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Thread: "Regaining momentum takes three times as much energy as sustaining momentum" - explanation please!

  1. #1 "Regaining momentum takes three times as much energy as sustaining momentum" - explanation please! 
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    Hello everyone

    Daniel Pink, who is a professional motivator/writer, likes using the statement "Regaining momentum takes three times as much energy as sustaining momentum", often to inspire (possibly apathetic) employees.

    I would like to know whether this is a general result, or only applicable to a specific scenario? And, also, how this can be proven?

    I am assuming that the regaining/sustaining of momentum must be compared over a set time/or distance. And that there must be a drag force or resistance of some sort, as well as a thrust or forward force.

    I have tried to play around with the equations for energy/momentum/force:

    E = p^2 / 2m

    E = F . d

    But, unfortunately, i seem to be going around in circles.

    Any help or thoughts would be much appreciated!

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  2. #2  
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    Quote Originally Posted by myriad View Post
    Hello everyone

    Daniel Pink, who is a professional motivator/writer, likes using the statement "Regaining momentum takes three times as much energy as sustaining momentum", often to inspire (possibly apathetic) employees.

    I would like to know whether this is a general result, or only applicable to a specific scenario? And, also, how this can be proven?

    I am assuming that the regaining/sustaining of momentum must be compared over a set time/or distance. And that there must be a drag force or resistance of some sort, as well as a thrust or forward force.

    I have tried to play around with the equations for energy/momentum/force:

    E = p^2 / 2m

    E = F . d

    But, unfortunately, i seem to be going around in circles.

    Any help or thoughts would be much appreciated!

    It is very simple, really.





    1. To get a particle "acquire" a momentum, it needs to get "up to speed", from 0, to . The energy expanded is

    2. If a particle moving at "loses" momentum, its speed goes from to 0. For the particle to "regain" the lost momentum , the speed needs to go from 0 back to .The total speed variation is in this case (from v to 0, from 0 back to v).
    Therefore, the variation of energy is , in this case
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  3. #3  
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    Would you get the best fuel consumption out a a car by letting it roll to a stop and then accelerating again? For sure not if you were applying the brakes, but otherwise I need to be convinced. A slower average velocity leads to less drag/friction after all.
    X0X I don't know how busy you are but we should explore this one with lovely equations as its fascinating. I am hopelessly busy until the weekend. Argh!
    Last edited by Jilan; 10-08-2014 at 06:18 AM.
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  4. #4  
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    Quote Originally Posted by Jilan View Post
    Would the best fuel consumption out a a car by letting it roll to a stop and then accelerating again?
    This is not a coherent sentence. Try rephrasing.


    For sure not if you were applying the brakes, but otherwise I need to be convinced. A slower average velocity leads to less drag/friction after all.
    X0X I don't know how busy you are but we should explore this one with lovely equations as its fascinating. I am hopelessly busy until the weekend. Argh!
    I have no idea what you are talking about, so I cannot put the above into equations. Collect your thoughts, ask again and I will answer.
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    Quote Originally Posted by Jilan View Post
    Would the best fuel consumption out a a car by letting it roll to a stop and then accelerating again? For sure not if you were applying the brakes, but otherwise I need to be convinced. A slower average velocity leads to less drag/friction after all.
    X0X I don't know how busy you are but we should explore this one with lovely equations as its fascinating. I am hopelessly busy until the weekend. Argh!
    Put it in neutral and let it slow but not to a complete stop if you can help it. It takes a lot of energy to get a mass moving from a complete stop. There are certain things to consider as to speed up too quickly may take a richer gas to oxygen ratio, and to go too slow may mean you have to be in a lower gear ratio especially in an auto tranny. Also of course we need to consider the air resistance as it increases with the square of the speed (I think) this may be debatable considering the shape of the vehicle. It is from an accounting viewpoint also dependent if getting to somewhere more quickly gets you more time on the clock at a higher wage than the cost of the extra fuel you may use, but should I think be tempered the more by how you like driving the most. It also helps a lot to travel at times of day when the traffic is light, so you can kind of play around as you wish to.
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  6. #6  
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    Quote Originally Posted by x0x View Post
    This is not a coherent sentence. Try rephrasing.




