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

Hi Syhprum

The speed of sound is only relevant regarding imperfect elasticity of the colliding balls. I'll come to that later - but first we need to look at the ideal case.

It is a complete misunderstanding that the small ball could go infinitely fast after its first bounce. If you separate the events, you can easily see what might happen (next paragraph). I have shown to my satisfaction (after Randall indicated that I'd got it wrong first time) that this will also happen for simultaneous collisions if the balls are perfectly elastic - but it's rather hard to communicate.

Suppose the lower ball hits the ground at speed V. As the collision is defined to be elastic, it rebounds upwards at speed V. The upper ball is going down at speed V, so the speed of approach between the two balls is 2.V. As the upper ball has negligible mass compared with the lower ball, the lower ball appears as if it is infinite. The speed of separation is the same as the speed of approach = 2.V. The lower ball is still going upwards at V, so the upper ball must be rising at 3.V. That is sufficient to allow it to rise to 9x the distance it fell. If there are no losses of any sort, and the balls remain perfectly aligned after each bounce, there are two possibilities - depending on whether "no losses" is the limit of the losses causing the lower ball to speed up more than the upper ball or vice versa. In the first case (assuming I keep my silly mug out of the way), the smaller ball will indeed increase its bounce speed by 2.V on each bounce, until it eventually reaches the Earth's escape velocity. In the second case, the upper ball's bounce speed will alternate between 3.V and V. (The reason being that they can be going in the same or opposite directions on alternate bounces of the upper ball).

Now to the effect of the speed of sound in the balls. I'll assume uniform solid balls, and that the material itself is lossless. What actually happens is that the surface compresses during the bounce, so the excitation of the sound wave is incomplete. If the time of the contact is comparable with the time of sound transmission across the ball, the kinetic energy will be converted into vibrations of the ball - no energy loss (assuming the material is lossless), but not contributing to bounce height. There are two ways around this: if the ball is very rigid, the sound transit time across the ball reduces far more rapidly than the contact time, so we can consider the contact area as a perfect spring and the ball as rigid - in this case the loss reduces to zero as the ball becomes infinitely rigid. At the other extreme of perfectly soft balls, the ball is moving faster than the sound wave, so you are looking at the balls becoming flattened at rebound. If it is infinitely soft, it becomes completely flat, and bounces like a uniform object - also potentially lossless. I'm not certain, but I suspect that the upper ball needs to be infinitely softer than the lower one if its collision is also to be elastic. So, the effect of the sound velocity is only to allow the bounces to approach the optimum height - not to arbitrarily increase the bounce height.

I hope that's helpful, rather than purely confusing.

Fyz

"Immediately after the big ball hits the ground, it moves upward with speed v. The small ball, however, is still moving downward at speed v"

At what point in this ideal world did the balls loose contact with each other ? For the no-deformation case they both bounce at the same instant.For the deformation case they both flatten at the same instant and rebound at the same instant (consider the vector of a particle in each ball at the interface).I prefer the latter notion since we are talking 'elastic'.The maths is compelling , but I think it's an attempt to explain what can be seen in a test , but ignoring the fact that the lighter ball will go off at a tangent with pretty dramatic effect (I can throw a ball a lot further horizontally than vertically).Whatever anyone makes of the responses to this thread it's certainly made me stretch my brain.Before anybody says so , no it doesn't take much.Now I'm going to sleep and dream about trisecting angles with Uncle Euclid.

Let us consider the two cases (infinite and zero thickness) separately - all other conditions ideal and Newtonian (earth perfectly spherical, stationed at the pole, so rotation is irrelevant, orbit around sun has no significant effect, the release mechanism has zero gravitational mass, and your face is far enough away to neglect it's gravitational effect...):

Zero stiffness: the lower ball is fully compressed before the upper ball starts to compress - the upper ball, being infinitely soft relative to the lower, as well infinitely light, only sees the lower ball as a rigid surface with an upwards velocity of V. => the upper ball's bounce speed is always 3V

Limit goes to infinite stiffness*: if the density of the lower ball is greater-than or equal-to the density of the Earth, gravity will provide a net attraction between the balls if they are contacting when released. If they are held contacting with zero force, they will bounce off each other. If the upper is resting on the lower, the stored energy will cause them to separate. Only if the initial force between the balls is extremely close to the net attraction will the balls be in contact when they bounce. So, the usual situation is non-contacting.

Now, the exceptional situation when hard balls remain in contact: the situation is equivalent to infinitely stiff balls with springs at the contact point (this narrowing contact point is the reason the hard spheres behave elastically). In this case we would expect the time constant of the small sphere to be short compared with the larger one, and the spheres will indeed bounce together. (We could theoretically make the relative time-constant of the smaller sphere relate to the larger one in any way we please - in which case the small-ball's bounce speed can be anywhere between V and 3.V.

Finally, the requirement is that the upper ball is on top of the lower one (again, ideally means exactly vertically stacked) and they are dropped (meaning exactly released). Therefore, even assuming you play cricket in mythical Etherville, your bowling/fielding talents don't seem to be relevant.

Fyz

*remember, the situation only has meaning as a limit. So, although the separation becomes "zero", if it is non-zero for the finitely stiff ball, then the zero-separated balls are still effectively separated.

Thanks . I'd made a hasty judgement on the balls flattening in the same instant (Your paragraph 2).

My horizontal reference was directed at what experimenters among us may have seen , though the term horizontal was not a good choice.

I think I get the argument , but does all of this trash my attempt at a tower of Galilean analogy that I posted a couple of hours back ?

Now you're talking Kris (or then you were). Take any triangle: trisect all three angles; the three intersection points where each pair of lines, closest to each original side, meet form an equilateral triangle. Prove it!

Yeah I know that one -Morley's triangle.Oddly enough I'd been looking at it to solve this impossible problem !.The proof of it's impossibility relies on a third order equation and transcendental numbers.There's a great paper on the web called "confessions of a weekend tri-sector".Learning does not have to involve an attainable goal I say.

You may find 'Langley's advantageous' angle amusing.Another great diagram to see is Simon Singhs illustration of (a+b)³. Once seen , never forgotten.Kris

Hi Fyz , indulge me with a last minute thought ;

Galileo is atop the tower of Pisa with an iron cylinder held horizontally.

He drops it observing the effect.

He bashes it about to create a neck near the top and repeats.

he narrows the neck to a hairs breadth and repeats

He cuts the neck and repeats.

OK , so he's now figured something about gravity

He enters our realm of "perfect elasticity"

The cylinder is held vertically ,and progressively narrowed at the neck and rounded.

What , in this somewhat borrowed thought experiment , prevents both balls from returning to initial hight ?

Hi Kris

I'm not entirely certain I am addressing your points here - be gentle, I'm doing my best.

I assume we are talking blunt cones throughout, not balls (whoops) as at the end of your paragraph. My cylinder was tapered, not necked. A necked cylinder could behave in all sorts of ways, depending on the shape above the neck.

Returning to the taper: the energy in the wave that was generated by the collision is concentrated over a smaller area than at the bottom; so even before the wave reaches the top the material doesn't simply stop, but starts moving upwards. In many ways, this is equivalent to what happens in a whip - though the reasons are different (the energy in a whip is concentrated because the speed of the wave reduces as the tip of the whip has nothing to apply the tension that generates the waves velocity).

Now we cut the cylinder near the top - the upper portion flies off at great speed upwards.

Why wouldn't the base cone bounce to its original height? Apart from the energy conveyed to the top section (assumed negligible), the requirement would be that all the energy remain in the single wave, and not get converted to other motions. I think this would require an infinitely slow taper. OK, perhaps you could do it if the cone has an infinitesimally narrow base...

BTW, you imply the top is rounded - I think that would definitely disperse the reflected wave into vibrations of the cone. We need the reflecting surface to be parallel to the incoming wave-front.

