the lighter ball after colliding with the heavier ball will be 3 times the velocity of the heavier ball when it hit the ground.

V1 = √(2.g.s₁)

V2 = V1 X 3 = √(2.g.s₂)

3√(2.g.s₁) = √(2.g.s₂)

9 x s₁ = s₂ = 9 x 0.5m = 4.5m

But the lighter ball's starting position for it's journey upwards was 0.1m above the ground since it bounced off the heavier ball. Therefore it can theoretically reach a height of 4.6m.

All this is assuming perfect everything, which we will no doubt debate for the rest of the week. Just wanted to clear up a basic misunderstanding.

Dear Sir

You happen to consider that the ball with negligible weight is travelling at the velocity of 3 times the falling velocity over the whole distance. But the fact is that when the larger ball is falling the smaller ball is also moving in downward direction. hence the point of contact of te 2 balls will detrmine the time at which the smaller ball will attain the new velocity and hence the distance it will travel at new speed.

Dear Guest, Please call me Davo. I meant to convey that the lighter ball will only be travelling at 3 times the velocity of the larger ball immediately after impact. I agree with your other statements, and would be interested to hear how you think they would affect the height to which the ball travels.

I believe that you will be hit in the face by the smaller ball.

The larger one can only bounce up to 0.5M, but

the sound wave initiated in the larger ball

can have sufficient energy / velocity, that upon reflection

can accellerate the small ball to sufficient velocity to hit you in the face.

The height that the small ball can attain is a function of how stiff the larger ball is.

Let's say that the person ducked out of the way and the ball missed. When the small ball came back down, would it reach the big ball at the same time at ground level?

Can we also build a test for that?

G

If everything is lossless, then the balls will reach the ground at the same time (3-bounces for large ball, 1 for small ball). If there are only air losses, the small ball will reach the large ball shortly before the third bounce, so the second bounce of the small ball will be slightly less than 0.6m. If the large ball loses just enough energy at each bounce on the ground (not part of the question, I think), it can bounce just before the small ball hits it, so the second bounce of the small ball would be slightly less than 12.6-m.

Fyz

Split the question into two parts- small ball- earth and

If we assume two system separately - small ball to large ball and large ball to earth.

large ball can only bounce by 50cm under elastic collission (neglecting small ball).

Small ball to large ball system- (Now let us assume large ball to earth and small ball as a football ) - How can it escape with the gravity of the large ball also coming in picture (in addition to gravity of largest ball- ie earth)?

All the collission are istantaneaous so at the moment of contact all are relatively stationary (ie large ball is still not moving outwards)

No the small ball can not hit you.

If the much smaller ball lags just a little behind the larger ball, it will strike the larger ball on its rebound, and its own rebound velocity will be two times the impact/rebound velocity of the larger ball ( 2v(sub I) ) plus its own impact velocity ( v(sub I) ), or 3v(sub I). It's kinetic energy will be 9 times greater, and (neglecting air resistance) it will rise to a height 9 times greater than that from which it had fallen ( KE = PE = mgh ). 9 x 50 cm + 10 cm = 460 cm = 4.6 meters.

Answer 1: Yes, by the smaller ball.

But because the two balls are released simultaneously, if they were originally in contact, they are in contact at the instant that the larger ball is instantaneously stopped (neglecting the slight difference in the force of gravity on the much smaller ball, which is only a little more than 5 cm farther from the center of the earth); the much smaller ball rebounds from a surface which is at that instant not moving, with only its own impact velocity, to reach its original height (neglecting air resistance) and also accompanied by the larger ball.

Answer 2: No.

Interesting. Two opposite answers based on alternative interpretations. What are your views on instantaneous events occuring simultaneously?

Davo:

"Interesting. Two opposite answers based on alternative interpretations. What are your views on instantaneous events occuring simultaneously?"

By "instantaneous" I suppose I mean something consistent with the paradigm formed of Newtonian infinitesimals of time, Dedekind cuts, and the Dirac delta function; less pretentiously, the point on a graph where the velocity of the larger ball (the whole ball all at once, perfectly rigid) is zero.

