Elastic Ball: CR4 Challenge (05/19)

If the ball bounces every time it hits the plane, then the gravitational force cannot be from the earth [or any other spherical mass] for if it was, then eventually the plane would cross the tangental point and it's apparent inclination would change to 'uphill'.

So, if gravity is not making this ball bounce, we must first determine what is!

I'm at a loss.

tom

=-==

The length of the plane would clearly determine how many times the bounced upon impact...before it was no longer on the plane. But lets not ignore what is being asked...

Certainly, gravity has to do it!

-Abe

If you post clever-clever comments like this one before a correct solution has been posted, I for one will assume that you found the problem as intended too difficult.

[Clearly, the uniform g approximation is reasonable for modest length of the first bounce. Wrt your comment: if you allow infinite bounces and no other sources of loss, the ball will bounce back and forth across the region where the plane is horizontal]

Based on the graphical representation, and the statement that "the figure shows the first two bounces":

The second bouncing distance is 8.077 feet.

The inclined plane is 19.27 feet long.

How did you get these numbers?

-Abe

By establishing a ratio to the sketch using my engineer's scale (ruler), based on the only known dimension of the first bounce, I then applied the ratio to determine the second bounce distance.

The text stated that the figure shows the first two bounces...

What else was there to go on?

OLD SCHOOL

What else was there to go on?

Trig, elastic collisions, reflection.

I haven't thought through the specifics, but I can see that the arc of the first bounce will be tangent to the line of reflection.

Oh my goodness, what on earth was I thinking? And your solution is????

Please enlighten the group on the values you are using, and how they were derived at, once you have arrived at the solution. It has been a long time since I graduated Cum Laude, and I have forgotten so very much.

K.E = P.E just before impact. What is the original height? How much energy is retained after the first bounce? H = 16t² where 16 = half acceleration due to gravity. What is t? Looking for a SIMPLE formula that excludes assumptions that have not yet been accepted as standards, I find myself still searching. If one takes the statement literally "the drawing shows the first two bounces" (or something like that) then the drawing should at least serve as a tool to check the solution. From what I have been reading in so many responses in almost all challenges, the text is always taken to literal extremes.

First, I would like to point out that I got up on the wrong side of bed this morning. I endured 102° F during the day yesterday, and tossed throught the night. Dust is blowing everywhere, and entering places not easily cleaned in public.

Having said that, you amuse me in your assumption that I hadn't considered the application of mathematics - as I see you fail to present any solution or suggest how to do so.

Your ability to throw words you don't understand other than as pixels or type, and your effort to suggest you know more than you do, would suggest to me that you might be better served seeking a career as a politician - that lot of scavengers for whom I have so little regard.

Please try to contribute as I have so humbly done, or at the very least, avoid embarrasing yourself by posting such absurd statements in the future.

Perhaps you should go back to bed today, too...

Yes, I agree and thank you for your eloquent suggestion. I have embarassed myself by venting on an innocent bystander. A vacation is in order

friction will make the ball start to rotate and it will store this energy and add to it and bound extra with more bounces, unless you assume frictionless ball/plane

Yes you are correct. It is also a necessary assumption that the ball has no rotational energy making it a point or the plane frictionless.

No Where In your Theory for solving this cunundrum do i seen you take into account the angle of trajectory or its associated effect on direction and force of impact.

The question states that the ball is dropped so the initial trajectory is in the direction of gravity until it hits the plane. The angle of the plane is irrelevant as the ratio between the first and second distance will be maintained regardless of the angle A (provided A is greater than 0 and less than 90 degrees).

To be exact, the inclination must be smaller than 45degree. Any greater, the deflection of the ball will be below the horizontal axis. The ball will be accelerating more as all of its motion will be downwards.

I haven't try computing it, but I think it will not be the same.

There is no problem if the ball is deflected below the horizontal. Sure it will be accelerating in the downward direction but it is anyway just to a lesser extent with smaller inclinations. The only inclination where the ball doesn't actually hit the plane is at 90 degrees as it travels parralel to it, for every other angle below this the ball bounces and continues to do so until it reaches end of the plane.

We have to consider the deceleration due to gravity if the ball deflect above the horizontal axis. The ball will not be able to travel parallel to the plane as there is an acceleration downwards and not horizontally, thus it will eventually touch the plane which will then change the constant horizontal force on the ball which will then depends on the angle which the ball hit the plane, which will perhaps create enough force to ensure a constant angle in which the ball will hit the plane each time. And also, the amount of kinetic energy will keep increasing ( no doubt ) due to conversion of potential energy from a fixed height that we don't know of.