Fyz

Hi Fyz , let me try to clarify my idea. The original thought experiment (don't know whose ) that I quoted starts with a single large ball.Through a sequence of test , the ball is gradually remoulded towards one small ball joined by a very narrow neck to a large ball held beside it .Both balls will be observed to fall at the same rate as the original single ball even when the neck is cut.Thus we conclude that objects fall at the same rate regardless of mass.

I was trying to extend this to the question posed. We start with a large elastic ball.This is progressively remoulded to something resembling a limbless Tellietubby .At the limiting point the poor tellietubby resembles a small sphere joined to a large one.

The crucial point is this : tellietubby would bounce back to original height just before his neck is cut , so why not when his head is severed and sits on top of his body.

BTW my best experiment returned the small sphere to 3 x original height. Varying the small ball gave best results when it was relatively rigid and dense !?!

I am essentially a believer in the answer to the question , but feel that something is missing - the 'correct' answer can not be physically demonstrated.Also the ice block challenge is getting some inane responses.These are "challenge Questions" and I feel that any of them merit dialogue rather than the pompous dismissal that many (especially Guests ) give them.More is learned on the journey to understanding a problem than knowing it's specific answer.

Wave propagation is new to me , but I appreciate your detailed explanations and will be printing them off for more leisurely study.Kris

If the small ball is insignificant enough would it escape the gravitational field of the larger ball??

Regardless of how high the balls bounce, it is not specified that the observer (you) are vertically above them: "looking directly" isn't sufficient. For example, you could be 4.5 m up and 1.5 m to the side and still "look directly". Hence there is no way to tell whether you will be hit "in the face".

Answer: You will not be hit in the face.

No, if the collision is perfectly elastic then the bigger ball will bounce back to an height of 50cm and the smaller ball remains sit on the top of the bigger one.

I agree with post #64.

The problem seems like a straightforward conservation of energy & momentum exercise, with the mass of the balls unknown (a small ball, M1, and a much larger ball, M2, with M1 riding on top M2). Thus, the mass of M2 is some multiple of M1's mass:

M2 = x* M1

Assuming the balls fall together from 0.5m in a vertical line & both hit at the same moment the system (M1 + M2) kinetic energy must equal the potential energy, which works out to a Velocity at impact (Vi): Vi = Sqrt(g).

Assuming, at impact, the larger (M2) ball stays on the ground and transfers all its energy to the smaller ball, conservation of momentum must hold:

(M1 +M2)*Vi = M1*Vrebound.

Subsituting for M2 & Vi, Vrebound is found to be: Vr = (1+x)*Sqrt(g).

Substituting Vr in the conservation of energy equations (with h = 4.45m, since M1 is rebounding off the top of the 5cm diameter M2):

M1*g*(4.45m) = 0.5*M1*([1+x]*g^0.5)^2

which yields for x that M2 must be almost twice the mass (1.983 x's the mass) of M1 for M1 to rebound to 4.5m off the floor (or 4.45m off M2). Since M1 is "negligible" in mass to M2 (say something like a marble on a billiard ball) one can assume M1 is on the order of one-tenth (probably less) the mass of M2, which indicates that M1 will not only rebound to 4.5m, it would go much higher -- indicating that aerodynamic drag can be safely ignored. Assuming the ball rebounds vertically, an observer monitoring the drop test from a seemingly safe vantage point four meters from the test start point would likely experience a memorable impact from M1.

This assumes M1 is comprised of some form of substantial mass (e.g. solid glass). If M1 were VERY light & small, something like a tiny pingpong ball, aerodynamic forces would predominate and it would not only not reach 4.5 m, it would likely "corkscrew" in flight somewhat randomly and slow dramatically fast (more analogous to a feather shot from an airgun).

"Assuming, at impact, the larger (M2) ball stays on the ground and transfers all its energy to the smaller ball, conservation of momentum must hold"

Therein lies the problem. Elastic collisions will conserve kinetic energy. A bounce with the ground will not conserve the momentum of the ball, however - what happens is that momentum is transferred to the (approximately infinite mass) ground, so that BOTH momentum and energy are conserved. Indeed, if conservation of momentum held for just the balls, the smaller one would continue into the ground at this increased velocity rather than heading upwards.

If we look at your "result" in terms of energy, the system has nearly 3x the kinetic energy immediately after the bounce than it had immediately before bouncing approximately (0.5*(3*V)^2/2) versus ((1+0.5)*(1*V)^2/2).

Fyz

The problem is essentially a special case of the conservation of momentum illustrated using a NEWTONIAN DEMONSTRATOR (one of those novelty devices having several steel balls hanging on strings); somone in the office probably has one nearby.

Simulate the problem by dropping two of the balls onto the remaining balls while holding them still or fixing them against a solid object (to simulate the "ground") and watch what happens: only one of the two dropped balls bounces back. With little improvisation effort a smaller ball can be added for more dramatic effect. One can see the effect much more crudely using simple objects (try stacking bottlecaps of different sizes & dropped a few inches; done cleanly the more energetic response of the smaller top cap along with a less resonsive [than if dropped singly] bottom cap, is readily observed).

What I observe is consistent with the equations presented (reference post #70). Why philosophize about what happens with such physics when a basic test is so easily done to observe the basics?

As for the alleged 3x increase in K.E. -- no idea how that can be derived from the equations presented as the velocities varied in relation to the masses involved (less mass--greater velocity & vice versa, but overall constant energy consistent with the stated assumptions).

I also assumed (unstated) that the ground (Earth & its surface) can essentially be treated as a fixed immovable object relative to both balls and that changes in the Earth's movement resulting from the impact of the balls is reasonably ignored (i.e. "<delta-V-Earth> = zero").

Let us simply write down the numbers for the Kinetic Energy (KE) based exactly on the "results" from #70:

Initial KE just before bounce = (M1+M2)*(sqrt(g))²/2 = M1*(1+1.983)*g/2

KE for stated velocity of ball2 just after bounces = M1*(sqrt(2*g*4.45))²/2 = M1*g*4.45

Ratio of KE after bounce to KE before bounces = 4.45*2/(1+1.983) = 2.984

Hopefully this is clear now. Perhaps you forgot that kinetic energy is proportional to the square of velocity?

Fyz

Ok--I see your point this time...and have a follow-up question; to summarize:

a) Cons of Energy suggests Velocity at impact, Vi: Vi = Sqrt(g)

b) Cons of Momentum applied to the rebound (assuming Vel of M2 =0 for simplification) leads to:

Vr = (1 + x)*Sqrt(g)

c) Applying Cons of Energy to M1 (small ball) using above Vr AND using a 'plug' of 4.45 meters for the required height (and ignoring the minor error of using the larger ball's, M2's, radius as its diameter) leads to a ball mass ratio of M2 about equal to or greater than twice M1 (rounded).

d) Which leads to your check & observation that for the small ball to reach the stipulated height more energy is required (about 3 x's more) than the system started with (a larger mass ratio more in line with the problem statement would increase the impossibility; I simply didn't bother to check the results for realism & made an implicit assumption it would work....).

Sooo: Given 'd' above (and that the larger ball will rebound to some extent, further reducing the energy transferred to M1) does this indicate (prove) that M1 willl not/cannot rebound to 4.5m (the real/overlooked "result")?

If the bounces are separate, up to 3x velocity after first bounce of the smaller ball is quite possible, which is 9x the ascent (+10-cm because it starts from the top of the other one).

Enough, already! Everyone back to work!

The Project Manager

The small ball.

Kelly

Simple "A" Level Physics:

The following assumes no air resistance, no friction:

Velocity when lower ball hits ground:

v² – u² = 2as (v= final velocity, u = initial velocity, a = acceleration, s = distance travelled), u=0 m/s, s=0.5m, a=g ms^-2 ,so

v = √(g) m/s

Elastic collision: means coefficient of restitution =1.