    I have no idea what you are talking about, so I cannot put the above into equations. Collect your thoughts, ask again and I will answer.
    I think Jilan is basically stating that it may take less overall energy to speed up and then coast than it does to continue at a sustained rate wheras the OP said something about it would take 3 times the energy to do that than to continue a sustaining rate. It may or not take 3 times the energy to regain the momentum, but there is a lot of energy saved in coasting as well. From a practical viepoint in the example of driving, if there is a red stoplight ahead, it certainly saves energy to coast to it and if you can time it well, and not have to stop, it also saves time. Most people drive stupid though.
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  7. #7  
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    Quote Originally Posted by mayflow View Post
    It may or not take 3 times the energy to regain the momentum,
    Basic physics says that it does. If you cannot follow the proof, this is your (repeated) problem.
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    I have given this some thought and the 3X stuff is nonesense (sorry X0X). The energy you have to put into a system to maintain the velocity is equal to the energy dissipated. If the energy dissipated is just a function of speed it does not matter whether this energy is put back in in short bursts or in a continuous matter. Cruising at a high velocity will consume more energy than cruising at a slow one, so if you get up to 100 mph then let it slow to 80 mph then speed back up again to 100 mph the energy would still be less than cruising at 100 mph. This is because the average speed will be less than 100mph so the energy dissipated will be lower.
    Last edited by Jilan; 10-08-2014 at 11:57 PM.
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    Basic physics says that it does
    It does not take 3 times the energy to regain vs sustain but 2. You proof account for gain and regain, and that is indeed 3.

    Quote Originally Posted by mayflow
    From a practical viewpoint in the example of driving,
    Extending this problem for car is interesting because in real life (that is: NOT drifting in space, which is a perpetual motion in its own right), anything that is sustained implies expenses of energy.
    It may even be applied to speed of work(ing class)...

    It's all about energy dissipation of course (see pikpobedy's link has usual) AND engine efficiency. Not only dissipation of drag, but all of the other heat/friction (breaking, steering, listening to radio), plus the fuel not transformed into work.

    In short, the turtle win over the hare, tell that to your boss.

    Aesop should get a Nobel prise, for that discovery
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    Quote Originally Posted by Jilan View Post
    I have given this some thought and the 3X stuff is nonesense (sorry X0X).
    You sound more and more like "mayflow".

    The energy you have to out into a system to maintain the velocity is equal to the energy dissipated.
    I'll have Ranch with this.
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    Quote Originally Posted by x0x View Post
    Basic physics says that it does. If you cannot follow the proof, this is your (repeated) problem.
    Your proof doesn't address the question asked. The question dealt with comparing the energy needed to "maintain" momentum, vs. that to regain lost momentum.

    So the first question becomes: "How much energy does it take to maintain momentum?" This is a bit of a open ended question, as it does not include under what conditions. In a frictionless, gravity free environment, it takes zero energy to maintain momentum. So if through some action momentum is lost, any amount of energy use to regain the momentum would be an indeterminate times greater than the energy needed to maintain it.

    So, to make any sense of the question, we have to assume that there is something that is acting against the motion. For example, an object rising upward against gravity. Thus some effort must be made to hold the momentum constant.

    How much energy does it take to maintain momentum under this situation?. Again, we are dealing with an open-ended question.

    We'll use the mass rising against gravity example. Basically, what is needed is to apply an upward force that exactly counters the downward pull of gravity, allowing the mass to rise at a constant velocity. Energy is force x distance, so the energy needed depends on distance the force is applied. Applying it as the object rises 1 meter takes half the energy as does applying it for 2 meters. Note that the value is independent of the actual upward velocity of the mass. It takes the exact same energy to maintain a constant upward velocity of 1 meter/sec for 1 meter as it does to maintain a constant upward velocity of 2 meter/sec for the same distance. As the distance over which we are maintaining momentum increases, so does the energy.

    In conclusion, there is no fixed amount of energy that "maintains" momentum. You have to set the height over which the momentum in maintained in order to get a set value for the energy.

    So to get an answer to this question we need the downward force of gravity and the height over which we are concerning ourselves f=ma and with gravity a=9.8m/sec/sec =g, so we end up with

    E=mgh (assuming h is small enough that g varies insignificantly over it.)


    Now let's deal with how much energy it takes to regain lost momentum. To make things comparable, we'll assume that the distance over which the momentum is regained is the the same as that over which we considered the momentum as being maintained.

    First off we know that it will be a minimum of mgh, the energy needed to lift the mass the height. In addition, you need the energy needed to regain the lost velocity. But that depends on the velocity of the mass, and we've already established that the energy needed to maintain velocity over a set distance is independent of the velocity.

    So we have one value that remains the same regardless of the velocity and the other that depends on the velocity, and we end up with the conclusion that over a fixed distance, the two values have no set relationship to each other.

    Gotta dash. If someone else wants to they can work it out for a fixed period of time. If no one has by the time I get back, I'll pick it up from here.
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    Quote Originally Posted by Janus View Post
    Your proof doesn't address the question asked. The question dealt with comparing the energy needed to "maintain" momentum, vs. that to regain lost momentum.

