A problem with KE = ½mv². Help.

[Qfwfq]

Now consider the situation where you and I are each on back-to back sleds. I fire my spring gun at your sled, where it is caught by a spring-loaded catcher. At what speeds do the two sleds move apart?

If you suppose equal masses, neglecting that of the bullet, then the answer is equal and opposite velocities.

 

Popular, at each shot the spring is exerting the same force on the sled for the same time but, each time, the velocity of the sled is greater so it moves through increasing distances during these equal times. Therefore the work is greater, each next time.

[CraigD]

Now consider the situation where you and I are each on back-to back sleds. I fire my spring gun at your sled, where it is caught by a spring-loaded catcher. At what speeds do the two sleds move apart?

The answer is straight forward (though not a nice, round number)

 

We can derive:

Momentum = 0 kg*m/s = MassSled1*VelocitySled1 + MassSled2*VelocitySled2

MassSled2*VelocitySled2 = -MassSled1*VelocitySled1

VelocitySled2 = -MassSled1*VelocitySled1/MassSled2

 

We have givens of:

MassSled1 = MassSled2 = 100 kg

Momentum = 0 kg*m/s

Energy = 0 J

Energy added by releasing spring 250.5 J

 

So:

VelocitySled2 = -VelocitySled1

 

So, after the spring is released:

Energy = 250.5 = MassSled1*VelocitySled1^2/2 + MassSled2*VelocitySled2^2/2

250.5 = 100*VelocitySled1^2

VelocitySled1 = (250.5/100)^.5 = ~ 1.58 m/s

 

And:

VelocitySled1 = -VelocitySled1 = -(250.5/100)^.5 = ~ -1.58 m/s

[Farsight]

Thanks for the feedback guys, I really appreciate it.

 

Craig, special thanks. I understand what you're saying, but I wanted to lead into perfectly inelastic collisions. Can we backtrack a little and recall that firing the puck propelled sled1 to 0.1 meters per second...

 

KE = ½mv² = 0.5 x 100 x 0.1 x 0.1 = 0.5 Joules

 

whilst the puck was propelled to 50 meters per second...

 

KE = ½mv² = 0.5 x 0.2 x 50 x 50 = 250 Joules

 

The puck then impacts sled2 with a perfectly inelastic collision. We know that the momentum is conserved so using p = mv

 

Puck p = 0.2 x 50 = 10 therefore

 

Sled2 v = 10 / 100.2 = 0.9998 => 0.1 meters per second.

 

The two sleds move apart at circa 0.2 meters per second. Now, it only took 0.5 Joules to accelerate sled1 to 0.1 meters per second. So sled2 must now have energy of 0.5 Joules, and we've lost 249.5 Joules. You could say that relative to one another the two sleds enjoy 2 Joules apiece, but there's still a lot of energy that has mysteriously disappeared into that catcher regardless of its construction.

 

Please check my arithmetic, apologies in advance if it's wrong, but where did the energy go?

 

If you're tempted to say "heat and noise" where would it go if we were talking about masses pulling masses via cables, or rockets burning constant fuel?

[Farsight]

By rockets I mean we put down our spring guns and strap one little rocket on to sled1, and another one on to the puck. We tie a long strong fishing line between the puck and sled2, then we light the blue touch paper and retire.

 

Since both rockets exert the same force F=ma, and the 100kg mass of sled1 is 500 times the 0.2kg mass of the puck, we know sled1 will accelerate at 1/500th the acceleration of the puck. So when the rockets have finished burning, sled1 is moving at 0.1 m/s whilst the puck is moving at 50 m/s. That's fine. Both rockets burned the same chemical energy to produce the same momentum, but the puck gets 250 times the kinetic energy of sled1. Then the the puck reaches the end of its line, and starts pulling sled2, which reaches 0.1 m/s exhibiting the same KE as sled1.

 

The explanation is that just about all of the kinetic energy of the puck has "been converted into heat and noise" because of a "perfectly inelastic collision". This is regardless of anything we might devise to engineer a smooth catch.

[arkain101]

I have to consider your point about KE and velocity comparisons..