Elastic collision with earth results in larger ball having upward velocity = √(g)

and at this instant, smaller ball (on top) has downward velocity of √(g)

Let mass of smaller ball = m, mass of larger ball = M. (And we know M>>m).

Elastic collision means both momentum and kinetic energy are conserved!!

so if v is velocity of smaller ball immediately after collision (upwards) and V is velocity of larger ball immediately after collision (upwards):

M√(g) - m√(g) = MV +mv (conservation of momentum)..................1

0.5mg +0.5Mg = 0.5mv² + 0.5MV² (conservation of kinetic energy).............2

From 1:

V = ((M-m) √(g) – mv)/M ............................3

Substitute 3 into 2:

(M+m)g = mv² + ((M-m) √(g) – mv)²/M

expanding and manipulating results in:

(M+m)v² - 2(M-m) √(g)v – (3M-m)g = 0

Divide throughout by (M+m):

v² - 2(M-m) √(g)v/(M+m) – (3M-m)g/(M+m) = 0

But M >> m, so (M-m)/(M+m) approx = 1, and (3M-m)/(M+m) approx = 3, so:

v² - 2√(g)v –3g = 0

solving quadratic:

v = (2√(g) ± √(16g))/2

v = √(g) ± 2√(g)

negative root is not physically realisable (ball passing through larger ball), so:

v = 3√(g) = 9.4 m/s

When collision takes place smallest ball is 10 cm (2x radius of larger ball) from ground, so

Using "v² – u² = 2as": (where v= 0 (at top of climb), a = -g, u=9.4m/s, s = height above collision)

Height above ground = 4.5 + 0.1 m = 4.6m

Small ball does hit observer.

For Large ball, V = √(g) = 3.13 m/s, so again using "v² – u² = 2as":

Height = 0.5m (approx).

Large ball does not hit observer.

Sorry,

Everyone seems to believe that the momentum of the large ball will transfer to the small one, but the transfer of p only reaches maximum then the masses are equal. The insignificant mass of the small ball means the bulk of the momentum will stay in the large ball.

Another way of looking at it is to compute the maximum velocity of the pair as they bounce. The small ball *strikes* the upper surface of the large ball at the same time it's velocity is zero. Therefore it bounces back up to a height of 0.5m

The only way your vision is in jeopardy is if the balls are made of Flubber, and that's not in the problem as stated.

I just took one ping pong ball over a basketball (both ball contact each other) and dropped them together (from around 0.5 m) and both bounces togheter.

So answer is No.

But I separated the ping pong ball from the basketball (maybe 5 or 6 mm) and the ping pong ball bounced and hit the ceiling (around 2.3 m) so... if they are not contacting each other it might be...

If they are perfectly elastic, and one rests on the other, then they will separate during the fall. If the smaller ball is too small, the separation will be inadequate. Using dense balls (similar to steel), air resistance would set a minimum size radius the smaller ball to be about 2-mm. Limited momentum transfer sets its maximum radius to about 16-mm. With the larger radius, there is little doubt that the separation would be adequate. I'm not certain what happens at the smallest end of the range. But if we use lighter balls (similar to glass marbles), the smaller ball's radius would need to be nearer 10-mm, and the separation should be adequate.

Both air resistance and bounce mean that the smaller ball must not be too small. The question says the smaller ball is small enough to "have negligible effect on the 50-mm ball" - so I suspect the question defines the upper ball is smaller than 2-mm diameter, and the answer must be "no". Either air resistance or lack of separation would be sufficient to ensure this...

Some contributors assume this - these are the ones who get huge velocities after the first bounce (and energy gain). Others take conservation of momentum and conservation of energy seriously.

Your argument is correct if the balls are touching each other at the moment that the lower ball strikes the ground, and the numbers are right it the upper ball is very light or very heavy, or if the balls have equal mass. I'm uncertain about other cases, because the movement of a sphere in a bounce is relatively complex, and there is relative movement throughout the sphere at all points of the bounce

If, however, they are allowed to separate during falling - as they would if the small ball was resting on the larger one before dropping - then it is more reasonable to assume that the lower ball completes its bounce from the ground just before the upper ball strikes it. So, when the two balls collide, the lower ball is going upwards at its maximum speed, the upper ball is approaching at the same speed. Approach speed is 2x => the separation speed will also be 2x, and the smaller balls speed after its bounce is 3x. The momentum of the lower ball is marginally reduced, the momentum of upper ball is small because it is very light. Exact numbers taking account of finite mass of upper ball have been given elsewhere in various forms, but can be expressed more simply for this purpose as:

V2 = (3-M2/M1)/(1+M2/M1)*V0
V1 = V0-(V0+V2).M2/M1

Where V0 is the speed of the balls immediately before impact, V1, V2, M1, M2 are velocities* and masses of the lower(1) and upper(2) balls respectively
*Upwards velocities immediately after the the balls have bounced off each other

Assuming that I have made no typing errors, if you put in the numbers, you will see that both overall momentum and energy are conserved. You will also see that the speed of separation between the balls is equal to their original speed of approach, and that equal weight balls (M2/M1=1) result in the upper ball going upwards at its original speed, as expected form observations of Newton's cradles everywhere.

ah, I think I see your point now- but it assumes that the balls seperate before impact, which I am not entirely convinced that they would. I think that if they seperate before impact or not is immaterial to the problem.

My (I posted #69 and #85) numbers come out astronomical because I use zero as the mass of the smaller ball.

I wrongly assumed that the full momentum of the 5cm ball is transfered to the very small ball (VSB), whereas I suppose the VSB would actually recieve only a portion of that momentum proportional to its mass- so in such a calculation, mass would cancel.

Excellent! Enlightenment! Thanks for describing it to a moron like myself :p

Had I thought you a moron, I would not have bothered. It's often a matter of knowing where to look. Sorry if it came over like that. I was about to write a separate reply to #85 - not necessary now?

Fyz

Using a softball which has a diameter of 10 cm. and a ping pong ball which has negligble mass comparatively as a sample, dropping the stacked balls from a height of 50 cm. or 20" seems to concur with the statement that the smaller ball the ping pong ball may be able to reach a height of 4.5 meters or 14.6'.

I am lead to believe the ping pong ball could almost take your head off!

Is this like rocket science or what?

I'm not an engineer but i think there are probably some perameters missing for a definitive answer to be given.

ie how elastic or bouncy is the 5r ball?

if the balls part company on their downward journey. Is it possible that the 5r ball could bounce, and due to it's small elastcity factor have reached its upper limit of its first bounce and is on its way down again when its passenger catches up causing the passenger ball to arrest its downward with a small upwards movement, to be hit at a much lower velocity by the 5r ball on its second bounce

.

if the balls don't part company on their downwards journey due to some partial vacuum, as the air flows nround the lower ball , then again would it not depend how much energy was absorbed by the ground on impact to decide how much energy remained to ping the tiny ball back up ahead of it.

or am i talking balls?

Come'on guys,

"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..."

1. Problem is ambiguous as to which ball's radius is 5 cm.

2. Problem does not mention material of balls, so, elasticity is unknown.

"The balls are dropped from a distance of 50 cm"

"Lying on a roof 4.5 meters from the ground, you observe..."

3. If it is assumed that this is done on earth (strike that--will not matter) the (nearly) order of magnitude difference between the distance the balls are free-falling (50 cm) and the distance of the observer (4.5 meters = 450 cm) means that the observer (looking almost straight down at the balls) is at a very safe distance from the event.

