A problem with KE = ½mv². Help.

[arkain101]

Right Craig, thanks.

 

Thats exactly what my logic was telling me.

 

I read directly in wikipedia that rockets run into mass dialtion issues. But, Even if they did, how could they when the fuel is 'keeping up'?

 

A little off topic but related to relativity. I noticed a little thing that helped show why (with equations) that things can not travel at C. You will be creating a wavelength more powerful than gamma, in fact the wavelength will be zero.

 

If you run the redshift equation/Doppler Effect and tell an object that is emitting to travel at C, the wavelengths end up as zero.

[Farsight]

Lsos: my problem is solved. I now understand why KE = ½mv². Thanks Craig.

 

The guy Gary Novak says "Energy misdefined in Physics", and what he says is interesting:

 

http://nov55.com/ener.html

 

I'm still struggling somewhat with what energy actually is, and whether the guy is talking any sense or is a charlatan. We treat Kinetic Energy as the total quantity of "pushing" a moving object is capable of doing to another object. By "pushing" I don't mean "pressing", as in force, but making something else move, and thereby doing "Work". The thing that troubles me is that there appears to be no real, actual, "energy" anywhere in the system, all there is is an object of mass x moving at velocity v capable of making another mass move to reach another velocity.

 

I'm also troubled by elastic and inelastic collisions. If I fire my puck from my sled into your sled, and can contrive a way to capture it in some form of elastic collision, the reaction on my sled is less than the action on your sled.

[arkain101]

Popular.

 

There is not Kinetic Energy. Its an equation created to explain how much fxd you can make happen out of moving mass.

 

It can be written like this aswell.

 

Ke=(M*V) (M*V) / (M+M)

 

 

It is to say that the object that interacts with our moving object will reflect the same M*V upon impact due to its inertia.

 

We then devide these two momentums so to speak (of equal mass, but only of the lesser mass) by the two equal lesser masses, for the purpose of aiming the result and product of the total momentum at only one of the TWO objects. We can only direct the full 'energy' of the equation into 1 out of the 2 objects. For you want A to move B or you want B to move A but that full conservation of momentum can only be 100% used in one or the other.

 

It is to say that any interaction of two free floating objects can only contain as much force as the lesser mass of the two objects involved (for free floating collisions). This is because if a big mass hits a small mass the small mass has lower inertia and easily bounces away and does not slow down the larger mass much, so it looks like they slowly seperate after they collide. If the small mass hits the larger mass it will barly move it and only contains the force of itself tied in with the always equal reaction, and here they will also appear to seperate slowly. It depends where you put you persective of which one is at rest.

[CraigD]

The guy Gary Novak says "Energy misdefined in Physics", and what he says is interesting: … http://nov55.com/ener.html …

I'm still struggling somewhat with what energy actually is, and whether the guy is talking any sense or is a charlatan.

I think it would be unkind to brand Novak a charlatan, as he doesn’t appear to be trying to exploit others for his own profit. In his brief web autobiography, he states that he is a trained biologist of some accomplishment, who apparently was unable to complete his PhD due to disability. He then read some popular Physics, and became convinced that it was fundamentally flawed from classical (Newtonian) mechanics on, inspiring his “Energy misdefined” website.

 

My impression is that he’s developed his ideas about fundamental flaws in classical mechanics in a haphazard way, without benefit of a sound basic education in it. He seems to have become emotionally invested in his conclusions being correct. Though it’s impolite to speculate publicly, I can’t help but wonder if his disability is of a psychological nature.

I'm also troubled by elastic and inelastic collisions. If I fire my puck from my sled into your sled, and can contrive a way to capture it in some form of elastic collision, the reaction on my sled is less than the action on your sled.

You can’t “capture” the puck with a single elastic collision.

 

In an elastic collision, both momentum and energy are conserved, that is:

[math]Momentum = Mass_1*Velocity_1 + Mass_2*Velocity_2[/math]

[math]Energy = {\frac{1}{2}}*Mass_1*Velocity_1^2 + {\frac{1}{2}}*Mass_2*Velocity_2^2[/math].

Following the initial firing of the first 0.2 kg puck from the 100 kg sled, the puck has [math]Velocity_{puck}=50 \text{m/s}[/math].

The second 100 kg sled has [math]Velocity_{sled}=0[/math].

So,

[math]Momentum = 0.2*50 + 100*0 = 10 \text{kg*m/s}[/math]

[math]Energy = \frac{1}{2}*0.2*50^2 + {\frac{1}{2}}*100*0^2 = 250 \text{J}[/math].