 

But, I was going to explain an example here but in the process realised that Ke=M*V^2 is incorrect for collision interactinos.

 

The example I call the "Perfect spring Kinetic Energy Experiment."

 

There is two objects with a mass (M) and inbetween these two objects is a spring with a mass (M) traveling back and forth between the two objects at a velocity (V).

 

The objects start off at rest with a distance between them greater than the spring, distance (d). The spring has a length (L) and a ratio (kg/mm). The spring is considered to be fired by a hypothetical force at a velocity (V). The perfect spring bounces of object 1 and delivers it energy (ke) and sends it moving at a velocity (V), then the spring bounces off the object and loses a value of momentum (p) and velocity (v).Then it hits the second object and sends it moving at a slow (v) and bounces off and again loses velocity. The spring keeps on doing this back and forth back and forth, untill eventually, the spring hits one of the objects at almost the exact same velocity the object is traveling, it then delivers one last bit of energy and finally comes to match the same velocity of the object and no longer is in motion to continue hitting the objects.

 

At this point we have object 1 at velocity (v) and object 2 at velocity (v) and then we have spring at velocity (v). Theoretically, if you were to run the numbers, the total velocity and momentum of these 3 objects at the end of the experiment should be equal to the originall energy and velocity that the spring had at the beginning.

 

I havnt worked it out. But I sure would apreciate someone writing this experiment out with real values. If the total energy left over at the end of the experiment exceeds the springs originall energy, then, do we not have a problem with the current 1/2(M*V^2) ?

 

Thanks.

[Farsight]

I'm not sure about that scenario, arkain. It sounds like a lot of work to compute. Is there a simpler way by presuming that the spring is attached to one of the masses?

 

Maybe I can demonstrate something with the sled and puck by reducing the weight of our puck from 0.2kg to 0.02kg. The same F=ma means the puck accelerates ten times as quickly as before, and will now be doing 500m/s. This then has kinetic energy of:

 

KE = ½mv² = 0.5 x 0.02 x 500 x 500 = 2500 Joules.

 

From the same spring or rocket? Help me out here somebody, please.

[CraigD]

The puck then impacts sled2 with a perfectly inelastic collision. …

 

So sled2 must now have energy of 0.5 Joules, and we've lost 249.5 Joules …

 

Please check my arithmetic, apologies in advance if it's wrong, but where did the energy go?

 

If you're tempted to say "heat and noise" where would it go if we were talking about masses pulling masses via cables, or rockets burning constant fuel?

Popular’s arithmetic looks correct to me.

 

As is often the case in inelastic (“no bounce”) collisions, the energy was indeed transformed from kinetic into mostly heat, a little bit of sound, and possibly some mechanical work, if, for instance, the puck dented the sled. Because taking all of these transformations are complicated co calculate, an experiment like this usually specifies that the puck collides with something quiet, dent-proof, and easy to measure the temperature of. A 1 liter plastic bag of water with a lab-quality electronic thermometer probe sealed inside would work well.

 

In this case, the 249.5 J

249.5 J *( ~ 1 calories/ 4.185 J) = ~59.6 cal

should increase the temperature in the bag by about

59.6 cal / 1000 cm^3 = .0596 ° C.

where would it go if we were talking about masses pulling masses via cables, or rockets burning constant fuel?

The cable example deserves description and analysis in more detail.

 

A rocket, however, is in principle not very different from pucks shot from a spring gun. If shoots many more much smaller particles (the rocket exhaust) at a much greater rate, but, like the sled and pucks system, those particles have equal and opposite momentum to the sled, but much higher kinetic energy.

[CraigD]

Maybe I can demonstrate something with the sled and puck by reducing the weight of our puck from 0.2kg to 0.02kg. The same F=ma means the puck accelerates ten times as quickly as before, and will now be doing 500m/s. This then has kinetic energy of:

 

KE = ½mv² = 0.5 x 0.02 x 500 x 500 = 2500 Joules.

 

From the same spring or rocket? Help me out here somebody, please.