It is self-evident that there is no way something dropped 50cm will bounce 450cm (9x higher than dropped) no matter what the materials or special conditions are (unless boosted)...aha!

dave

dnewberry@aseapower.com

Contributions to these discussions seem to fall into 5 categories:

1. ~5% give well reasoned and similar answers.

2. ~30% debate the finer points from the 5% mentioned above.

3. ~30% give the same answer as the 5% mentioned above, but with no reason stated.

4. ~30% give a very different answer to the 5% mentioned above with no explanation.

5. ~5% misread the question or another person's comment and shoot off on a fruitless tangent, often claiming that their view point is supported by something vague like "basic physics", "introductory physics", "common sense" or even that it is "self evident" rather than refering to a recognised law such as conservation of momentum or Hooke's Law.

Welcome to category 5, Dave Newberry!

"2. Problem does not mention material of balls, so, elasticity is unknown." The question states "If all collisions are elastic"

You say that there is 'no way something dropped 50cm will bounce 450cm (9x higher than dropped)'. I think we would all agree with you on that one, but in this case TWO things were dropped. One ball bounces off the other ball.

Well put Davo,

I was about to jump in and say something along the lines of you post but you managed to get in first and do far more eloquently than I.

Turbulence of the air surrounding the balls will cause them to be misaligned at the time the bigger ball rises to impact the smaller one, causing the smaller one to careen off in a random direction, therefore missing the observer above the event.

For the desired answer, consider the big one hitting the ground, then the big one hitting the small one on its way back up. Energies, therefore magnitudes of relative velocities, are conserved, and momenta, and velocity, of the larger item in collisions are conserved. When the big ball hits the ground, the ground's velocity is unchanged, and the big ball moves upward at it's impact speed. The big ball continues to move upward with velocity unchanged when it contacts the small ball. The small ball is approaching the big ball with twice the velocity it is approaching the ground. It recedes from that collision with the same relative speed upward, or three times the speed it approached the ground, and therefore nine times the energy, so it could reach a height of 460cm, since air resistance is neglected.

Clearly this procedure is meant to be conducted under ideal conditions. The experiment will be conducted in a vacuum ( since we were given no information about altitude, air density, etc. ).

This creates two cases.

1) the observer is not wearing a space suit. The observer suffers a seizure and falls downward where he is struck by both balls. (assuming his head deforms so that both balls can touch his face at the same time )

2) the observer is wearing a spacesuit. The small ball impacts his faceplate with relatively little force. All the same, the observer flinches, falls off the roof and ends up with his face in contact with both balls.

I know this is true because it happened to my brother-in-law.

POD

I agree

We assume no air friction, ideal elastic collisons (means no kinetic energy-to-heat convertion). 1. Maximum velocity V of the larger ball dropped from a height H1=50cm is reached just before collision with the ground: V = sqrt(2gH1), where g = 9.8m/s^2 Collision can be considered as a) collision of the larger ball with the ground and after this b) collision of two balls. 2. After collision with the ground, velocity of the larger ball changes direction at the same absolute value. 3. After collision of two balls, velocity Vs of smaller ball equals: Vs = 2V -> (typical example: tennis ball hit by the racket has velocity twice as big as the racket velocity) 4. Maximum height Hs of the smaller ball can be calculated as Hs = 2r + Vs^2/(2g) = 2r + 4H1 = 2.1m, where r is a radius of the larger ball. Answer: neither ball reaches height of 4.5m

Can you explain your stage 3? My version of stage3 would be:

After the lower ball bounces, it is rising at speed V, and the smaller ball is still falling at speed V. This gives a relative speed of approach of 2V. When they bounce off each other, the relative speed of separation will also be 2V. Because the small ball has little effect on the speed of the larger ball, the larger ball continues to move upwards at V relative to the ground - so the total speed of the small ball relative to the ground is 3V.

If this is wrong, please say where

I think you are right. - This was my mistake. - I forgot that the small ball has downward velocity V. In inertial system moving upward vith velocity V, the big ball doesn't move(after collision); the small ball has velocity -2V before collision and +2V after collision with heavy ball. This means that upward velocity of the small ball is 3V relative to the ground. Therefore it will reach height 9 times larger then initial 50cm. => (50cm)x9 = 4.5m.

Not being a member I cannot see or add to my last post #34 but I would like to consider a hypothetical idea where the larger ball strikes the short end of a non elastic and weightless lever which is occupied on the other end far enough away from the fulcrum point to balance the weight of the larger ball. I think the smaller ball may burn up by friction with the air before it gets out of our atmosphere.

Don

Assuming that the lever has no wind resistance and is perfectly rigid - wouldn't the fulcrum break first?

After reading through the numerous comments, solutions and observations it's obvious a lot of you have forgotten Engineering 101. First a definition to refresh your memory and then a few assumptions:
1. Elastic collision means NO LOSS of ENERGY. There is, absolutely, no deformation allowed in an elastic collision. All momentum and energy is transferred.
2. For such short distances we can safely neglect air resistance.
3. To use real numbers let's assume that "neglible" means the mass of the larger ball m1 is 100 times larger than the mass of the very small ball m2. This is very important as we'll see later.
4. We will assume a proper mechanism will release the balls in contact with each other.
5. The impacted surface is sufficiently rigid and massive to provide for the elastic collision given.

Now we have a nice simple free fall momemtum problem. But we need to define our variables and givens:
m1 = mass of 0.1m diameter ball.
m2 = mass of very small ball, the diameter not important but the ratio of masses is, remember we previously assumed m2 = m1/100 to achieve neglible status.
r1 = radius of larger ball, m1, Given r1 = 0.05 meters.
d1 = diameter of larger ball, m1, 2*r1 = 0.10 meters.
s0 = Initial height from surface to base of larger ball (set at datum - this will become clear later when we will need to introduce vector directions). Given s0 = 0.5 meters.
s1 = Surface.
s2 = Bottom of very small ball at impact above surface.
s3 = Height of roof.
s4 = Height of face looking at very small ball.
v0 = Initial velocity at s0 = 0.5 meter height to bottom of larger ball.
v1 = Instantaneous velocity at impact.
v2 = Instantaneous velocity of very small ball at elastic collision transfer.
v3 = Instantaneous velocity at face.
v4 = Average velocity of very small ball from collision to face.
t0 = Time at s0 and v0.
t1 = Time at s1 and v1.
t2 = Time at elastic collision.
t3 = Time to face.
Ao = Constant acceleration, for Earth's gravity this is 9.81 m/sec^2 (nominal value).

Initial values:
s0 = 0
t0 = 0
v0 = 0
The general distance equation for a free falling body is:

s(t) = (Ao*(t(t) - t0)^2)/2 + v0*(t(t)-t0) + s0

Find the time to fall from 50 centimeters (0.5 meters).
For (t)=1 and (s)=1 we have:
s1 = Ao*(t1)^2 + 0 + 0
0.5 = (Ao/2)*t1^2, solving for t1;
t1 = (2*0.5/9.81)^.5,
t1 = 0.319275 seconds from release to impact.

The general velocity equation for a free falling body is:

v(t) = Ao*(t1-t0) + v0
Find the velocity at impact.
For t1 = 0.319275, t0=0, and v0 = 0 we have:

v1 = 9.81*.319275 + 0 = 3.13209 meters/second

Momentum:
Remember F = ma? It is also written F = m * dv/dt and since we're doing elastic collisions the force of impact must equal the force of reaction.
F1=F2
F1=m1v1/(t(t)-t(t-1))
F2=m2v2/(t(t)-t(t-1))
m1v1/(t(t)-t(t-1)) = m2v2/(t(t)-t(t-1))

Zero deformation equates to equal time so time cancels out and we then have:

m1v1=m2v2

Recall that we assumed that m2 = m1/100 as being neglible? We now have:
m1*v1 = m1*100*v2, NOTE the masses cancel out! Solving for v2.

v2 = 100*v1, Very important! No matter what material you use it's the ratio of masses which becomes the multiplier for the initial velocity of the very small ball.

v2= 100*3.13209 = 313.209 meters/second. WOW! And I don't mean World of Warcraft.