    So the first question becomes: "How much energy does it take to maintain momentum?" This is a bit of a open ended question, as it does not include under what conditions. In a frictionless, gravity free environment, it takes zero energy to maintain momentum. So if through some action momentum is lost, any amount of energy use to regain the momentum would be an indeterminate times greater than the energy needed to maintain it.
    Precisely. This is why I reframed it as the energy necessary to "acquire" the momentum . I think this is what the OP wanted, this is what I answered. See my post.
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    Quote Originally Posted by x0x View Post
    You sound more and more like "mayflow".



    I'll have Ranch with this.
    There, I amended my post by one letter.... Does it make sense now?
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    Quote Originally Posted by Jilan View Post
    I have given this some thought and the 3X stuff is nonesense (sorry X0X).
    It is a basic exercise in physics, I am sorry that you have so much difficulty with it.

    The energy you have to put into a system to maintain the velocity is equal to the energy dissipated.
    Adding obvious platitudes does add any content to solving the exercise.

    If the energy dissipated is just a function of speed it does not matter whether this energy is put back in in short bursts or in a continuous matter.
    What basis do you have for this claim? Put your claim in mathematical formalism and we'll see if it is true or bogus.



    Cruising at a high velocity will consume more energy than cruising at a slow one, so if you get up to 100 mph then let it slow to 80 mph then speed back up again to 100 mph the energy would still be less than cruising at 100 mph.
    This is because the average speed will be less than 100mph so the energy dissipated will be lower.
    Obviously. What is your point? Has nothing to do with the simple exercise in the OP.
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    Okay, let's pick up where we left off.

    Instead of a fixed distance, we'll assume a fixed time.

    To maintain momentum in our example for a given time, we need to work out what distance is traveled in that time. This is simply h=vt. We can then substitute vt for h in mgh and get:

    E=mgvt.

    The faster the object is moving, the greater distance it climbs in any given time, and thus the energy needed to maintain velocity increases.

    To regain lost momentum, we have to regain lost velocity so this part works out to

    ,

    Here we will assume that v =the v from above and the object has lost all momentum.

    We also have to include the energy needed to lift the mass the distance it travels as it regains the lost velocity. Since the object is accelerating from 0 to v, this works out to be



    We also know that v=at so we get



    Work this into the equation for energy needed to lift a mass a given height and we get



    Which leads to a final total energy needed of



    our original questioned dealt with the ratio between this and the energy needed to maintain momentum (mgvt), so we will divide and reduce the problem which simplifies to:



    note that unless the fraction v/gt>1, the energy needed to regain lost momentum in a given time is less than that to maintain that velocity in the same time.

    This might seem odd at first, but remember, a part of the energy required is that part needed to lift the mass the height it travels while the mass accelerates up to v. A mass accelerating from 0 to v travels a shorter distance than a mass moving at v moves in the same time. The average speed of the object is less over the time period. So then we look to the other part of the total. It's the energy needed to accelerate up to v (mv^2/2) Unless this is large enough to overcome the above deficit, the maintain energy is greater than the regain energy.

    For example to maintain the upward velocity of a 1 kg mass moving at 1 m/s for 1 sec against gravity takes 9.8 joules of energy.

    a 1kg mass accelerating from 0 to 1m/s in 1 sec accelerates at 1m/sec/sec. in one second it travels 1/2 m. to move up 1/2 meter against gravity takes 4.9 joules. To accelerate a 1 kg mass from 0 to 1 m/s takes 0.5 joules 4.9+0.5 = 5.4 joules which is less than what it took to maintain the upward velocity.

    However the same problem with an object moving at 100m/s give 980 joules for the maintaining energy,
    490 joules to lift the accelerated mass the height climbed accelerating from 0 to 100m/sec
    and 5000 joules to accelerate it from 0 to 100 m/s
    for a total of 5490 joules, which is greater than that needed to maintain the upward velocity.

    So in the end, neither using a set distance or time in which the momentum is maintained and regained yields a fixed ratio for the energies needed.
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    Quote Originally Posted by Janus View Post

    So, to make any sense of the question, we have to assume that there is something that is acting against the motion. For example, an object rising upward against gravity. Thus some effort must be made to hold the momentum constant.