 

Given [math]Momentum[/math] and [math]Energy[/math], we can solve the 2 equations for [math]Velocity_{puck}[/math] and [math]Velocity_{sled}[/math], as follows:

Given:

[math]0.2*Velocity_{puck} + 100*Velocity_{sled} = 10 \text{kg*m/s}[/math]

[math]Energy = 0.1*Velocity_{puck}^2 + 50*Velocity_{sled}^2 = 250 \text{J}[/math]

 

we can solve the 2 equations for [math]Velocity_{puck}[/math] and [math]Velocity_{sled}[/math]:

[math]Velocity_{sled} = \frac{10 - 0.2*Velocity_{puck}}{100}[/math]

[math]0.1*Velocity_{puck}^2 + 50*(\frac{10 - 0.2*Velocity_{puck}}{100})^2 -250 = 0[/math]

 

Expanding, arranging, and substituting variables A, B and C, we find:

[math]A*Velocity_{puck}^2+B*Velocity_{puck}+C =0[/math]

[math]A= \frac{0.2 +0.2^2}{100} = 0.2004[/math]

[math]B= \frac{-2*10*0.2}{100} = -0.04[/math]

[math]C= \frac{10^2}{100} -2*250 =-499[/math]

Solving via the quadratic formula:

[math]Velocity_{puck}= ( -B +(B^2 -4*A*C)^.5 )/(2*A) = 50 \text{m/s}[/math]

[math]Velocity_{sled} = \frac{10 - 0.2*50}{100} =0[/math]

(one of the 2 solutions gives the initial velocities)

 

[math]Velocity_{puck}= ( -B -(B^2 -4*A*C)^.5 )/(2*A) = ~ -49.8 \text{m/s}[/math]

[math]Velocity_{sled} = \frac{10 + 0.2*49.8}{100} = 0.1996 \text{m/s}[/math]

(the other, the velocities after an elastic collision).

[arkain101]

I was wondering if about a few things.

 

Has the speed of light (the value of its velocity) been explained why it is that particular value?

 

Has the reason for the constant velocity of light for an observer been explained how and why?

 

Has the redshift been explained how and why light changes in frequency with the velocity of the source? In logical terms?

 

I wonder because I thought I had made some discoveries tonight with some things.

 

For example, I found that the speed of light can be calculated with a very simple concept. Even without lazers or any kind of measuring. It can be shown to be how and why that number or rate is what it is.

[arkain101]

I was hoping someone could quickly help me confirm these questions.

 

Scenario:

We have a rocket traveling in space. We are observering from a point at rest in space.

rocket 1000kg.

the rocket can exert a force of 2000N.

To keep the work simple. We ill assume this rocket loses no mass when it pushes a force with its fuel. We can pretend its just a force on the rocket.

 

We turn on the rocket and accelerate it.

a = F / M

a = 2000 / 1000

a = 2

a = 2m/s^2

 

We accelerate the rocket up to a velocity of 100m/s.

 

If we reverse the thrust of the rocket it should come to a stop in the same mount of time it took to accelerate and cover the same distance to do so.

Its momentum is its M*V. 1000*100= 100,000 P

 

Now if we speed the rocket back up to 100m/s again, then, after a moment we begin to double the velocity of the rocket to 200m/s. Now the momentum is 1000*200=200,000 P.

 

If we reverse the burn direction it should stop in only double the distance than when at 100m/s. Is this true?

 

Can Kinetic energy really apply when we are not dealing with any other objects in collision path? No work can be done to anything but the rocket. Lets assume the force the rocket uses is hypothetical and there are no gasses exiting.

 

Does the rocket in fact take 4 times the distance to stop like a car does in this scenario?

or,

Does it only act like an object under a force? and will double the distance to stop if it doubles its speed.

 

If I had the equations I would happily work it out for myself. But I dont and its late, and I will have to look around.

 

btw, this has nothing to do with my above post.

[CraigD]

Has the speed of light (the value of its velocity) been explained why it is that particular value?

Though there are many rigorous theories showing how the speed of light can be derived from other fundamental physical constants of the universe, or from complicated dynamic systems beyond direct measurement (eg: string theory), to my knowledge, none have show why any of these constants must have the value it does. The best explanation for why the speed of light is what it is appear to me to some version of the anthropic principle, which suggest that there’s no especially good reason the speed of light and other fundamental physical constants are what they are, other than if they were different, it wouldn’t be possible for the universe to contain such phenomena as animals like us wondering about such questions.