Because the problem statement used elastic collisions all the kinetic energy was transferred to the very small ball. Thus the larger ball remains motionless and in contact with the surface. And yes, the very small ball will hit you in the face if directly over the balls.
Of course in real life that is dependent on the actual mass and hardness of the ball materials and considering the coefficient of restitution and air resistance. But for hard materials it won't be much of a decrease.

Check the velocity at your face in case you want to see how much effect gravity has on slowing the very small ball's vertical velocity.

Distance from very small ball to face is:
s4 = s3 - d1
s4 = 4.5 - 0.1 = 4.4 meters.

Solve for time (using general equation again), and remembering that v2 is in v0 position AND the direction vector is now upward so Ao is now negative as we are working against gravity.:

s(t) = (-Ao*(t(t) - t0)^2)/2 + v0*(t(t)-t0) + s0

s4 = -Ao*t(3)^2/2 + v2*t(3) + 0, insert values.

-4.905*t(3)^2 + 313.209*t(3) - 4.4 = 0, basic quadratic equation, solve for roots.

Roots = (-b +/- (b^2-4*a*c)^.5)/2*a

First root:
(-313.092 + ((313.209^2-4*(-4.905)*(-4.4))^.5)/2*(-4.905) = 0.01405 seconds.
Second root:
(-313.092 - ((313.209^2-4*(-4.905)*(-4.4))^.5)/2*(-4.905) = 63.841 seconds.
Since second root is entirely unreasonable and the first root is reasonable.
t3 = 0.01405 seconds

Velocity at face (remembering again that Ao is negative, working against gravity)

v3 = -Ao*t3 + v2

v3 = -9.81*0.01405 + 313.209

v3 = 312.954 meters/second instantaneous velocity at your face (if you're really looking down at it).

If you choose to do attempt this experiment for real at home or in a lab, please be careful and take needed precautions to prevent injury and possible death.

For example a steel ball 10 cm in diameter is about 4.11 kg, 1/100 of that mass equates to a steel ball approximately 2.1 cm in diameter.
At 300 meters/second it's going to hurt real bad.

You can more easily do this with a basketball and a tennis ball.

Approximate mass of a basketball is 600 gm and is 24 cm in diameter.
Approximate mass of a tennis ball is 57 gm and is 6.5 cm in diameter.
Or roughly a ten and a half to one ratio of masses.
Changing the diameters and heights, the exit velocity of the tennis ball is about 30 meters per second, that's impressive. Include the coefficients of restitution and a little air drag and the velocity should drop into a range that you can experiment with in your driveway and confirm the equations.

Reading your assertion under 1 "there is, absolutely, no deformation allowed in an elastic collision. All momentum and energy is transferred." it is clear you have either misremembered or were incorrectly taught. The constraint is that there is no net transfer to unwanted modes of vibration when the collision process finishes. Note that momentum is conserved irrespective of elasticity.

Similarly, under 2, you say that "for such short distances you can safely neglect air resistance". How so, when calculations say you can't necessarily do so. (E.g. 4.5-metres is not such a short distance for a ball of 8-mm radius - it will disturb air with a mass of 0.9-gm while rising that distance.)

I have skipped the majority of the intervening stuff, because it's late here and my temper won't take it. But, jumping to the end, your suggested speed of 300-metres/second is nonsense - this conserves upward momentum, but acquires a factor of nearly 9 excess kinetic energy; as you say yourself, you have to conserve both energy and momentum, and your solution clearly fails to do this.

Fyz

If you assume perfect elasticity (this does not necessarily imply zero times) there is no need to go into forces - momentum and energy changes in each ball are far more practical. However, the source of the problem result can still be fund in this complex presentation:

It arises from the following equations:
F1=m1v1/(t(t)-t(t-1))
F2=m2v2/(t(t)-t(t-1))

You are equating the velocities. But the balls were previously moving, so you will need to equate the changes in the velocities. It is also helpful to make the equal and opposite action and reactions explicit - i.e. show the opposite signs (I appreciate that they are implied in your treatment). I.e.

F1=m1(v1initial-v1eventual)/(t(t)-t(t-1))
-F2=m2(v2initial-v2eventual)/(t(t)-t(t-1))

Also, t(t)-t(t-1) can cancel because they are intrinsically equal, not because they are zero (in itself, that would give no information - equating infinities is as helpful as equating zeros - effective as a limiting value of a finite process, but otherwise devoid of information)

Fyz

I have an elastic band that does not deform . I call it a rope.

If you are lying on a roof 4.5 meters (4.5 meters x 3.28084 = 14.76378 feet x 12 = 177 inches) watching two balls falling 50 cm (50 cm x 0.3937 = 19.685 inches) to strike an elastic surface, will one of the balls (both of which are elastic in nature also)be able to bounce all the way back to you to hit you in the face? Sorry, not in this life time.

The problem assumes that by some mysterious means, the two balls, one that is smaller sitting on a larger one, will generate enough NEW energy to propel one of the balls back up to you. This is a factor of over 9 : 1 (177 divided by 19 inches) of NEW energy create by the dropping balls. Unless my physics teacher in high school was dead wrong, you lose energy every time your object touches another object. Lost energy means that you cannot have your object even get back up to the original starting point.

Some of the discussions that I have read discussed the NEW found energy being generated by the increased distance between the heavier and the lighter balls, well it would seem to me that: Number 1 -- Objects of the same shape fall at the same rate of speed. Ball = ball regardless of size or did the experiment on that Tower of Pisa not mean a thing to the world of science. Number 2 -- Remember the two balls are only falling 19 inches, so how much separation could there be even if you do not take into consideration the vacuum formed by the larger ball in front of the smaller one and the attraction of the masses. Number 3 -- Enough said.

Unless I am reading the problem wrong, it is physically impossible to be hit by one of these balls. NASA would love the elastic substance that would keep going higher and higher, especially when you could bounce a rocket payload into space.

I really think this is just a trick question put out by Chris Leonard to see if we are really thinking about the problem as presented or are we putting more into it than we need to. If there is a substance with the characteristics needed to make this a valid experiment, I will gladly invest in it.

Hi Procrastin,

You have obviously never tried this as I and the Guest in post #82 have. Believe me the smaller upper ball take off like a rocket. If you don't believe me go at get a basketball and a tennis ball, drop them as the question stipulates and look directly down from above. I suggest however you wear a pair of safety glasses as you will get hit in the face by the tennis ball.

Would you be surprised to see a light object on the longer arm of a lever rise above the level from which a heavier object was dropped? Hopefully, you would not - but your argument applies equally to that case. If you are still dubious, follow Masu's advice and try the experiment of post #82 for yourself. Hopefully, the evidence of your own eyes will then convince you to look at the maths with a better educated eye.

Fyz

backing this up, and making the test slightly simpler:

use coins (I used a US quarter and dime) a quarter is about 1.5x the diameter of the dime, about 1.5x as thick, and both made of similar materials- most importantly, you probably have both on you right now (I don't keep a basketball and tennis ball in my office, I don't know about you guys)

drop them seperately, then set the dime on the quarter and drop

sure, losses are high, but the dime will rise at least 3x higher on the 2-coin test than it will alone- and, you will notice that the quarter still rises some in the 2-coin test (not as high as it does dropping alone, but it does still rise) as well, proving that full momentum transfer to the dime is not the case (momentum is transfered to the dime proportional to the dime's mass)

Guest: thanks for the thought. It would be a terrific idea if the smaller coin could be made to bounce above the original drop height. I've tried this with some local (non-US) coins, and although I see about a factor of two improvement in bounce height, I can't get any near this. If it worked that would be great - but as it failed for me, it might not quite be the thing to convince the sceptics

Fyz

Sorry to be so late to the party but I have fresh material. First I'd like to say that I agree with Davo in post 5 that it's a trick question and balls hit the ground, not the observer, at the time they bounce. Ignoring that case, I agree with Randall in post 6 that the small ball hits the observer after the first bounce. I chose to enter my comment in response to Physicist in honor of the amount of effort he's put into this topic and for what I've learned from him today.