I doubt the angle in which the ball hit the plane will be able to be a constant.

You wrote during the first bounce, it will take t time to reach the same height. What same height are you refering to? It will take another t time to reach the bottom of the height, which gives you 2t, but at the 2nd bounce, the ball would be hitting a lower part of the ramp, which will give you (2t + H), where H is unknown.

A inclined ramp resting on a horizontal plane due to gravity.

The 'height' is the distance from the ball to the plane in a straight line that is perpendicular to the plane. This is different to the height above the ground. To see this just tilt the original bouncing ball drawing until the plane is horizontal and the ground is on an angle. Gravity now acts in a direction that is somewhat sideways and can be broken up into a vertical and horizontal component. The ball bounces to the same 'height' above the plane for each bounce which is equal to the 'height' (remembering that the 'height' in this definition is the perpendicular distance from the ball to the plane) at which it was released.

Mmm, I assumed gravity to be perpendicular to the ramp, as if otherwise, it will look like I threw the ball at an angle against a flat plane.

But I got your idea of seperating gravity into 2 different components. Accelerations towards 2 different directions that adds up into the resultant of gravity downwards pull.

Your answer seems close enough, yet there is some parts that I feel that that it might be wrong. But I'm stuck figuring a way to explain or find the answer.

I will be interested in a "correct" answer. It seems to me that we need one more variable to be fixed, such as the angle of the ramp or the "spring" constant. Please let me know if I am wrong.

The trajectory must take into account the angle of the plane relative to imperfections in its surface and that of the ball as neither will ever allow an undeflected bounce. As to the angle is there a transverse angle to the plane.

Do I See It.

Hi, BobD!

So lovely when somebody else does all the work.

GA.

Mark

Congratulations to you for hitting the ball out of the park. (so to speak)

d1 = d2 since action and reaction are equal and opposite

For elastic d1 = d2.

d1 will not be equal to d2. You are missing something. :D

As per BOBD calculations; d2= ( initial velocity * Time)+ 1/2 acceleration x time **2.

Iinitial velocity shoudl be g sinA x 3t. but it is mentioned different.

Metamorphosis, see posts #22 and #24 (for your information.)

I agree that "d1 is not equal to d2". Others mention "d1 does not equal d2".

Do you agree that the whole problem as originally stated is significantly lacking in a key fact (the lack of the height above the plane when the ball is first dropped?)

You may have overlooked the direction of travel is different before d1 started. d1 results from no air resistance and dropping the ball vertically.

Except for the case when the plane is horizontal (making d1 = 5 impossible), the direction of travel before d2 is not the direction of travel before d1. Thus I conclude, unless I also overlooked something, d1 never equals d2.

The distance the ball drops is not given. This could be a situation of more variables than facts.

d1 = 5 could be true for an infinite number of plane angles, no matter how many assumptions are made to simplify the problem.

The following is an illustration:
If the ball were dropped 0.3 inches above the plane, d1 = 5 implies the plane's angle is far less than horizantal. If the ball were dropped 20 feet above the plane, d1 = 5 implies a plane very close to horizontal, so the ball can bounce very high without traveling horizontally very far.

This is another illustration, taken from junior high math.
The problem would be illustrated by a test question asking us to calculate the value of A. We are told two things: A+B=2 and B=1. We see A has a unique answer.

If B=1 is not told to us, there are an infinite number of answers for A. In fact, I can not think of any answer that is not a valid answer for A if B=1 is not told to us.

To get back to the question, the teacher MAY have forgotten to drink enough coffee to wake up before writing the challenge question. Since I am a Christian, perhaps they are the enemy trying to waste much of my time by a small investment of their time? This was the way history's "5th columists worked, as well as today's terrorists.

Yeah, I would like to have more variables to play with. I can only make that many assumptions without making me the God of the world in which that question happened.

"I assume that there is no air resistance, the constant elasticity of the ball even after n times of bouncing, the efficiency which potential energy converts to kinetic energy remains at 100%, no ion winds, no radiation, no magnetism, no height constraint, no specification on the ramp, no ....... gravity, perfectly smooth surface, nil tribo charging, zero induction etc etc. Alright. I conclude that the ball will stay in the air vacuum motionless with no contact with the ramp thus voiding the question. until I assume otherwise/when the answer is released"

You have to note that they did not specify the size of the ball or ramp. It could be a microscopic ball & ramp. Ion wind will be able to affect it.

Yup. Material isn't stated, so it could be any elastic ... metal? :D I do not have to dope it either, as all kinds of materials can be charged up through tribocharging and create electrostatic attraction. Alright. 5 points for off topic!