 

I suspect that no explanation of the values of fundamental physical constants will ever be as logically satisfying as the explanations of the values of fundamental mathematical constants such as e and Pi.

Has the reason for the constant velocity of light for an observer been explained how and why?

IMO, yes, elegantly, convincingly, and superbly experimental confirmed. I’m referring, or course, to Special Relativity, which explains the counterintuitive relative-motion-independent constant speed of light by allowing time and distance to vary between observational frames at motion relative to one another. I think Popular gives a very good perspective on this in ”Re: A thought on relativity”.

 

Has the redshift been explained how and why light changes in frequency with the velocity of the source? In logical terms?

Most people, I think, find the analogy between Doppler shift in light and in sound a convincing, though informal, explanation. Other good natural language-based explanations exist – one of my favorite is to imaging the mechanics of light as represented by a moving conveyer belt, the frequency of a particular source of photons as the rate at which a person places pebbles on the belt, and the frequency that an observer perceives as the rate that another person picks them up. It’s easy to diagram this scenario on paper to show how movement by the emitter or receiver person causes a shift in frequency, redshift or blueshift.

I wonder because I thought I had made some discoveries tonight with some things.

Do tell – but best start a new thread, as this one has drifted pretty far from its original subject :doh:

[arkain101]

Craig, thank you for the replies!

 

always excellent as usual. I had the same idea, but I wanted to ask a second opinion to make sure I was not going to be wasting my time.

 

If anyone could answer my question on the rocket question it would be great its kind of important.

 

To summerize. Does a rocket in space have the same action as a car on the ground; Where, if you double the initial velocity to a now called "latter velocity", the distance to stop becomes 4 times as great than at the initial velocity.

The car on the earth experiences a collision type of interaction when its in motion with the ground.

The rocket in space is a free floating object with no collisions, but only a force the equally accelerates and decelerates it.

[Farsight]

Re post 72, thanks Craig. And thanks for the math. OK, I get it. In rough terms, the puck has changed velocity by circa 100m/s and the sled by .2m/s. There's something of interest to me here about Momentum and Kinetic Energy, but I wish I knew what!

 

Arkain: your question sounds similar to what you were saying about a dropped object. After 1 second it's travelling at 9.8m/s and its average speed over that second is about 5m/s, so it's fallen about 5m. After 2 seconds it's travelling at 19.6m/s and its average speed over that second is about 15m/s. So it travels a lot further in the second second. It's the same with a decelerating rocket, only the other way around.

 

The whole thing about the rocket is that it's basically a sled firing a whole pile of pucks, and when you're looking at it from an at-rest reference frame the amount of "work" or "force times distance" or Kinetic Energy given to the rocket by a gallon of the fuel varies hugely with the speed of the rocket. But all you've actually got is a mass and a velocity, which equates to a momentum which equates to "force times time". I for one need to think some more.

[CraigD]

Does a rocket in space have the same action as a car on the ground; Where, if you double the initial velocity to a now called "latter velocity", the distance to stop becomes 4 times as great than at the initial velocity.

With respect to distance traveled, yes. The position of any object under constant acceleration is described by:

[math]Distance = \frac{1}{2}*Acceleration*time^2 +Velocity_{initial}*time+Distance_{initial}[/math]

regardless of what produces the acceleration.

 

With respects to Work, a ground car and a rocket are different.

For the rocket, [math]Work = Power*time[/math]

For a car, [math]Work = Force*Distance = Mass*Acceleration*Distance[/math]

 

So, for example a 1000 kg car on a smooth road on an airless planet, and a 1000 kg rocket in space with an exhaust speed of 5000 m/s (a bit better than the Space Shuttle), both accelerating at a constant -5 m/s^2 from an initial speed of 100 m/s, would stop after traveling [math]\frac{1}{2}*5*20^2 +100*20 = 3000[/math] m.

The car’s brakes would do [math]1000*5*3000 = 15000000[/math] J of work.

For the rocket to accelerate 5 m/s^2, it needs to user energy at a rate of

[math]Power = \frac{1}{2}*Acceleration^2*Mass_{rocket} +\frac{1}{2}*Velocity_{exhaust}^2*\frac{Mass_{rocket}*Acceleration}{Velocity_{exhaust}} = 12512500 \text{ W}[/math]

The rocket’s motor would do [math]12512500*20 = 250250000[/math] J of work.