It looks like I'm the first to run a computer simulation. Using the free, evaluation version of "Working Model," I created ground and the two balls with elasticity = 1 and the small ball = 1/1000 the mass of the bottom ball. I added a beam at 4.5m as the observer with default elasticity = 0.5. If the balls are in contact when dropped, they bounce as a single unit. The program doesn't consider deformation of the large ball so this result can be argued. The fun happens if the balls are slightly separated when released. The small ball "shoots like a rocket" off the larger one and has just enough push to tap the observer! The large ball is just starting a third descent when they collide again so the second bounce is short of the observer. The large ball is traveling upward when they next collide. The observer will lose an eye if he doesn't get out of the way of that rebound!

It's been fun guys,

Bob S.

Hi Bob

Thanks for the post - very constructive. You will understand that I'm only doing to recover things that I once knew - or thought I did, so it's me that should be grateful to everyone else for introducing new aspects (particularly as you say Davo and, more tersely, Randall)

Anywhere else, it would probably be a trick question - but there have been some overly simple questions that only made sense as tricks - and they turned out to be merely naive, so I'm working on the assumption that this was not deliberate.

Now, I think we've established that the condition for uniform spheres to be elastic is that the surface compression and the rigidity of the body do not interact - that is, we can treat each sphere as a point mass on an ideal spring. That should be reasonably simple to model. If these simple resonators are in contact initially, I believe you will find that the reaction of the very light ball depends on the relationship between the resonant frequencies of the spheres - the lower one resonating against a massive ground, the upper one resonating against a relatively massive lower sphere. If the resonant frequency of the smaller mass is relatively high, I think the spheres will tend to bounce as one. If it is relatively low, the larger mass will have completed its bounce before the smaller one is affected, so we would see the classic three-times velocity. Intermediate relationships would need numerical analysis (or simulation).

Returning to the musak of the spheres - naively, we would expect the smaller sphere to have higher resonant frequencies in all respects (proportional to the radius), so I think that "bouncing together" will not be too bad an approximation if the small sphere is effectively of negligible size. So the residual question is whether the balls will be close enough at ground impact for the bounces to be treated as one.

The other thing worth considering is what happens if we have uniform bodies that could theoretically rebound elastically. "Right prisms" (including cylinders and tubes) could satisfy this if the material's Poisson ratio was zero and the material itself were lossless. The other requirement for the collision between two specific prisms to be perfectly elastic is that the times for the compressional waves to propagate through each prism are identical. If we follow that through, we can see that contacting cylinders would always cause the upper cylinder to rebound at the 3V-delta, rather than rebounding together.

I enjoyed your illustration of the ball bouncing off my(?) perfectly flat face. However, as the ball is supposedly rather light, I would not be too concerned if it were coming upwards at 6.25-m/s (about the maximum possible if it had hit my face on the immediately preceding bounce). That's only 14-mph, after all - nothing compared with a hard-ball (or cricket if you prefer)

You may (or not) be interested in the following extension - can we envisage an arrangement where the upper object rises faster than 3x after the first bounce? It seems the answer is yes - if we make the lower object a blunt cone, the change in velocity at any height will be proportional to the square-root of the cross-sectional area. Place a suitable object on top of a sharp enough cone, and the sky's no limit (excepting air resistance, of course).

Fyz

Be careful about taking one on the chin for science. That small ball only needs to be negligible compared to the large ball. :-)

-Bob S.

It's the same speed as dropping it from <2-m (~6-ft). I agree that I would have cause for concern if the large ball was made of osmium.

m₁ , m₂ mass of balls

v₀ velocity of either ball just before impact

v₁ , v₂ velocity of balls immediately after impact

conservation of momentum:

m₁v₀ + m₂v₀ = m₁v₁ + m₂v₂

v₂ = (m₁v₀ + m₂v₀ - m₁v₁)/m₂

v₂² = (m₁v₀ + m₂v₀ - m₁v₁)²/m₂² (1)

conservation of kinetic energy:

0.5m₁v₀² + 0.5m₂v₀² = 0.5m₁v₁² + 0.5m₂v₂²

m₁v₀² + m₂v₀² = m₁v₁² + m₂v₂²

v₂² = (m₁v₀² + m₂v₀² - m₁v₁²)/m₂ (2)

combining (1) and (2) and simplifying:

(m₁² + m₁m₂)v₀² - 2(m₁² + m₁m₂)v₀v₁ + (m₁² + m₁m₂)v₁² = 0

v₀² - 2v₀v₁ + v₁² = 0

(v₀ - v₁) = 0

If I read this aright, your balls still appear to be falling. Could you have omitted the impact with the stationary ground?

Oops. Let v_BI = - v₀ . (-v_BI - v₁) = 0.

If one collides with the other after it rebounds:

conservation of momentum:

m₁v₀ - m₂v₀ = m₁v₁ + m₂v₂

v₂ = (m₁v₀ + m₂v₀ - m₁v₁)/m₂

v₂² = (m₁v₀ - m₂v₀ - m₁v₁)²/m₂² (3)

conservation of kinetic energy:

0.5m₁v₀² + 0.5m₂v₀² = 0.5m₁v₁² + 0.5m₂v₂²

m₁v₀² + m₂v₀² = m₁v₁² + m₂v₂²

v₂² = (m₁v₀² + m₂v₀² - m₁v₁²)/m₂ (4) (same as (2) above)

combining (3) and (4) and simplifying:

(m₁² + m₁m₂)v₀² - 2(m₁² + m₁m₂)v₀v₁ + (m₁² + m₁m₂)v₁² - 4m₁m₂v₀² + 4m₁m₂v₀v₁ = 0

m₁(v₀ - v₁)((m₁ - 3m₂)v₀ - (m₁ + m₂)v₁) = 0

m₁ = 0, no ball

(v₀ - v₁) = 0, no collission

((m₁ - 3m₂)v₀ - (m₁ + m₂)v₁) = 0

(m₁ + m₂)v₁ = (m₁ - 3m₂)v₀

v₁ = v₀ - v₀( 4m₂ / (m₁ + m₂))

lim(m₂→0) v₁ = v₀ , m₁ is the much more massive ball

m₁ = 3m₂ , v₁ = 0

m₁ = m₂ , v₁ = -v₀ , equal masses

m₁ = (1/3)m₂ , v₁ = -2v₀

lim(m₁→0) v₁ = -3v₀ , m₁ is the much less massive ball

both , 32 ft per sec per sec

Is this an answer to some other post?

I'm thinking we can't know the answer without more information. Wave action through the large ball would need to cause the part of the large ball in contact with the small ball to rebound faster than the average rebound velocity for the large ball. My gut feeling says that a slinglike rebound would be more likely if the smaller ball had enough weight to deform the surface of the large ball beneath it so as to be in contact with the surface of the large ball for a longer time as it momentarily expands beyond its normal y dimension. This is more weight than "negligible" would suggest. Perhaps the viscosity of the large ball will also come into play.

Another thing that troubles me is that waves of different kinds are liable to propagate around the ball, some across the surface and some through the middle like in an earthquake. If p waves, which should arrive first (strength and speed dependent on the balls "filling"), dislodge the small ball before s waves arrive, the s waves will not sling the small ball away as efficiently or possibly at all. I can't back it up, but again my gut feeling is that the s waves are more likely to be locally amplified in a concentrated way and provide enough thrust to launch the smaller ball a great distance.