Although simultanaeity is more difficult, perhaps of a phenomenological and hermeneutical category, in this case I suppose my working notion is based on actual physical contact of the objects obviating the need to consider synchronicity problems arising from spatial separation.

In this (simple) case, physical continuity and a tacit assumption of perfect rigidity (!) seem to be the basis for instantaeous simultanaeity.

But before anybody takes me to court about any of this, you might want to read this first (it's really not just humerous): http://www.physicsforums.com/archive/index.php/t-38649.html

Hi Davo

Here we are, well entrenched in your category 2, I think. I wouldn't personally choose the term instantaneous here, but my 2-penny worth are that the balls effectively strike at the same instant if the smaller ball is anywhere between touching the lower ball at impact, and separated from that by a falling distance that corresponds to two transitions of a sound wave across the lower ball. Nicely demonstrated by guest's experiment in post #82

Fyz

Hi Guest, your: "... much smaller ball rebounds from a surface which is at that instant not moving, with only its own impact velocity, to reach its original height (neglecting air resistance) and also accompanied by the larger ball." rings another alarm bell...

I think one can assume that the small ball is indeed in contact with the large ball at the moment the large one hits the floor (air resistance being less on the smaller one in the wake of the larger one).

Since nothing is perfectly rigid, it will take some time for the floor's effect (moving at the speed of sound through the large ball) to reach the top of the ball. Till that instant, the contact surface between the balls still move down unrestrained. Now elastic compression will begin at the top of the big ball, while its bottom is already compressed to some degree.

Given no losses (perfectly elastic), I now have an "in-between" gut-feeling for the height: more than once, yet less than thrice the drop heights - until the sums (and more assumptions) are made, I simply don't know!

Regards, Jorrie

Play with a Newton's cradle for a while: the "macro" effects are always the same regardless of the "micro" setup. I can't see how to prove it when the small ball stays in contact with the large one but I'd be prepared to bet that the result would be very close to the one where the large ball has completed its bounce before the small one hits it.

One way of easily simulating the experiment is to place a ping pong ball on top of a golf ball in a small tube. Release the golf ball from a height of about 1 foot and the ping pong ball shoots up well over six feet. Theoretically (ignoring wind resistance and coefficient of restitution): because the golf ball is about 18½ times as heavy as the ping pong ball the max height should be about 7.8 times the "drop height".

Exactly unity, I think. Partly the wave propagating round the larger ball makes the top stationary for the small ball to bounce from, partly the restricted number of solutions that conserve both energy and momentum (lossless bounces)

Fyz

Jorrie:

"Since nothing is perfectly rigid, it will take some time for the floor's effect (moving at the speed of sound through the large ball) to reach the top of the ball. Till that instant, the contact surface between the balls still move down unrestrained. Now elastic compression will begin at the top of the big ball, while its bottom is already compressed to some degree."

There is a much greater compressile force on the bottom of the larger ball than on it's top, but unless the bottom of the ball is infinitly compressible, both compressions occur at the same time. Suppose that the larger ball had simply been at rest all along, with some compression at it's base due to it's weight; the smaller ball would rebound from the larger ball's upper surface "normally," with a depression depth and contact time proportional to the compressile and restorative forces of the larger ball (kind of like a little trampoline). Then in the case of the larger ball impacting the surface while the smaller ball is on top, the downward momentum of the larger ball compresses it's bottom somewhat, so that the top moves downward under the little ball somewhat, delaying the time to maximum compression of the top of the larger ball. Then the restorative forces on the bottom of the larger ball move that little trampolene area on top back up. It seems to me that at this point some additonal momentum could be transferred to the little ball; the smaller ball may not complete it's rebound off of the surface of the larger ball until after the larger ball has begun it's rebound.

Hi Guest, you said: "There is a much greater compressile force on the bottom of the larger ball than on it's top, but unless the bottom of the ball is infinitly compressible, both compressions occur at the same time."

You're right for all practical purposes, but since the speed of sound cannot be infinite in any ball, there will be a few μs delay between the bottom of the ball hitting the floor and the top of the ball even "knowing" that it has happened.