 

So, for this specific maneuver, the car is about 17 times as energy-efficient as the rocket. Since distance increases proportionally to time[math]^2[/math], the greater the initial velocity, the more energy-efficient the rocket becomes compared to the car. I haven’t carefully checked my work, but I calculate the velocity at which the rocket becomes more energy-efficient than the car as 1669 m/s.

[arkain101]

Re post 72, thanks Craig. And thanks for the math. OK, I get it. In rough terms, the puck has changed velocity by circa 100m/s and the sled by .2m/s. There's something of interest to me here about Momentum and Kinetic Energy, but I wish I knew what!

Does this explanation help?

 

Kinetic energy does not exist untill you have TWO sets of momentum. When you do have two sets of objects so to speak then kinetic energy can happen. Since all energy versions are interchangeable this should apply to all version of energy.

 

The question is which of the two sets is which. Ke=(M*V) (M*V) / (2*M)

 

This is what makes me wonder; Can a moving object be said to gain energy as it speeds up? What if it is in a part of the universe where no object will ever cross its path. Or a hypethetical place where its the only object, how can you determine how much work it can do when there is nothing to compare it with or measure from.

 

I looked at some equations for quantum physics and found a familiar interesting thing.

looking at Ke=(M*V) (M*V) / (2*M)

 

and

 

(planks constant reduced) h = h (regular)/ 2*Pi

 

 

 

What is familiar here is a constant needs two things to happen. and so does energy (M*V) (M*V). There is a comman devider of 2*__ . Two things. So for planks constant there is two fundamentals of pi. and in kentic energy there is two fundementals of mass (made of constants of pi)

[arkain101]

reply to post #78 - http://hypography.com/forums/showpost.php?p=108884&postcount=78

 

Okay Craig thanks.

I think I see.

 

If a rocket can acclerate 0 to 100m/s in 10seconds and in the process cover a distance of X meters. It can also stop in the same X distance in the same time.

Then from 100m/s it can accelerate from 100m/s to 200m/s in the same amount of time as from 0 to 100, although it is already covering a distance of X meters each second and so that distance gets carried with the new distance and they are combined for a new total distance of (X+X) + (X+X) so to speak. 4x. and so, reversing the thrust from 200m/s to 0 will cover X(original) + 3X

 

Although how this says it gains an entity called energy is still not settled with me. For a rocket is a sharing of momentum device, it sends fuel in motion too. So the "energy" is only half used in rockets acceleration(s), or in other words, the momentum is split in two.

So if the energy or force (energy) could be put into full use in deaceleration, the rocket should not take 4x the distance to stop.

[arkain101]

Oh and I was looking for someone to check my work in this post.

 

http://hypography.com/forums/108692-post9.html

 

I was calculating the energy produced in the tidal Hyrdaulic gravity lift device I designed and got some mind blowing figures.

 

I think its because I used a 1000:1 ration. However It is possible to design it to have that.

[Lsos]

I don't see why the rocket wouldn't take 4x the distance to stop. The same eqautions apply whether the force comes from the ground or propellant or gravity or anything. The same reasoning to why it would take 4x the distance to stop apply as well...it takes it only 2x the time to stop when travelling twice as fast, but since its average velocity is greater, it has to have travelled more than 2x the distance. 2x the velocity and 2x the time = 4x the distance.

[CraigD]

…This is what makes me wonder; Can a moving object be said to gain energy as it speeds up? What if it is in a part of the universe where no object will ever cross its path. Or a hypethetical place where its the only object, how can you determine how much work it can do when there is nothing to compare it with or measure from.

I think there’s a neat answer to the semi-philosophical conundrum of the speed of an object that is the only object in the universe. If there’s only one object in the universe, it can’t be accelerated, as there’s nothing with which it to have an equal-and-opposite reaction.

 

So, yes, if an object can speed up, it can be said to gain energy as it does.

 

If the universe contains 2 objects, interaction is not only possible, but required and continuous, because gravity, though very weak compared to the other 3 fundamental forces, is an attractive only force acting over unlimited distance. Even in a hypothetical universe with mass but no gravity, if at least 2 objects exist, all of classical mechanics are meaningful and defined.

(planks constant reduced) h = h (regular)/ 2*Pi

 

What is familiar here is a constant needs two things to happen. and so does energy (M*V) (M*V). There is a comman devider of 2*__ . Two things. So for planks constant there is two fundamentals of pi. and in kentic energy there is two fundementals of mass (made of constants of pi)

[math]\hbar[/math] is simply a shorthand for [math]2 \pi h[/math]. It’s purpose is to make equations like the uncertainty principle’s

[math]\Delta x \Delta p \ge \begin{matrix}\frac{1}{2}\end{matrix} \hbar[/math]

Easier to read and write than

[math]\Delta x \Delta p \ge \pi h[/math]

 

[math]h[/math] can’t be derived from [math]\pi[/math], or vice versa. [math]h[/math] is just a fundamental constant of physical reality, estimated by experimental measurement, and, according to some theories, not actually constant, but possibly changing over time or under different conditions.