The same (250.5 J) spring wouldn’t give a 0.02 kg puck 10 time the speed of a 0.2 kg puck. It would give it

0.02 kg *V^2/2 = 250.5 J

V = (2*250.5/0.02)^.5 = ~ 158.3 m/s.

 

:) What’s wrong with the reasoning that a puck that the spring accelerates a 10 times the rate of another will end up with 10 times the speed? This is another good exercise for the reader. Hint: Velocity = Acceleration*Time. How long was the 0.02 kg puck accelerated, compared to the 0.2 kg one? (the spring’s extended length is constant)

[Farsight]

OK, thanks Craig, I got it, the spring goes BOING rather than B-O-I-N-G.

 

How about the rockets?

[Farsight]

Uhhhrrrrn. OK. I got it at last.

 

If I'm a rocket it takes me an amount of work W to accelerate the puck to 1m/s.

 

If I'm moving along with the puck, it takes me the same amount of work to accelerate the puck from 1m/s to 2 m/s.

 

But I had to do work on me to get to 1m/s. Hence the integral.

 

I've now got the picture. Energy is like money. I can use it to buy a holiday, whereafter it's spent, gone, dissipated, wasted. A spent force. Like the rocket power wasted trying to move a mountain. Or I can use my money to buy land, which I can sell later, so it's kind of not spent. It's kind of saved, invested, converted from one asset to another. Like some of the rocket's chemical energy spent moving the puck has been invested in the puck. And of course sled1 weighs in somewhere between the mountain and the puck.

 

So, if Energy is like money and I can either waste it buying a holiday or I can keep hold of it by buying land.... it's like, if I buy time, it's wasted. If I buy space, it isn't. Maybe there's something profound here. But maybe not.

 

Sorry to have wasted your time.. and energy ,guys.

[arkain101]

its not so much like that.

 

You know, im sure,

 

Kinetic Energy increases 4 times each doubling of the velocity. Your always have 4 times the energy then then you do at half your current speed. There is rest, then there is the first minute amount of motion possible, at that point you have 4 times the energy you would then at half that speed etc.

 

Any object that speeds up increases in energy perpotional to the square of its velocity, which is really quited strange. Where as momentum will remain will continue to increase directly with P=M*V .

[Farsight]

So if velocity is metres per second, or distance divided by time, and acceleration is distance divided by time divided by time...

 

p=mv means momentum is mass multiplied by distance divided by time.

 

F=ma means force is mass multiplied by distance divided by time divided by time.

 

So momentum is force times time. The same animal as impulse. And when I think of my gyroscope and it's angular momentum, it feels like it's a mass. Is momentum what mass is all said and done?

 

KE=½mv² means Energy is mass multiplied by spacetime space area divided by a spacetime time area. Huh? It sounds like a tension between space and time. Jeez, don't say KE is just another wound-up spring.

 

Uhhhn. My head hurts. I'll have a look on Google. But later.

 

http://www.rwc.uc.edu/koehler/biophys/2c.html

 

 

Dynamics

 

The study of the causes of motion is called "dynamics". We will introduce five new "dynamical" variables: momentum, force, potential energy, pressure and power. Each of these quantities will directly or indirectly involve the mass of the object. In fact, mass is defined dynamically in terms of force and acceleration, as we will see shortly.

 

First, let us define the momentum P...

[arkain101]

Popular, and others, I may have solved this little boggle here.

 

First off I would like to mention that what CraigD and others have stated about Kinetic Energy being equal to 1/2(M*V^2) is of course correct.

 

I realised the problem what people see in this is not The Energy, or really the equation. Kinetic Energy is a representation of how things work, and so is not incorrect. The problem that the minds who find something wrong here is actually in how things work, and not the representation.

 

First I want to show a simple way to show 1/2(M*V^2) is right, and how it is not a determinator, but is only a reporter.

(pls check my work)

 

Example a.

 

We have a box of clay on the ground. We are going to be dropping a 1kg ball from a height of 2m for the first and initial drop and measure how far the ball sinks into the clay.

We know that the acceleration of an object in earths gravity near the surface is 9.8m/s^2. This is our acceleration.

We also know that each time you raise an object double its previous height, it doubles its velocity. (leading to x4 of the energy).