I'll say it again theoretically d1=d2 (and do not forget Newton)

Consider this

If the ball was bouncing down a very long inclined plane does the length of the bounce increase. NO! if it did the ball would start bouncing 5ft and end up bouncing a mile.

Remember each time it bounces gravity reverses itself to zero since it goes up, stops, and them comes down again.

On the other hand if a ball starts bouncing down a plane at 5ft does it bounce less?

Think

If A=0 then d1=d2=0 i.e. even spacing - why does the ball eventually stop ?

If A=90 d1=d2=0 because it misses the plane entirely !

What happens in between ? Why should d1>d2 or d1<d2 ?

Objects will accelerate to what speed ?

The ball starts at zero, accelerates, reaches a constant speed, stops at the bounce, go up (or across), stops due to gravity, starts again, accelerates, reaches a constant speed, stops at the bounce etc......

Also feel it should be d1 = d2.

In any trajectile motion distance covered would be same if the starting velocity is same with same angle.

In this case intial trajectile velocity is same in both bounces.

The velocit at whch the ball hit back will be same as that of speed at which it is thrown.

Also speed of bounce is same as that speed of hit( considering conservation of momentum)

Assumptions:

1) No air friction

2) Complete elastic collission.

3) No revolution of ball.

But given the assumptions of no friction, no air resistance, and a perfectly elastic collision the ball would continuously bounce. So if dropped exactly vertically on a flat plane it would bounce up and down to the exact same height infinitely. When it is dropped on to a sloped plane the first bounce will cause it to move some distance x (say 5 feet) in the horizontal direction and so now some portion of the energy is converted from vertical bouncing to horizontal bouncing thus changing the angle at which it strikes the plane on the second bounce. This is how we get a "longer" second bounce. Although I do not have the time right now to go through the math I don't believe we can determine a specific horizontal length for the second bounce, only a length in proportion to the first bounce as a function of the angle A.

I don't think the angle of imapct would be different assuming no energy lossed and friction

Angle of impact does matter, because although there are no energy losses, there are energy gains. Every time the ball bounces, it should return to its original starting height (no energy loss). However on the return flight it will fall further, because the plane is inclined. On the second bounce it should again rise to the original drop height, but then fall even further, because of the plane inclination. Because it continues to fall farther the distance must be greater. d1 can not equal d2.

The angle of impact does not matter as it remains constant. The ball will not return to it's original starting height as it has lost some of it's "up / down " energy, which has now been converted to it's forward motion energy. The angle of impact to the plane will always be directly with gravity if we are ignoring the "rotational / air resistance" properties. This would result in D1=D2.

I think the angle is very important. If it isn't then the time is needed.

That first hit, which if assumed is dropped from stand still, the angle will be horizontal to the "flat ground", and not horizontal to the ramp plane. The second hit will receive an additional component added to the velocity as was done from the elastic impact of the first hit. This velocity that is added for each bounce will be g*cos(x)+v-- the velocity in the x direction after the previous bounce. Thus v2 = 0+gcos(x) + gcos(x) == 2*g*cos(x). Now to get d2 we need the time, which we can't find without t, and to find t we need the angle(x) to plug to find the time for the first bounce. Thus we can find the velocity in the x direction after the second bounce but not d2, due to lack of information.

If you were to try to use potential or kinetic energy the same would be true, either the max distance in y direction after first bounce would be needed, or the time--in either case we need one more piece of info to be able to solve it.

The reason is because though gravity is counteracting itself in the y direction some of the kinetic energy is diverted into moving the ball in the x direction. Thus if the ball doesn't return to the same height from the plane as the previous bounce, the velocity is increasing in the x direction and the energy in the y direction is increasing. To find how fast it is increasing and thus be able to find the distance the second bounce goes down the ramp, the angle of the ramp (x) is needed/time/ some other variable is needed.

biggjoshie, see #22 and #24 for general comments.

I think you are right (I can not be sure until I do the calculations). If any of the missing and unguessable facts were stated, then all other information could be calculated.

The missing information is: the plane's angle (your idea), the trajectory's angle after impact (your idea), the time of flight during d1 (your idea). Of all the missing information, the most easily used is the height of the ball when it was released. (The equivalent information would be the ball's speed when it first hit the plane.)

As I said in #24, the height of the ball before released is critical.

When a ball is dropped on a flat surface all of the kinetic/potential energy is converted back and forth. On a plane it is changing each time due to conversion of some vertical movement into horizontal movement due to change in direction of momentum from elastic rebound.