 

[math]\pi[/math], on the other hand, is one of those “jump out at you from everywhere” mathematical constants that emerges from even very simple number play. For example, the infinite series

[math]\frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \frac{4}{9} - \cdots = \pi[/math]

 

The integer “2” does seem to have a sort of magical significance in Math and Philosophy. It’s the smallest integer that can be used as the base of a positional numeration system – no numeral can represented a number more fundamentally than in binary. Networks of any number of nodes can be decomposed into an ordered list of 2 nodes, but not of 1 or zero nodes. It takes 2 to tango. And so on…

 

I expect that, the more you contemplate mathematical physics, the more you’ll be struck by the ubiquity of the number “2”.

[CraigD]

I don't see why the rocket wouldn't take 4x the distance to stop. The same eqautions apply whether the force comes from the ground or propellant or gravity or anything. The same reasoning to why it would take 4x the distance to stop apply as well...it takes it only 2x the time to stop when travelling twice as fast, but since its average velocity is greater, it has to have travelled more than 2x the distance. 2x the velocity and 2x the time = 4x the distance.

Lsos is correct – the car and the rocket, having the same initial velocity and constant acceleration, stop in the same distance.

 

What I was pointing out in post #78 is that, while the car and rocket have identical motion, they require different amounts of energy to have it. The intuitive reason for this is that, unlike the car, the rocket is littering space with energy, in the form of the fast-moving matter of its exhaust.

:QuestionM An observant reader will wonder how, then the rocket can become more energy efficient than the car at about 1669 m/s. The answer is: it can’t. I’ve omitted a very important factor of how rockets work from my calculation, creating a “phantom source” of energy. Finding this is left as an exercise for the reader. :QuestionM

 

Lsos is also correct in his observation about velocity and time. Put algebraically, we can show this, starting the usual formulae for Distance and Velocity under uniform Acceleration:

[1] [math]D = \frac{1}{2}At^2 +V_{initial} t[/math]

[2] [math]V_{final} = At + V_{initial}[/math]

 

We can rearrange [1] into

[3] [math]D = \frac{1}{2}(V_{initial} + At + V_{initial})t[/math]

 

substitute [math]V_{final}[/math] for [math]At + V_{initial}[/math], getting

[4] [math]D = \frac{1}{2}(V_{initial} + V_{final})t[/math]

 

or, in ordinary language, “distance equals average velocity times time”.

[arkain101]

I think there’s a neat answer to the semi-philosophical conundrum of the speed of an object that is the only object in the universe. If there’s only one object in the universe, it can’t be accelerated, as there’s nothing with which it to have an equal-and-opposite reaction.

 

Right. So wouldnt this suggest each and every equation or force is a result of a minimum of two objects?

When you look at how everything operates there appears to be nothing that is of itself, it needs some kind of relationship.

 

 

So, yes, if an object can speed up, it can be said to gain energy as it does.

Okay. Speed up only means change between minimum of two things right? Thus energy can be gained. Or, according to our observation point work can be done on the target of choice.

 

If the universe contains 2 objects, interaction is not only possible, but required and continuous, because gravity, though very weak compared to the other 3 fundamental forces, is an attractive only force acting over unlimited distance. Even in a hypothetical universe with mass but no gravity, if at least 2 objects exist, all of classical mechanics are meaningful and defined.

 

I do agree. Except for one thing. I am troubled to accept gravity is a force. It acts like a force in every way but it could be a form of dimensional motion. Things like to obtain low energy. Gravity is a form of this I think.

All forces can create acceleration. When an object accelerates it feels a force act upon it. Gravity has no force. An object simply just moves in unity.

 

One way I look at it is dropping an object compared to pushing an object.

 

If we have a force behind an object that accelerates an object to 9.8m/s in one second and we upon that velocity it runs into a impact measuring device like clay, or a scale of sorts.

 

then compare that work and force with the same object dropping from a height that allows it to impact the same measureing device at the same velocity it may turn out to be less.

 

I say this because gravity acts on an object like an object free floating in space at a certain velocity. Where as, a force acting on a object contains the energy of the moving mass + the force driving it.