Velocity at the bottom will be 4.43 + 1 = 5.43 m/s.

Therefore Energy = 1/2 mv^2 = mgh

E = 1/2 x 1 x 5.43^2 = 1 x 9.81 x h

h = 1.502 m

If you add the two kinetic energies (at the bottom position) you get E=10,312J, so you get again: h=1,05m...

So what's the problem???...

The difference is that the energies are not equvalent. Adding 1 m/s to an object at rest is different than adding 1 m/s to an object already travelling 4.43 m/s, because velocity is squared in energy.

Enegry change at rest to 1 m/s = 1/2 x 1 x 1^2 = 0.5 Joules
Energy change at velocity = 1/2 x 1 x (5.43^2 -4.43^2) = 4.93 Joules

No... It's nothing like that...

The correct method is to add the two kinetic energies at the point of our interest... If you just add the two velicities and you calculate the total kinetic energy (considering this sum) leads you to false result...

Hi Poisson. You wrote: "The difference is that the energies are not equivalent."

You are right and your sums are correct. This brings us to the real 'heart' of the puzzle. The horizontal kick-mechanism transferred more energy to the sphere than the vertical one and hence used more energy as well, correct?

Now suppose the kick-mechanism was attached to the ball and not to the structure, in other words, it was moving with the ball and designed to give a 1 m/s kick to the ball. If done in the original vertical position, one would expect it to expend a net 0.5 J on the ball, just like before (it will use more energy in order to accelerate itself, but that is not part of the question).

When the kick is delivered horizontally at 0 m height, will it now also need the 4.93 J of energy that you calculated, or will it only need 0.5 J, or something in-between? (Again consider only the energy imparted upon the ball, not on the kick mechanism itself.)

Jorrie

Hi Jorrie,

Thanks for the confirmation. In order to increase the ball's velocity by 1 m/s when its already travelling up to speed, it will need the 4.93 J not the 0.5 J.

This is an interesting result, if I'm correct. Does it not follow that as velocity gets increasingly large, the energy needed to increase an object's velocity also becomes increasingly large, even for incremental increases of only 1 m/s?

Hi again Poison. You asked: "Does it not follow that as velocity gets increasingly large, the energy needed to increase an object's velocity also becomes increasingly large, ..."

Yes, there are two effects, the one Newtonian and the other relativistic. What we calculate here is purely Newtonian due to the low velocities. It boils down to the fact that in the 'fixed' reference frame, the force must still be applied for the same time to get the same impulse. In that time the ball moves a larger distance during the horizontal 'kick' than was the case for the vertical 'kick'. The same force for a longer distance means more work, meaning more energy transfer and so on...

In the reference frame of the ball, things may be different. For all practical purposes, it's moving inertially at the bottom. It receives a force for a specific time (the same time as in the 'fixed' frame, because the speeds are not relativistic). In its own frame, the ball only 'sees' 0.5 J of energy added to it. Since the kick-assembly travels with the ball, it delivers only 0.5 J to the ball.

However, we agreed that the ball will reach 1.5 m height in the fixed frame, so it must have more additional energy than the 0.5 J. Where does this 'extra' additional energy come from?

Jorrie

Whether the mechanism is in the ball or outside of it, I agree with Poison. To increase the ball's velocity from 4.43m/s to 5.43m/s requires 4.93 Joules.

Hi Maths_Physics_Maniac, (info Poison). You wrote: "Whether the mechanism is in the ball or outside of it, I agree with Poison. To increase the ball's velocity from 4.43m/s to 5.43m/s requires 4.93 Joules."

Yes and no. Set up an inertial frame x',h' that is moving at -4.43 m/s (i.e., to the left) relative to the U frame. At the lowest point the ball will be at rest relative to frame x',h'. It gets the 1 m/s speed kick to the left. What is the ball's increase in kinetic energy in frame x',h'?

Kinetic and potential energies are completely frame of reference dependant and I propose that we can only make claims as to how much energy is actually expended to accelerate the ball by working in a comoving frame (like x',h'). Even the "ball's own frame of reference" that I used before make no sense, because the ball's kinetic energy in its own frame of reference is always zero and its potential energy constant.

It becomes a bit 'hairy' to calculate the total energy for this setup relative to frame x',h' - still working on that and I will appreciate any help...

Jorrie

Well, looking at that there picture, it looks like that ball is gonna roll a bit. So in order to figure the velocity at the bottom I think we might want to consider the energy that was converted to rotational energy, 1/2Iω² . Seems like maybe mgh = 1/2mv² + 1/2 Iω². So it looks like we don't have enough information.

Since you gave us its mass, all we need before we can calculate I (moment of inertial) is its radius.