"Only the component of the ball's velocity perpendicular to the plane will result in the bounce and since we have assumed a perfectly elastic interaction between the ball and the plane the bounce will be at the same height (perpendicular to the plane) for all bounces. In addition the time that the ball takes between bounces will be the same because this time is determined by the initial distance of the ball perpendicular the plane (gravitational potential energy perpendicular to the plane) which is converted to kinetic energy perpendicular to the plane and then to potential energy and so on (since no energy is lost)."

This is not true, because the angle of impact continually changes. If there is an angle that is not similar to the horizontal, thus the plane would not exist. Each bounce there will be less height reached from the plane because the ball will now have a horizontal component. Thus either time is needed to see the amount of energy that is converted to horizontal movement from vertical movement, or the angle of the plane is needed to see how much energy is converted from vertical to horizontal movement.

(the only way the ball could bounce to the same height would be if you gave it energy...KE total == PE total. KE==KEy+KEx...thus PE=KEy+KEx..

in beginning PE == KEy1+0, second bounce PE =KEy2+KEx)

Please read #24.

Do you feel the ball's speed is unnecessary to know? The ball's speed will change the unique angle resulting in d1 = 5 feet. That angle will change the horizontal component of the ball's speed during d2.

Is it possible that a slower speed (and thus a more horizontal surface) results in the dame d2 as a faster speed (and thus a more vertical surface?)

If the correct answers are "Yes" to my above questions, then you just prompted my to see the error of my ways. I won't have time to try physics for 2 weeks, when everyone will have abandoned the question. So, please answer with math or opinion.

BobD is right - it is not necessary to know the ball's initial height or speed or the angle of the plane.
[Subject to the necessary assumptions that:
the bounce and travel are both lossless, and
that no energy is coupled to rotation of the ball, and
that gravity is constant in magnitude and direction]

If you are so minded, you can of course assign algebraic parameters to all the variables (bounce height, time between bounces, angle, velocity at first bounce...) and calculate using vertical and horizontal coordinates. The times of the bounces, the vertical height of each bounce above the plane, and the horizontal distance of each bounce will all change - but these variables all cancel out for the length of the bounces along the direction of the plane, so we don't need that information.

BTW, subject to the simplifying assumptions, the general solution (for the Nth bounce) can be calculated as follows (different but equivalent to the methods given elsewhere):

Average velocity along plane during bounce number N =
t_b.gn.cos(A).{[(N-0.5)+(N+0.5)]/2} =
t_b.gn.cos(A).N
. where t_b is the (constant) time between bounces
But we are told that the distance for the first bounce is 5' - i.e.
t_b.gn.cos(A) = 5'

Substituting, we see that the distance for each bounce is
5.N feet

BTW, we can calculate any of the "cancelled" variables in terms of three of the others - i.e.
the 'height' of the bounce in the direction perpendicular to the plane is:
t_b².gn.sin(A)/2
The velocity perpendicular to the plane at first bounce is:
t_b.gn.sin(A)
etc.

Thanks CR4 for an elegant problem - and also to BobD for the matching elegant solution.
Finally - for those with rulers - the diagram is clearly not to scale - indeed the variable 'heights' of the bounces as drawn is clearly intended to require independent thought...

Hi, Physicist? !

Haw! Very neat. GA. Does this mean that the height from which the ball was originally dropped was 2.5 ft (a full pseudo-previous bounce), or 1.25 ft (the mean of the curve)?

Also, since the ball's dimensions and mass are not given, is it possible from the bounce to determine the coefficient of restitution between the plane and the ball?

GA again, I say.

Mark (Where'd that mint julep disappear to???)

There was an omission in my equation for the distance of the first bounce - it should have read t_b².gn.cos(A). I failed to notice because it cancelled with the same omission in the desired result.

That means that the height of the drop is dependent only on the angle of the plane.
The linear distance of the initial drop (vertical height) is 5'.sin(A).
Which means that the maximum distance from the plane in each bounce is 5'.tan(A)

The equations give the horizontal length of a full pre-initial bounce (bounce #0) as zero. This correlates nicely with the 'ball' being dropped vertically in the first instance.

Hi, BobD!

"The vertical height it reaches in successive bounces will be the same because it is perfectly elastic and so all of its vertical kinetic energy is converted to potential energy (height) and then back to kinetic energy"

Overkill, m'friend. Perfect elasticity is not required to get the same bounce. In fact, perfect elasticity would just muddle the question (because then where would the five-foot first bounce come from?). All that's needed is constant elasticity (the same ol' coeff. of rest'n. bounce after bounce of both ball and plane), which given the fact that the degree of elasticity's not a dimension mentioned in the challenge _{(the thing is just described as 'elastic' so we know it'll bounce!)}, is easily assumed.