It is hard to imagine that there are not combinations of weight and qualities of the large ball's material that would produce yes and no answers. However, if "negligible" means something like a grain of rice on a basketball and the large ball is as stiff as a basketball, the answer would seem to be no.

Keith

Just saw this on (don't laugh) Beakman's World. A goofy Saturday morning science show for kids. He put a tennis ball on top of a basketball and dropped them at chest height. The tennis ball was launched like you couldn't believe. The energy stored by the basketball from it's collision with the ground was transferred to the tennis ball. Due to the mass difference, the energy excels the tennis ball at a much higher rate of speed in comparison to the basketball's bounce.
You would think with that with this explanation you would be hit in the face by the ball if you were directly above, but just as the energy transfer is "amplified" so is the position of the tangent intersect of the two balls. The centers of the two balls would have to be perfectly vertical for the small ball to be launched vertically. Since you have two circular objects touching, you don't have much room for error. Theoretically, if you could do this experiment with two square shaped objects, it would bounce straight up (as long as the ground was flat). Andrew Williams

Did they really say in the programme that "the energy stored by the basketball was transferred"? If you'd been watching carefully, you'd have seen that the basket ball bounced back nearly to its original height - so most of the original energy in the basketball stayed right there. Tsk tsk

What can't speak can't lie. If the heavier ball returned to a liitle bit less than original hight , the loss of it's potantial energy would have a very significant effect on a much lighter ball above it. For what it's worth I'm puzzled by the 'correct' answer - the large ball can not instantly start moving upward (upon ground contact ) because it is elastic and would take time to deform before returning to shape and rise off the ground."words mean what I intend them to mean" ! (Alice in Wonderland ) .

I fear that in practice it would be even more tricky with the square shaped objects, as the base of the lower one would have to be aligned to the ground - else much of the energy would be converted to vibration. I think that is why coins don't usually bounce too well. Lozenges, perhaps?

Fyz

The 2 balls at most can bounce back to their original positions if there is no friction and it is completely a reversible process. Otherwise the 2 balls bounce back to lower positions due to energy loss.

If the big ball bounced back to its original position while the small one bounced higher than its original position, the system would have had more energy than when they are in their initial positions. You break the energy conservation law.

YM

I suggest you actually try it and see for yourself, I think you will be surprised with the results. As for the conservation of energy the large ball doesn't quiet bounce as high as it would have if the second smaller ball wasn't there.

In reply to 140 the idea of using a blunt cone (presumably with a hemispherical base to concentrate the stored energy of compression to the small object at the top seems a great idea, is this not the same principal that is used in Plutonium bombs to compress the core?

Yes, the principles are the same, as are some versions of fusion bomb (I believe that magnetic confinement is currently winning the competition for fusion reactors, however) - though in both those cases the contribution of interest is the physical compression rather than the velocity of the material. The crack of a whip has similar origins, though the reason for the concentration of energy is change in acoustic velocity rather than reducing cross-sectional area. Regarding the shape of the base - if we assume a flat ground, I think we would be looking at the spherical shape being concave, so the wave leaves the edges first and ends up spherical. I think there would at best be problems with non-uniformity of the intensity if we did it this way, so some of the energy would end up as vibration of the cone - but even with a flat base the top would go off at great speed. If we want to minimise the loss to vibrations, I suspect a lens in the base of material with a different acoustic velocity (but with similar acoustic impedance) would be suitable.

But perhaps this is trying to push a somewhat pointless technology a bit far?

Fyz

Someone needs to learn a little about the fact that velocity is a vector and the direction of the velocity is important in this discussion. As the large ball hits the ground it's velocity changes direction, the small ball's direction is still the same, the relative velocity between the two balls is v-v= 0 not 2v. And how could the little ball pick up another v of velocity simply by changing. Besides, this is a conservation of momentum problem, not conseration of energy.

Perhaps that someone is yourself. Before the first bounce, the velocities are both v measured downwards, so the difference (I'll take little ball minus big ball in each case) is v-v. After the big ball has bounced, and before the little ball hits it, its velocity is -v, so the difference is v - -v = 2v. Go figure

Such statements need to be considered with more care. Perfect elasticity means conservation of kinetic energy so K.E. is the one property of the pair of balls that is conserved throughout the bounces. Also remember that the balls' momenta are not conserved during bounces from the ground - the lower ball's momentum is reversed in sign, and twice its original momentum transferred to the (assumed infinite mass) ground. (Total momentum is conserved, of course).

Fyz

Cool, only 54 years old and I see my first documented explaination of a perpetual motion machine!

Please explain - conservation of energy is not perpetual motion. Or do you just mean the legitimate "perfectly elastic" approximation?

Fyz

....except that the "big" ball was never introduced--except in the solution; there being no visible reason to conclude that the very small ball was not, in fact, (what would be) the large ball--owing to careless syntax of the challenge. Properly stated the challenge would have read:

A very small ball is on the top of another[, larger] ball with a radius of 5 cm; the mass of the smaller ball is negligible compared to the larger ball. The balls are dropped from a distance of 50 cm between the ground and the bottom of the larger ball. Lying on a roof 4.5 meters from the ground, you observe this experiment looking directly to the falling balls. If all collisions are elastic, will you be hit in the face when the balls bounce? If so, {immaterial & irrelevant>>[by which ball?]}>>>[...by which, either, or both]?

Such as it is, without evaluating the alternative case (where "very small" is the "big" ball), and demonstrating that (then) the big ball would not strike the face (and without also publishing that result), the solution given cannot be said to be unique; hence the challenge itself must be said to be invalid.

No need - unless you consider a 10-cm diameter ball to be "very small"

This solution is not correct.

Conservarion of energy means that both the small and the big ball bounce back and reach their original height at zero speed. If the smaller ball would get higher as suggested in the answer, it means the big ball would not reach it's original height 0,5meter but has given some of its own energy to the small ball.

I am neither a theoretician or well versed in mathematics but I would like to suggest an experimental setup that that would approximate the conditions laid down in the question.

suspend three steel balls of mass 100Kg, 1Kg and 10 grams from 10 meter cords so that they are just in contact.

Move the two lighter balls away from the heavier one to such a distance that when released together they attain a velocity of 1/ms by the time they strike the heavy one (I leave it to others to calculate the appropriate distance).

when the lighter ball is ejected on rebound its velocity can be calculated by noting the angle its suspension cord makes with the vertical and various mass ratios between the two lighter balls can be tested to establish the effect this has on the velocity.

Disclaimer

The specified condition are not met precisely.

The proposed Newtonian demonstrator experiment - but 100-kg sounds a bit steep. Perhaps it would be better to simulate this with a second 1-kg ball, and release the two 1-kg balls at the same time (e.g.cut a supporting string?) to simulate an infinite mass ground?

Why I specified such a large mass to approximate to the infinte mass specified in the question was so that the mass ratios between the balls could remain high without the mass of the lightest becoming too low.

There is of course no need for the heavy ball to be either spherical or supended it could even rest on the ground as long as it had a vertical face aligned to the axis of the two small balls.

This is not what is wrong with the answer, as the question said the mass of the smaller ball was negligible compared to that of the larger ball. The 'official' answer merely applies that approximation. The solution corresponds to the question in that respect.

However, the question clearly implies the presence either of air or of an evacuated enclosure - you could not place your face above the experiment without one or other. The enclosure would stop the ball hitting your face, and the answer clearly neglects air resistance.

The small ball is not traveling at 3v. It travels at 2v. The relative speed of the balls is v. So it does not hit you in the head, although maybe it should?