Now once we've done all that, we can start to make the basic calculations needed to know how much energy it will take to get that ball a rolling 1 m/s faster. Its gonna take a bit more than just the increase in the linear component of its kinetic energy. Now we can look at this from any ol frame of reference we care to, but the inherent energy required to accelerate that ball is gonna be the same.

Gavilan

Hi Gavilan, you said: "Well, looking at that there picture, it looks like that ball is gonna roll a bit."

In a friction-free environment?

Introducing friction will complicate things considerably, but feel free to assume some values and start calculating! It will however obscure the point this puzzle is trying to make.

Jorrie

Hi, Jorrie... If Poison is right the result doesn't make sense... It doesn't matter for how long a force is applied on the ball (as you said before)... Afterall the result must be the same in both cases: an increment of the velocity which is Δu=1m/sec... So, in both cases the amount of the (extra) kinetic energy that is obtained by the ball is the same: ΔE=0,5J... In the second case, if the h is larger (e.g. 1,5m) this means that the final potential energy is higher than the sum of the two kinetic energies (E+ΔE) which must be 10,312J... This is against the law of energy conservation... Where this "extra" energy came from???...

Hi George, you asked: "Where this "extra" energy came from???..."

This is the point of this puzzle, so to speak.

Jorrie

So, that's why I said that the Poison's method must be wrong... Because it seems that suggests that the ball obtains more energy than is permitted... Afterall, what's the actual result???... Is it h=1,5m or h=1,05m???... e.g. what result we could get in a real experiment???...

Hi again George. Poison and myself believe it is 1.5 m.

I think we are correct, because NASA has demonstrated the principle hundreds of times in space. Our problem is to find the mechanism, the explanation...

Jorrie

O.k. I accept this as a fact...

But does this mean that the method that I followed above (e.g. the sum of the two kinetic energies and then calculation of the h) is wrong?????... and why???... (also, why the two methods, i.e. Poison's and mine, give different results???)... I only need a clarification...

Hmmm... ... Now I realise that the extra kinetic energy that I considered is not ΔE=1/2.m.(0,5)²... To add the two kinetic energies was right but the "ΔE" was wrong...

Hi again George, ""To add the two kinetic energies was right but the "ΔE" was wrong...

You cannot add two kinetic energies linearly. You must add the velocities and then calculate the new kinetic energy, i.e., ½m(v1+v2)².

I will provide analytical 'proof' of the whole scenario in due course.

Jorrie

You are right...

Well here is what happenS:

W=F.S=m.γ.S=m.(Δu/Δt).[(2u₀+Δu)/2].Δt=m.Δu.[(2u₀+Δu)/2]

(where (2u₀+Δu)/2 is the mean velocity @ Δt where we apply F or γ)

results: case1: 0,5J & case2: 4,93J

The above equation shows the dependence of the work (W) that we offer to the ball(or the energy which is obtained by the ball) with the u₀ (the velocity that the ball has at the moment when we start to apply the force)... This means that the larger is the u₀ the larger is the work that we offer to the ball (in order to get the same increment of the speed of the ball)...

If we carry this experiment up to the speed of light minus 1 m/s, does the energy required to acheive the speed of light become so large that it becomes an impossiblity?

Hi Poison, yes, as soon as speeds go relativistic, things get more complex. The energy needed to 'kick' something to the speed of light would be infinite, no matter how close it was to light-speed before the kick.

Jorrie

Hi George, you wrote: This means that the larger is the u₀ the larger is the work that we offer to the ball (in order to get the same increment of the speed of the ball)...""

Yes, that's true in 'our frame', where the ball has speed u₀. However, in a frame that is comoving with the ball prior to the 'work offer' being made, i.e. u'₀ = 0, the work will be less, as can be easily verified. In this frame the force is applied over a shorter distance and hence the energy required is less.

The catch in this is that the energy source must be in the comoving frame already and hence it did cost a lot of energy to get it there from 'our' frame. Unless of course, we 'mine' the energy right there in the comoving frame. Hmm...

Jorrie

The puzzle is about where the extra energy of the ball come from when the kick is given horizontally at the bottom of the U rather than when the ball is at rest at 1 m height. The difference is 1.5 – 1.05 = 0.45 m height gain or 4.43 J of 'free' potential energy, which sounds paradoxical.

The simplest way to explain the 'paradox' is by means of Newtonian reaction mass propulsion. Let the 'kick mechanism' be a 0.226 kg compressed spring that is located in a hole in the 1 kg ball. The spring can be released to serve as a 'kick motor'. When released, the spring flies backwards, releasing its potential energy of compression. Let that energy be enough to give 1 m/s speed change in the direction of movement of the ball and 4.43 m/s speed change of the spring in the opposite direction, hence conserving momentum. The spring mass was chosen do that after the kick the spring will be static at the bottom of the U frame. It's the simplest to analyze and gives the optimal energy transfer.