[ Or, by perfect elasticity, did you just mean constant elasticity?]

In any case, your elegant solution doesn't call for perfect, or any grade, of elasticity except a consistant one, and one that delivers a five-foot first bounce. Relax! You don't need perfection, and you already got it right!

Mark

Hi Mark,

By perfect elasticity I mean that all of the kinetic energy (perpendicular to the plane) is converted back into gravitational potential energy (perpendicular to the plane) after each bounce with none being lost to heating of the ball or some permanent deformation or numerous other areas where real life bouncing objects tend to lose energy. It is perfect because of this. Constant (but not perfect in the sense I have defined it above) elesticity will not get the same result because the ball will not bounce back to the same perpendicular distance above the plane each time but to some % of the previous distance. If we assume a constant elasticity where 50% of the kinetic energy is lost the ball will only spring back to 50% of the height it was dropped from after the first contact and then to 25% after the second, 12.5% after third and so on so the time between bounces will reduce and so the length between bounces will also not follow the same relationship as if the elasticity was perfect.

I went out and dropped a golf ball and a basket ball on an inclined plane - 30 degrees and 45 degrees.

D2=1/4 D1 or 15 inches.

The vertical drop speed increases at 16*t² so the ball only has 1/4 the time for the second bounce!

It does not say the drawing is to scale nor that gravity did not help the ball to bounce.

I myself spent entirely too much time bouncing the problem around, looked at too many formulas and searched for others - thinking that perhaps a standard energy loss equation would explain the missing variables. So, I have been monitoring the progress of you who are still wading hip deep in numbers and formulas, as it has been ages since I have been crunching numbers in some other-than-habitual routine manner.

Still waiting for a distance to be offered; except for BobD, no one seems ready to take the leap...could it be possible that the sketch itself holds the answer as a "scaled" diagram, as was pointed out by that beligerant, groughy old fart with the disposition of a caged badger?? oh wait... that sounds like me...my apoligies to the group and especially any individuals. But what do yu think about using the drawing - or is that unthinkably simple? 8.077 ft.

currently, until some formula that I also have not been able to find, or have ever seen/ heard of for the energy conversion, I agree with your 8.077 ft, although I would like to place some error percentage on it, such as 6% for sand, and 6% for being grouchy d2==8.077' +/- 1' .

On the other hand, I say that d2=2gcos(x)*tb

where:

x is the angle of the plane that is perpendicular to the force of the gravity.

Gravity is a constant.

tb is derived from manipulation of PE and KE equations, and is dependent on either (x), or the height of the drop, time of ta of flight from bounce one to bounce two., or max height reached after first bounce.

also, another component that i just realized is from each bounce an amount of energy is added to the PE due to declining plane, but is also unknown due to the missing data mentioned before(angle, height of drop or time, or PE)

I agree, and think I might have previously eluded to that as well...in some fashion or other...

Anyway, the only practical solution I could come up with was to scale the drawing.

However, the brainstrainer, when constructed in a manner similar to the sketch, using a RUBBER ball, shows less of a consecutive distance...

http://www.globalspec.com/BrainStrainer/

checked it out--not sure that the ball is considered 100% elastic.

Hi, biggjoshie!

Been through all this. There's no room for either interpretation or common sense in these challenges (not even allowed to call them 'riddles'!).

Ya gotta go with the bare bones of the generator. If the thing says 'elasticity' (or whatever), then that's what you've got. Might as well assume that the thing and its incline have perfect elasticity, in which case there may not even be a second bounce. But there is a second bounce, so they have somewhat imperfect elasticity, and their coefficient of restitution is a non-issue as a mathematic construct, so long as it remains constant. It's the incline that give us the distance differential; and since the incline's length is not given, it's endless (which is environmentally unfortunate since we only need it for two bounces'-worth of distance!)

Mark

the ball is not even 50% elastic...--doesn't bounce to 50% of height after dropping from horizontal, therefore if our ball is considered 100% elastic, it doesn't have much worth when comparing to this question, the brain strainer follows common sense, whereas this question skirts common sense with the whole no friction, no air resistance, no energy is lost added(other than additional falling distance as you move down the ramp) etc mentioned from a couple other posts, that are typically assumed true in physics questions where this info is not provided.