In those few μs, the top surface moves downward undisturbed. Then the "sonic wave" from the bottom hits the top surface and it starts to slow down. Only then can the small ball begin to make an impression.

I'm not sure if this has any bearing on the outcome of the experiment, but it is something that can easily be demonstrated with a suitable (very) fast camera. I suspect that there will be oscillations in both balls, complicating a prediction even further!

Regards, Jorrie

It does - with no gap and perfect elasticity, bounce height is unity because the top surface exactly stops on the first acoustic transit. Two transits later (down and up), the top surface is going upwards at the original velocity. Described theoretically, and then demonstrated in post #82

Fyz

Can this not be solved by an alternative approach - an external observer with no frame of reference could see the level surface approach and impact the two balls (as if it were moving upward ! ). I think momentum would transfer to the upper ball. how much so I don't know , but modifying a Newtons Cradle would give an idea (replace one end ball with a ball of 'insignificant' mass) of the mechanics. If my deduction is right , that's one-in-the-eye for conventional thinking.

Hi Guest, hmmm... "an external observer with no frame of reference..."?

I think you meant to say "from the frame of reference of an observer riding on one of the two balls".

Exactly the same principles will apply as in the reference frame of the floor. The shock wave still travels up and down through the lower ball, as Physicist has analyzed before. The result will still depend on whether the small ball is in contact with the large ball by the time the shock wave reaches the upper limit of the large ball.

Changing reference frames sometimes makes problems easier; in this case I think it complicates things!

Regards, Jorrie

Hi Jorrie

Finally, Randall's comments and a good night's sleep have convinced me that near-zero-force contact between the balls (at the time of collision) will make no difference. I forgot the effect of the "free" surface of the top of the lower ball on the surface velocity - when the wave reaches the top surface, the change in velocity will be twice the change in the travelling wave. So the upward wave will first stop the ball, and the reflected downward wave will start the upwards motion. Unfortunately, none of the provided emoticons is adequate to convey my embarrassment. I will put more in my supplement to my reply to Randall

Fyz

Hello Jorrie , Your phrase does seem better. I was trying to see what occurs when the initial shock wave travels up through the lower ball and reaches the upper ball. I see the lower ball remaining fixed until its shock wave returns to ground causing the ball to ping upward exerting force on the upper ball (?).I guess it depends on how the lower ball deforms (ie does its top surface move before the ball moves as a whole).

The quetion is open to interpretation , I am going to see a mate later and pinch his Newtons cradle for some testing !

Regards Kris (previous posted as guest)

Hi Jorrie,

There's an interesting discussion of Newton's cradle here:- http://www.lhup.edu/~dsimanek/scenario/cradle.htm

Which contains the following comment and reference:-

Reality check 2: We assumed instantaneous propagation of compression, even though we knew better. Why does this seem to work? Hermann and Schmälzle [1986] note that the collision time for 5 cm diameter steel balls is 0.2ms. This is much longer than the propagation time of a compression pulse through steel, which is about 10^-2ms.

Hi Randall

That's very interesting; it's nicely expressed, and includes one item of measurement and model data that we have been sadly lacking. I have a feeling that this is the curved contact region behaving as a spring (as we previously speculated). In that case, I expect the loss will reduce as the ratio of collision time to acoustic transition time diverges from unity (in whichever direction). That is to say, the collision can be elastic if the surface deformation time is very short compared to the acoustic velocity, or if it is very large. If so, for their practical case (and ours if we postulate steel balls) the loss reduces as the magnitude of impact increases.

I noticed that the experiment/model of Flansburg and Hodnut corresponded to an energy loss of about 5%. Can it it be just a coincidence that this corresponds to the ratio of the times? But I think it means that, unlike rods and cylinders, the impact of uniform spheres cannot be truly elastic - something I suspected, but was uncertain about.

Fyz

There seem to be agreement that the velocity of sound in the balls is very relevant to the the velocity with which the light ball is ejected from the pair after it reaches the solid ground.

The question postulates balls of perfect elasticity although most discussion and experimentation has used steel balls, what is the velocity of sound in a perfect ball? is it infinite, if so a zero mass 'small' ball would be expelled at infinite velocity (Einstein permitting)