The spring uses 0.5 J to accelerate the 1 kg mass by 1 m/s and E = ½mv² = 2.215 J to accelerate itself from rest to 4.43 m/s. The sum of these energies (4.93 J plus spring losses) was supplied when the spring was compressed. The energy required for accelerating the spring is only half the 'free energy' - where is the other half hiding? Well, we had to hoist the 0.226 kg spring to 1 m height before the experiment, using E = mgh = 2.215 J of energy, which is the other half. So, in the end, energy is conserved and the energy input is the same for both vertical and horizontal kick-scenarios, yet the horizontal kick gives more maximum height to the ball.

Jorrie

PS: Perhaps Gavilan can show us the same U-frame scenario for his electromagnetic propulsion so that we can compare results and more importantly, settle the issue of horizontal versus vertical kick with his propulsion scheme. It should be simple to actually set up a thought/practical(?) experiment to verify it on Earth's surface.

Jorrie,

I make this analysis by looking at the energies before and after the kick. The mass of the ball is 1 kg which includes the kicker. The kinetic energy of the 1 kg ball at the bottom of the 1 meter free fall is mgh = ½ m V² = 9.81 Joules.

Since it is stated that a .226 kg spring is used as a kicker, the ball will have a mass of 1kg - .226 kg = .774 kg after the kick.

The velocity at the bottom of the free fall was given as 4.43 m/s and the final velocity was to be 1 m/s greater or 5.43 m/s.

The KE final = ½ *.774 *5.43*5.43 = 11.41 Joules.

The change in energy is 11.41 Joules – 9.81 Joules = 1.6 Joules.

The velocity of the .226 kg spring relative to the now .774 kg ball would have to be 2.03 m/s in order to impart 1.6 Joules of energy. As you can see, I cannot yet agree as to the amount of energy required to increase the speed of the ball or the speed of the spring relative to the ball. The ball will however, move through ≈ 1.5 meters on the other leg of the U.

I again return to my original premises that KE=1/2mV² and that the energy needed to impart a change in speed where the speed is much less than C is |KE initial - KE final|.

Jorrie stated – "The spring uses 0.5 J to accelerate the 1 kg mass by 1 m/s".

This is true ONLY when the initial speed of the ball is 0. When the ball is moving at 4.43 m/s it would require an input of 1.6 Joules to increase the speed of the .744kg ball to 5.43 m/s.

Gavilan

"Faith begins where reason exhausted collapses." - Albert Pike

Hi Gavilan, you wrote:

"Since it is stated that a .226 kg spring is used as a kicker, the ball will have a mass of 1kg - .226 kg = .774 kg after the kick. "

It doesn't make much difference, but I made the ball 1 kg and the spring 0.226 kg, for a total of 1.226 kg. For the specific setup, that makes the calculations easier.

"I again return to my original premises that KE=1/2mV² and that the energy needed to impart a change in speed where the speed is much less than C is |KE initial - KE final|".

Again, this is frame dependant. In the inertial frame of the ball-spring combination before the kick, the ball was at rest and it gains only 0.5 J. As I've shown, there is however no total energy gained or lost.

The crux is, with the exact same spring and the exact same starting point (i.e. the exact same total energy input), the horizontal kick gives the ball a higher 'apogee' than the vertical kick.

This may however only be true in a Newtonian reaction mass scenario. I would like to know if your electromagnetic propulsion exhibits the same altitude difference between a vertical and horizontal kick. Maybe not, so why not calculate it for your scheme?

Jorrie

Jorrie;

Before we continue, can we agree that the approximate energy transfered by a kick motor will approximate 1/2Δm_fuelV²_exhaust.? Where m is the fuel mass and V is the rocket exhaust velocity.

Gavilan

Hi again Gavilan.

If you mean energy transferred to the rocket, I'm afraid I do not agree with "... the approximate energy transfered by a kick motor will approximate 1/2Δm_fuelV²_exhaust.?"

This only gives the kinetic energy change of the exhaust gasses relative to the rocket inertial frame before the kick. In my scenario the spring ("exhaust gas") gets a kick of 4.43 m/s, translating to 2.215 J in the frame before the kick, while the 1 kg ball only gets 0.5 J. How do you relate this to your quoted statement?

Remember, it's only momentum that is conserved in every interaction, irrespective of frame of reference.

Jorrie

Jorrie,

Looking at both our figures. How do we rationalize the conservation of momentum at the expense of conservation of energy? Remember our hypothetical systems are 100 percent efficient and without frictional loss.

Can you help me with this?

Do you have access to an air table or air slide?

Thank You

Gavilan