Consider a coordinate system with x along the surface of the plane and y perpendicular upward from the surface. The ball is initially moving with a speed of V0 when it hits the surface. This has velocity components sinA*V0 in the x direction and -cosA*V0 in the y direction. After the first impact, the x velocity is unchanged since I've assumed a frictionless surface. The y velocity just changes sign and is then cosA*V0. Since there is no friction, there is no torque on the ball and it doesn't spin. The acceleration of the ball in the x direction is g*sinA and the acceleration in the y direction is -g*cosA. We can integrate the accelerations and get equations for the velocities. One more set of integrations and we get the distances. I get x(t) = V0*sinA*t + 1/2*g*sinA*t^2 and y(t) = V0*cosA*t -1/2*g*cosA*t^2. Solving the y equation for t when y = 0, the next impact is at t = 2*V0/g. At this time x = 4*V0^2*sinA/g. Since the acceleration in the y direction is constant and the impact velocity is the same, the next bounce will take the same amount of time. At t = 4*V0/6, X = 12*V0*sinA/g = three times the initial of 5 feet or 15 feet. 15 feet minus the initial 5 feet leaves 10 feet. The next impact would be at 24*V0*sinA/g = six time the initial distance or 15 feet for the bounce.

Jim

Doesn't add up here...care to show your work?

"the x velocity is unchanged since I've assumed a frictionless surface."

x velocity changes after each bounce regardless of friction. Also found a physics simulator to emulate what is happening and to help visualize, its called Phun.

http://www.phun.at/

you can make a plane, fixate it, put bounciness to 1.000 on a ball, and friction to 0 on ball and plane, take away air resistance, and simulate it. I don't have time now, or lack of interest because I think there is missing info. One could theoretically create some sort of a ruler and measure out distance ... it appears to be about 2X the first bounce: which would fit your d1=5, d2= 10 answer. comments...

make the angles a bit more different to prove your point,

also what is that simulation software, if formulas are inputed, what did you input, etc... a little background on this random graph is needed...

I'm not trying to prove my point. I'm just trying to give you a little insight. Equations are the same as in my post #57. Software was Excel spreadsheet. Details left for your development.

I'm "inclined" (get it? HAR!) to go along with BobD and jim35848, who appear to be saying the same thing in different ways. Many have been ignoring the fact that with each bounce, the ball gets an additional impulse in the horizontal direction, and will therefore accelerate horizontally. That's why d1 can't equal d2, unless the plane is horizontal and the horizontal acceleration is zero. It also makes sense that the distance *perpendicular* to the plane (which BobD confusingly referred to as the "height" because of his rotated frame of reference which so many didn't get) would be constant with each bounce. BobD helped me see this by postulating a "horizontal" plane in which gravity acted at some angle not normal to the plane. Although this is an inherent contradiction ("horizontal and "vertical" have no meaning other than that dictated by gravity), it did make it easier for me to see that the gravitational component normal to the plane is constant, so the ball always returns to the same "height" in a perfectly elastic bounce. This is borne out by jim35848's Excel plot as well, or so it appears to my eye.

I'm way too lazy to attempt this mathematically, so I'm gonna go with the guys who did. (The drawing can't be to scale, because the "heights" of the bounces normal to the plane are not constant. Sorry grouchy guy with sand in his ass-crack, but that screws your proportionally-derived answer. Besides, 2d1 is so elegant, is HAS to be right! In fact, it looks like it's nd1, which is even cooler.)

The formula is indeed nd1.

The force of gravity acting on the ball can be broken up into two orthogonal vectors one acting parallel to the plane and the other perpendicular to it. Since these vectors are orthogonal their effects on the ball's motion are independent of each other and the resultant motion of the ball is the sum of these effects. So considering the effect of each of these separately, the perpendicular gravity vector just makes the ball bounce up and down on the same spot going to the same perpendicular distance above the plane each time since is perfectly elastic interaction. This also means that the time it takes between bounces is the same and if we take t as the time for it to first hit the plane from where it was dropped the time between successive bounces is 2t.

Now consider the gravity vector acting parallel to the plane. This is just accelerating the ball with constant acceleration parallel to the plane starting from the point on the plane that is perpendicular to where the ball is first dropped from.

Now combining the effects of these two vectors we get that the ball hits the plane at times t, 3t, 5t, 7t, etc (where t is time for it to first hit the plane from where it was dropped). Now at time = t the ball has also moved horizontally along the plane by an amount = (1/2.horizontal component of gravity).t². lets call this distance D. The second contact happens at distance (1/2.horizontal component of gravity).(3t)² = 9D the third at 25D the fourth at 49D, etc. Since d1 was the distance between the first and second contact (length of first bounce) its length is 9D-1D=8D, while the distance between the second and third contact (length of second bounce) is 25D-9D=16D=2d1. Now the distance between the third and fourth contact (third bounce) is 49D-25D=24D=3d1, and so on.

Therefore the distance between successive bounces is nd1 where n is the number of the bounce.