 

Our first drop is from 2m. With a 1kg ball. The ball sinks 0.1m into the clay (soft clay).

 

The energy of the first drop is equal to 1/2(M*V^2). The velocity of the ball is 19.6m/s, when dropped from 2m.

so,

1/2(1*19.6^2)= 192.08 J

 

The second drop we drop the ball from 4m (twice the height). Velocity is 39.2m/s upon impact.

The energy is, 1/2(1*39.2^2)= 768.32J

 

Lets compare drop 1 with drop 2.

 

768.32 / 192.08 = 4

 

So yes, it appears that doubling the velocity of an object increases the energy by 4 times.

 

But, Energy does not do things. The energy equation is reporting what nature is doing and what it can be capable of. The energy can do work, and work is Force X Distance.

 

The ability to apply a force over a distance, increases x4 when the velocity increases by x2.

 

Lets look at this with the work.

 

W=F*D

 

drop 1. The energy to do work, was 192.08J, over a distance of 0.1m into the clay.

w=f*d

 

192.08=F*0.1m

 

192.08/0.1=F

 

F=1920.8

 

drop 2.

 

W=F*D

 

768.32=F*0.4m ---> F=768.32/0.4m ---> F=1920.8

 

We find that even though we drop the ball from different heights, the same force is involved, only over different distances. In the first case there was 192.08J , 1920.8N of force over 0.1m . In the second case, there was 768.32J, 1920.8N of force capable of 0.4m into the clay.

 

So we arrive energy is not wrong. Energy is only displaying how work can be done from the energy produced from a moving object.

 

This is where we arrive at the part that is in question. Excluding math, nature works in a way that all measurements of motion have 4 times the energy of the measurement at half that speed forever on down slower and slower. (you will never get to a destination if you keep going half way, and you will never stop if you keep going half the velocity.)

 

Now we have to go back into time a ways where discoveries of mass, inertia, and equal and opposite reaction were discovered and further elaborate on these terms.

 

Yes these laws and terms are quite real, but these terms do not explain as much as they state an observation.

 

I am now suggesting something quite new to explain this behaviour of nature.

 

We can consider mass and inertia to be the same thing. When we look at two objects that collide, one at rest, the other in motion. One has mass, with velocity, and we call it momentum. The other has mass and inertia. When they hit they both equally and oppositely react. So vice versa, we consider the object at rest in motion, and the one in motion at rest. The momentum of the once moving object is now refered to as inertia, because it now looks at rest. The object once at rest, is now in motion, and its inertia is now momentum, from velocity and mass.

 

Now this is saying that both objects are moving, and both objects are at rest, and either object is moving and either object is at rest, all at the same time, due to relative observation.

 

Of course we run into some issues here when we think of both objects moving at the same time.

 

But here is where I suggest even stranger thoughts.

We know that this universe is under the operation of laws, and constants, and numbers and mathamatics. In the grand operation we find these things. Now, If we consider all things simply a mathamatical operation, we can get away with having all interacting objects being the one in motion or rest.

 

Mass causes a curve in space time completely measureable by mathamatics. Although, I had even made a post that said explain mass, and explain inertia, it is not exactly such an explainable thing. However, if mass is also no more than a controlled opertation of mathamatics and is not really there, like other things in physics, then it CAN be explained. Then we have light/photons, in the same way as we look at gravity, and how I suggest with mass, we exclude photons as something tangible, and call it a operation of mathamatical design.

And so, in this way, we arrive at matter, being operation in mathamatics and not really a thing. The inertia of matter (when we thing of an object at rest) would be the clue, or evidence of the consideration that, that particular operation in the space-time is argualby the object in motion.

 

An object in motion can be considered to be the one with energy, and the one with momentum.

An object at rest can be considered the one inertia.

yet we can flip all this on opposite sides of the equal sign.

The inertia of an object can be considered to be the unrealised momentum of the object in motion from another observation fram.

and an object with energy in momentum can be considered at rest in which it would act as inertia from another observation frame.