I think that this problem is more complex than it appears at first site, when the large ball hits the solid surface the kinetic energy that it has acquired goes into a wave of deformation that moves at the speed of sound from the point of contact to the trailing face, if the small ball was not there it would be reflected at the trailing face back to the leading face and propel the large ball back the way it came.

When the small ball is present some of the energy in this wave of compression (travelling at about 2km/sec in a steel ball) will go into propelling the small ball at quite a high velocity the actual velocity in proportion to the mass ratio of the balls but the limit would be the velocity of sound in the large ball

Wow, what planet does this logic come from that you apply? You have just created energy from nothing! I thought that was reserved for the Heavenly Creater. You should patent it. Please do not work on anything that involves the safety of others.

Two things; First the Small Ball can only go higher than where it started if the Large Ball imparts some of it's energy to it. If this happens, then the Large Ball cannot return to the same height as it's starting position as you have stated. In fact, you need additional information of the weight ratios to determine the actual response. Second, I think you must have been struck in the head with something as a child.

Dan H.

Please read some of the detailed material in the thread. Perfectly elastic is the limiting condition of low loss, negligible weight is the limiting condition of some weight. So, following the question, three-times bounce velocityt for the smaller upper ball and no loss in bounce velocity for the lower ball is the appropriate answer - but no air loss is not.

The "answer" is predicated on the assumption that there had occured during the fall some degree of seperation. Assuming that there was no seperation;

While the stated [yet undefined quantity of] elasticity may impart some fraction of the 2v stated as the larger balls velocity in the "answer" and therefore cause the smaller ball to gain a small bit above the larger, any number of factors could over come this small fraction and keep the two balls joinded until entropy ceased there motion;

Static charge collected during the movement through the air.

The 'negligibl'-ility of the small ball could be such that the gravitational force of the larger ball might overcome gain imparted by elasticity.

And of course, the question properly phrased to avoid confusion, would have ruled out the possibility of either of the balls being made of Flubber...

Hi Physicist

I find you arguments in 174 convincing, I hope someone can produce a video of a simulation using a Newtons cradle.

I still have a niggling doubt as you seem to assume that there is some separation between the balls as they strike the ground, I do not see any valid reason why such a separation should arise as there are at least three ways in which the balls are attracted to each other

Please see my reply on this to Kris - post #179. You will see that there is indeed a conceivable condition where the balls are in contact at bounce, and that the speed of bounce of the upper ball can then be anywhere between V and 3V, depending on the relative hardness of the balls.

(But, is infinity/zero a load thereof?)

Surely , if the larger ball, having hit the ground where energy must have been lost and then hit the small ball where more energy, however insignificant,must have been lost, is capable of attaining the same height as it was originally dropped from, this represents perpetual energy!

The small ball may well bounce higher than 4.5m, but without additional input of energy, the large ball must bounce less than 50cm, unless of course the large ball was made of flubber.

See the terms of the question - "negligible" means the effect is small enough to be treated as such. The problem with the official answer is that, if the relative mass of the smaller ball is truly negligible, the effect of air resistance on that ball is huge.

What is 'flubber'?, some strange substance to which the laws of physics do not apply.

Yes ! It's in a film with Robbin Williams I think .Kris

From the way I interpet this the small ball is on top. If the small ball was on bottom there would be no chance that a small ball would ever be enough to send a larger/heavier ball to any height let alone 9h .................

The question here seems to be that if the small ball on top was accelerated by the collison with the larger/heavier ball after it collides with the bottom and is on its resulting upward travels.

I am not so sure that the small ball would reach even the height that the balls were dropped from because there is something that was not noted that would make a big difference in the outcome !!!!!

There is nothing said about this was or was not to take place on an airless world. As this is assumed to take place here on Tera Firma, then fluid dinamics must play a role in the outcome !!!

I Question " Will the balls still be in contact with each other and if this is so then what would the outcome be hitting the bottom while still in contact with each other ?? ".

There exist a possibilty that they will rebound as one and the small ball will not achieve the height that they were dropped from !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Why do so many want to change the rules of the question? Which position the larger ball is in is not specified. What is specified is that the author percieves the top ball to be very small, and that it is on top of another ball with a radius of 5 cm. The "very small ball" could be larger than 5 cm. "On the top of" means touching and centered unless defined otherwise. "Elastic" does not specify a mathmatical equation. Influences of anything other than what are specified by the question should not be considered when searching for a solution. "Words mean things". (Rush Limbaugh)

It seems that you are objecting to answering more than one interpretation of the question - and then insisting on an eccentric one of your own. On the equivalent basis that "on top of" should mean touching and centred unless otherwise specified, if the same sentence defines the size of one ball and states the other is very small, the correct interpretation is that the "very small ball" is very small compared with the defined ball - unless otherwise defined. Plus, in your interpretation, you state that the very small ball has radius greater than 5-cm -and on the basis of "negligible mass" for the 5-cm ball, very much greater than 5-cm (unless the questioner states something about densities). Do you really know anyone who would describe something larger than 10-cm diameter as "very small" in the context of a 50-cm drop?
If the balls are contacting at release, is the top one resting on the lower one or is it held in pressure-free contact? In the absence of additional definition, I would assume the upper ball is resting on the lower one - so they will be separated by the time they bounce. Does the question state that the balls are released simultaneously?

Given the thread from which the questions come, and the answers given, it is legitimate to assume that they are written on the basis of the most natural interpretation, rather than as a wordplay exercise. But it is still legitimate to consider physical practicality whaen answering - even if the questioner didn't

Centred on the rules of the question, my words "could be larger" are not dependant on who I know, how well I know them, or how they would describe anything. "On" equals contact unless otherwise defined. "On top of" when used in regard to a sphere must be centred as anything else would be eccentric.

I did not object or insist. I asked.

A question is not really a question if you know the answer too! (John Prine)

The above (badly drawn ) images represent the gradual morphing of a shape (from the bottom up ). Volume/mass is the same as the 2 balls in question are formed. When the small ball is eventually formed does it really take on new properties in an instant ? I am referencing an analogy I made earlier about Galileo , but perhaps did not describe well.

Hi Kris

Sorry it took so long to get back to you. I think there is a distinct difference between the necked balls and the separated balls - the lack of tensile connection. Just before the sections are constructed as separate touching balls, you have the condition that a bounce could break the join. This is why I don't think it is realistic to consider two balls as a limiting case of a shaped ball - except insofar as narrower necks will break with smaller shocks.

Fyz

Thanks Fyz

The neck I envisage is infinitesimally narrow and short before separation . When the neck is cut the behaviour of both balls suddenly changes (wow ! - , before separation the shape had bounced to original height. I don't think anyone has done this with balls of the same material - using different material can produce spectacular results. Nobody can have tested this Q as described . I can see this becoming a 1 person crusade (or folly ! ). Since the last 2 Challenges have vexed me I shall continue to 'flog a dead horse' . Appreciate the reply . Interestingly nobody else has clearly been able to rebut my conjecture .

Kris

It's not exactly a sudden change when the neck goes to zero - the first change you see is when the neck is only just broken, so the balls can still bounce almost as one (though vibration may cause some additional separation. As the neck gets thinner, less of the energy is dissipated in breaking the neck, until the balls are entirely separate. At that point, the bounce depends on the properties of the balls.

However, if they are of the same stiff solid material and very different in size, I think they will bounce as one - because the bounce time of the lower ball will be long compared with the upper ball, so the upper ball never really sees an impact. Only if they are able to separate (during the fall) by a distance that corresponds to the bounce time of the lower ball will we see the 3x speed of rebound - and the separation distance (neglecting air as ever) depends on the size of the smaller ball. I.e. if a smaller ball of negligible relative mass and the same materials as a larger ball is resting on top of the larger ball when the larger ball is released, they will effectively bounce as one. (No need for the join).

Fyz