It appears we have a few solutions for the problem described if the impact is frictionless and the ball does not spin. However, the ball would pick up some spin if there was friction at the interface. Even if there was no energy loss, the ball would be spinning after impact and the translational speed after impact woud be less than before impact. I think this would slow the linear velocity component along the incline but not change the component perpendicular to the surface as compared to the frictionless case. At the next bounce, something similar would occur but the ball would already be spinning but probably not as fast as after the second impact.

Would this effect change the answer for the stated problem or would the answer still be the same if friction is included???

The answer is d2 = 10ft

Let the initial speed of the ball as it leaves the incline after the first bounce be V

let suffix n refer to the direction normal to the incline, and s to that parallel to the incline, Thus resolving V we get Vn=V*cos(A), Vs=V*sin(A): similarly resolving the gravitational acceleration gn=gcos(A), gs=gsin(A).

NORMAL TO THE INCLINE

The time for the ball to reach its maximum "height" fron the incline is given from 0 = Vn-gn*t_top

so the time between leaving the incline and returning to it is t=2*Vn/gn which is t=2*V/g

Assuming perfect elasticity, the ball keeps bouncing in this manner hitting the incline every t seconds

PARALLEL TO THE INCLINE

distance of the first landing of the ball d1=Vs*t+1/2*gs*t^2

distance of the second landing x2=Vs*(2t*+1/2*gs*(2t)^2

and the second bounce d2= (x2-d1) =Vs*t+3/2*gs*t^2

The ratio d2/d2 is (Vs*t+3/2*gs*t^2)/(Vs*t+1/2*gs*t^2)

and this reduces to the value 2 when we substitute the values

Vs=V*sin(A): gs=g*sin(A) and t=2*V/g

therefore d2 = 2*5ft = 10ft

Note that the answer is independent of both V (the original ball velocity) and A the angle of incline

Hi, SlideRuler!

First, a great big welcome to CR4! I can see by your answer to the puzzle that you will be a wonderful contributor to the CR4 website blogs in future. It's great to have you aboard!

As to your solution, it was very kind of you to revisit the solution in blog #8. As far as I and my hammock are concerned, the result is just as good. It's just wonderful to have someone else to confirm why I get to enjoy the world.

Mark

Thanks for the welcome

Since i'm new to this; what's the protocol? If someone posts the correct answer, we don't post? I'm not likely to be the first to answer, my typing is too slow.

That brings up another point. What editor can I use? I see people including graphics etc.

Hi, SlideRuler!

It turns out that Smiley Central deposits a link for Adware. Just use I.M. Smiley for the smiley graphics, as they have no hidden qualities. They just get their income from shared Google searches from their search line.

Mark

Hi Everyone,

Firstly, well done BobD on your very succinct analysis of this challenge.

Secondly, I'm sure most of you are aware of the PHUN software ( A physics simulation sandbox) which was introduced during Fyz's falling ladders challenge.

For those of you who do not have it installed in your computer, you may download it from here:-

http://phun.cs.umu.se/wiki

It is a great way to simulate lots of different physics problems regarding falling and colliding bodies in particular and is ideally suited to this challenge.

I have created a simple scenario of a ball falling on a plane ( Both ball and plane need to have friction set to zero and bounciness set to 1.00).
I have created a second upside down plane just above the ball's initial drop height to show how the ball always returns to the same perpendicular
distance from the plane as was explained by BobD in his answer.

Once the software is loaded on your computer, simply copy the text below into a file called "Elastic Ball.phn" and place this file in the directory
C:\Program Files\Phun\scenes

You should be able to then run the program and load this scene and see visually exactly what happens.

You can even put a ruler along the direction of the plane on your screen and verify that BobD's answer is indeed correct - Have PHUN everyone :)

********** Start of text file called "Elastic Ball.phn" - DO NOT include this line in the file!!!! ***********
version = 1;
{
angle = 0.0;
body = 0;
color = [0.5744562149047852, 0.4262505173683167, 0.6796445250511169];
dist = 4.183838367462158;
friction = 0.0;
id = 302;
normal = [0.1366370618343353, 0.9906211495399475];
pos = [-0.07333333790302277, -1.033332943916321];
restitution = 1.0;
type = "plane"
};
{
angle = 0.0;
body = 0;
color = [0.4113966524600983, 0.5336093306541443, 0.9409500360488892];
dist = -4.494370460510254;
friction = 0.5;
id = 382;
normal = [-0.1414213478565216, -0.9899494647979736];
pos = [0.0, 0.7733328938484192];
restitution = 0.5;
type = "plane"
};
{
angle = -2.854032039642334;
body = 12;
collide = true;
color = [0.5145063996315002, 0.7980877757072449, 0.2328007966279984];
density = 2.0;
friction = 0.5;
id = 398;
pos = [14.09203720092773, -7.163225173950195];
radius = 0.04807396978139877;
restitution = 0.5;
type = "circle"
};
{
angle = 0.0;
body = 13;
collide = true;
color = [0.8359873294830322, 0.6629816293716431, 0.6519746780395508];
density = 2.0;
friction = 0.0;
id = 402;
pos = [0.8600000143051147, -3.960000038146973];
radius = 0.04853402078151703;
restitution = 1.0;
type = "circle"
}