 

The statement, ALL ACTIONS HAVE AN EQUAL AND OPPOSITE REACTION, states exactly what I did above, depending on point of observation. It sounds strange to think this way, but when accepted, It says all things are not real, but are calculations like all things in space-time which only come to existence with the inside of the source of ones consciousness. Then as we look at how nature gains 4x the ability to do work (energy) if an object doubles in velocity, and explain it with, all motions with the two reference frames of an action are mathamitcal processes which both reside on either side of an equal sign.

 

Inertia=(momentum)mass X velocity

 

When we look at all the othe unexplained mathamatical constants and rules that govern the universe, I wondered, lets not stop there, go back in time and apply such mathamatical explanations to things like mass, and inertia etc.

 

I dont know what kind of responses I might get on this one. If needed, I will more clearly and consicely put it together.

[CraigD]

We also know that each time you raise an object double its previous height, it doubles its velocity. (leading to x4 of the energy).

This is incorrect. :lightbulb

 

The basic definitions for kinetic energy and objects under under uniform acceleration are:

Energy = (1/2)*Mass*Velocity^2

Velocity = Acceleration*time + Velocityinitial

Height = (1/2)*Acceleration*time^2 + Velocityinitial +Heightinitial

 

Solving for time, Velocity, and Energy when Height = 0, Velocityinitial=0, Heightinitial = 2 m, Acceleration = 9.8 m/s/s, and Mass = 1 kg,

0 m = -4.9 m/s/s * time^2 + 2 m

time = ((2 m)/(4.9 m/s/s))^.5 = ~ 0.639 s

Velocity = -9.8 m/s/s * time = ~ 6.26 m/s

Energy = .5 kg *Velocity^2 = ~ 19.6 J

 

 

Solving Solving for time, Velocity, and Energy when Height = 0, Velocityinitial=0, Heightinitial = 4 m, and Acceleration = 9.8 m/s/s, , and Mass = 1 kg,

time = ((4 m)/(4.9 m/s/s))^.5 = ~ 0.904 s

Velocity = -9.8 m/s/s * time = ~ 8.85 m/s

Energy = .5 kg *Velocity^2 = ~ 39.2 J

 

So doubling height increases velocity by 2^.5 (~ 1.414), and doubles kinetic energy.

 

From this, we can derive the equation for potential energy for an object under uniform acceleration, which is:

Energy = Acceleration*Mass*Height

[Farsight]

I like this arkain:

 

An object in motion can be considered to be the one with energy, and the one with momentum. An object at rest can be considered the one inertia,

yet we can flip all this on opposite sides of the equal sign.

 

Craig, I confess I didn't pick up on that double the distance. Thanks for pointing it out.

 

As a musing, if the whole sled/puck system was travelling at 1000 m/s and we fired the spring gun, the puck's velocity would increase by the same old 50 m/s, so it gains 10,250 Joules. The sled's velocity decreases by the same old 0.1 m/s which means it loses 10,000 Joules. The difference is the 250 Joules in the spring. I wonder if there's some way to forget about the "sled" and make use of the extra 10,250 Joules in the "puck"?

[arkain101]

Quote:

Originally Posted by arkain101

We also know that each time you raise an object double its previous height, it doubles its velocity. (leading to x4 of the energy).

This is incorrect.

 

The basic definitions for kinetic energy and objects under under uniform acceleration are:

Energy = (1/2)*Mass*Velocity^2

Velocity = Acceleration*time + Velocityinitial

Height = (1/2)*Acceleration*time^2 + Velocityinitial +Heightinitial

 

Which part is not correct?

 

1)An object dropped from hieght A will impact at certain velocity, and the same object dropped from twice the height of A will impact at twice the velocity

 

or

 

an object that doubles its velocity, gains 4x the energy than it previously had (at half the Vfinal).

[arkain101]

So I summerize, the problem that person you linked us to and the problem you and I also saw with Kinetic Energy, is not the equation, but rather the explaination as to why an object that doubles its velocity, increases its energy by x4. It comes down to saying, because what it impacts with has inertia, and has an equal and opposite reaction. But these are simply statements of the obvious, and do not explain why mass has its mass and mass has its inertia.