HEIGHT DECREASES LENGTH INCREASE.FIGURES DEPEND ON FORCE.

If you take the ball and drop it it first starts to slide, then it starts to roll and continues to roll until it reaches enough compression to overcome gravity. AS it rolls it spreads the gravity over a distance and as it rolls the ball is decompressing in the area already rolled.

I read the frictionless board and completely elastic ball, but that is assumption not basis on any of the facts given.

So even if you could assume the ball is completely elastic, I don't see how you can assume that the incline does not have friction.

So when the ball starts to come in contact with the board it starts to slide then it starts to roll. Being completely elastic it will take longer to reach maximum compression, and during this time it has been rolling, and the compressed spot is moving around the ball and as it rolls the ball is decompressing even as it compresses in a new spot.

The ball will never reach the distance on the second or third bounce as it did on the first!

In reality, a harder ball bounces further than the more elastic ball since it reaches maximum compression much faster and will slide further before it starts to roll.

The term "elastic" when used without qualification is usually taken to imply that the ball itself is "perfectly elastic" - and indeed that the situation can be treated as lossless except where specifically stated otherwise. Even if the medium is lossless, the ball will only bounce perfectly elastically under the limiting conditions that
a) the surface deformation is small, and
b) the contact time is long compared with frequencies of vibration of the ball
- otherwise there will be conversion of energy from the bounce to vibrations in the ball.

If you allow generation of spin due to friction, you need to know how much of the energy is converted to spin - i.e. the size of the ball, and the proportions of the momentum that is converted to rotation and to vibration. If we assume that the ball rotates as a single entity, I doubt that you can simultaneously balance momentum and kinetic energy (i.e. the situation has to be lossy).
In any event, allowing conversion to spin would contradict the second convention for such challenges - that the solver should make the minimum additional simplifying assumptions that allow a unique solution without additional numerical information.

In practice, if the coefficient of restitution is greater than about 80%, we may expect the second and third bounces to be longer than the first - regardless of whether the ball is large enough for conversion to spin to be a significant issue.

Interesting problem indeed, but I saw it only after the answer was posted. Anyway, I'd like to follow a graphical approach to reach the result, as it provides a better visualisation compared to equation solving. We only need to use the fact that the area under a velocity-time graph is the distance travelled.

My figure below shows only bare essentials due to my limitations in coping with computer graphics. It is so easily done by free-hand sketching on square-ruled paper (similar approach as in my post in the River Crossing thread).

I'll try to keep my text brief. The X and Y directions positive are as shown, inclined at α. Acceleration components are g.sin α down the plane, and g.cos α normal to it. They may have any ratio to each other (positive of course) depending on the inclination between zero and 90 degrees. These values are the slopes of the two velocity components, and the two motions in X and Y directions may be treated independently.

At zero time the ball is dropped from some arbitrary height 'h', and it makes the first bounce after a time 't', both values being of no interest to us. In the Y direction the ball keeps bouncing back to height 'h' at points A, B, C, etc. which are separated by time intervals of 2t. In the X direction the ball is steadily accelerating, and we are interested in the distances down the plane for the first two bounces.

This is merely a matter of 'measuring' the area under the X-velocity segments AB and CD (i.e. to the time axis). Before the first bounce, the area under OA is the small triangle which we can treat as a half-unit of area. Simple counting of rectangles (and triangles converted to equivalent rectangles) will yield the area under AB as 8 units, under BC as 16 units, under CD as 24 units, and so on in arithmetic progression. Since we are told that the 8 units under AB is d₁=5 feet (or whatever), the distance for the next bounce will be just double that or d₂=10 feet (and subsequently 15, 20, etc.) All other numerical information is irrelevant, or cancels out as already demonstrated in other posts. The same argument will hold good for a very steep inclination α in which the slopes of v_x and v_y are interchanged, though the figure may look somewhat different.

I think one of the gurus will be able to 'explain' my method more lucidly with a better diagram perhaps.

=TeeSquare=