    How much energy does it take to maintain momentum under this situation?. Again, we are dealing with an open-ended question. .
    Interesting approach, Janus

    Notwithstanding that the OP asked for something really simple where the answer prompts the solution, let's look at the interesting "complication" that you provided. If you like watches, like I do, you will know that "complication" means a really interesting mechanism, like a repeater. Anyway, you have chosen to "maintain" constant momentum while climbing against a gravitational field. Such a feat is much more complicated (excuse the pun) than described in your solution. What you need is to solve this ODE:



    One needs to find the expression of the unknown propulsion force that will guarantee . The only acceptable solution is for obvious reasons. This gives with the immediate solution . Now, the total energy becomes .
    The energy required to regain the momentum is even simpler, if the object is thrown with the initial speed against the gravitational field then its momentum will vary from to 0 and back to while the energy transforms from kinetic into potential and back to kinetic.
    Last edited by x0x; 10-09-2014 at 11:18 PM.
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    Quote Originally Posted by x0x View Post


    What basis do you have for this claim?
    It's just the conservation of energy. How can the method of putting in the energy affect the energy required? However the energy is added, the energy put in must equal the energy dissipated. So unless there is some reason to think that short bursts of effort somehow dissipate more energy this should be quite apparent.
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    Quote Originally Posted by myriad
    Daniel Pink, who is a professional motivator/writer, likes using the statement "Regaining momentum takes three times as much energy as sustaining momentum", often to inspire (possibly apathetic) employees.
    Welcome to the forum.

    Unless there are forces at work momentum is constant. However since he's talking about momentum being sustained it means that there are no forces at work here. Since there are no forces at work the momentum is constant and so is the energy and as such it takes no energy to sustain momentum so I can't understand what he means. Sounds like technobabble to me.
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    Jilan: I have given this some thought and the 3X stuff is nonesense (sorry X0X).

    xOx: It is a basic exercise in physics, I am sorry that you have so much difficulty with it.

    cinci: I think the problem with the calculation is v-to-0-to-v calculation having the energy m(2v)^2. In this case v is a vector quantity and cannot be added as the calculation implies. Also, the part from v-to-0 isn't something one would generally do; that is, work against oneself. Even if that was the assumption, the energy would be mv^2 to get it back where it was.

    A different solution might be found in terms of work. Assuming that the momentum you desire is for a body with a constant force opposite to the direction of motion. If you stop supplying the force, the velocity of the object will eventually become zero. Restoring the object to the desired velocity requires whatever energy was required the first time you got it up to speed.

    Having done some project work, I would say the estimate of 3 is probably low for human interactions. Once a team loses its focus it's devilishly hard to get it back.
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    Quote Originally Posted by Jilan View Post
    It's just the conservation of energy. How can the method of putting in the energy affect the energy required? However the energy is added, the energy put in must equal the energy dissipated.
    I see that you are repeating the same platitude.

    So unless there is some reason to think that short bursts of effort somehow dissipate more energy this should be quite apparent.
    That was not the question I asked you, thanks for playing.
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    Quote Originally Posted by cincirob View Post
    Jilan: I have given this some thought and the 3X stuff is nonesense (sorry X0X).

    xOx: It is a basic exercise in physics, I am sorry that you have so much difficulty with it.

    [B]cinci: I think the problem with the calculation is v-to-0-to-v calculation having the energy m(2v)^2. In this case v is a vector quantity and cannot be added as the calculation implies.
    The velocity changes from to so the total impulse change is and the total speed change is
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    Quote Originally Posted by Physicist View Post
    Welcome to the forum.

    Unless there are forces at work momentum is constant. However since he's talking about momentum being sustained it means that there are no forces at work here. Since there are no forces at work the momentum is constant and so is the energy and as such it takes no energy to sustain momentum so I can't understand what he means. Sounds like technobabble to me.
    For once you are correct. This is why I changed the problem statement from "maintain" to "acquire".
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    Quote Originally Posted by x0x View Post
    Interesting approach, Janus

    Notwithstanding that the OP asked for something really simple where the answer prompts the solution, let's look at the interesting "complication" that you provided. If you like watches, like I do, you will know that "complication" means a really interesting mechanism, like a repeater. Anyway, you have chosen to "maintain" constant momentum while climbing against a gravitational field. Such a feat is much more complicated (excuse the pun) than described in your solution. What you need is to solve this ODE:



    One needs to find the expression of the unknown propulsion force that will guarantee . The only acceptable solution is for obvious reasons. This gives with the immediate solution . Now, the total energy becomes .
    This is the expression for the total energy for the mass with respect to the Earth at a given r and v. To work out the energy needed to maintain v(and momentum) as the mass climbs from r to r+h we need the difference in this energy for the two heights. This just ends up being the difference between gravitational potential between r and r+h:


    For a given value of h

    If maintaining momentum fora set time, we can again replace h with vt.
    [/quote]


    The energy required to regain the momentum is even simpler, if the object is thrown with the initial speed against the gravitational field then its momentum will vary from to 0 and back to while the energy transforms from kinetic into potential and back to kinetic.[/QUOTE]

    Again, in my example, we are talking about regaining momentum that is lost while at the same time climbing against the same gravity field. In this case I assumed that the mass lost all its upward momentum. So the energy needed to regain the lost velocity is the same, but we need to change how we figure the energy used while climbing up against gravity while regaining that speed. As above, when using the same distance traveled for both maintaining the momentum and regaining it, one answer depends on v and the other doesn't so there is no real way to compare the two.

    That again leaves us with only regarding time as the shared factor.

    But now we are left with a choice. Before, when g was considered a constant, the force needed to accelerate the mass upwards for a given upwards acceleration remained constant. Now, with g decreasing, a constant upwards force generates a increasing acceleration. Or conversely, a constant upwards acceleration can be maintained by a decreasing force.

    Of the two, keeping the acceleration constant is the easiest to work with, because d, the distance upwards the mass travels in time t is easy to work out and 'a' can still equal v/t.
    you'll end up with a messier ratio, but a solvable one for v and t.

    Working it out for a constant upward force is a quite a bit hairier, And I think I'll skip it. I'm not even sure if there is a direct equation solution for finding the distance traveled in a given time when acceleration changes over distance. It is basically like trying to solve for the distance traveled by an object in an elliptical orbit over a given time. The only direct solutions are for the period of the whole orbit or the half-orbits between periapis and apoapis. Any other solution has to be found by iteration.

    This is one reason I added the caveat of using a value for h that allows us to ignore the change in g over h. It gives a "cleaner solution" which makes it easier to see if any direct ratio falls out. The other reason is that I wanted a solution that was not just limited to the example at hand, but could be used in any general case where the force tending to rob momentum from the mass was a constant or near constant. I know this isn't strictly true in a real world case, but "spherical cows" and all that.
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    Quote Originally Posted by Janus View Post
    This is the expression for the total energy for the mass with respect to the Earth at a given r and v.
    Yes, this is exactlty what I said: "E is the total energy". The term is what imparted the particle the initial speed and the term is simply the mechanical work exercised by the force cancelling the gravitational attraction: (sorry, I inadvertently dropped the term ) such that the particle maintains its speed. If you want, you can consider the mechanical work as the "contribution" that "maintains" the speed.
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    Quote Originally Posted by x0x View Post
    I see that you are repeating the same platitude.



    That was not the question I asked you, thanks for playing.
    I cannot follow your comments. I would not regard the principle of the conservation of energy as a platitude.
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    Quote Originally Posted by Jilan View Post
    I cannot follow your comments. I would not regard the principle of the conservation of energy as a platitude.
    He can probably give you a series of irrelevant equations, but it mystified me too.
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    Quote Originally Posted by mayflow View Post
    He can probably give you a series of irrelevant equations, but it mystified me too.
    You and Jilan are interchangeable. Have fun basking in your respective ignorance(s).
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    Quote Originally Posted by x0x View Post
    You and Jilan are interchangeable. Have fun basking in your respective ignorance(s).
    So are you are really believing that is takes 3 times the energy to recover momentum than to sustain it?
    Or were you just playing?
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    It's a figure of speech. A metaphor for motivation. A bunch of buzzwords. Also known as BS.
    Maintaining momentum can be anywhere from nothing to almost everything.
    Freefall in a vacuum, walking through air, swimming in water, crawling through quicksand. Vastly different drag and fluid flow characteristics.
    So everything depends on the assumptions made for the comparison.

    And is maintaining momentum in a business always good? Who needs inertia in the wrong direction?
    What about nimbleness, agility and the ability to change direction and adapt quickly?
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    xOx: The velocity changes from -v to +v so the total impulse change is 2mv and the total speed change is 2v.

    cinci: Impulse and energy are two different animals. Impulse is linear with speed, energy is proportional to the square of velocity.

    For instance, using the classical equation for energy, if you have two particles of equal mass and speed on a collision course the energy of each is

    E = .5mv^2

    and the energy release by their collision is

    E = 2 X (.5mv^2) = mv^2.

    It is not is not

    E = .5m(2v)^2 = 2mv^2

    as your analysis suggests.
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  31. #31  
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    Quote Originally Posted by Jilan View Post
    So are you are really believing that is takes 3 times the energy to recover momentum than to sustain it?
    Or were you just playing?
    It is an exercise often found in high school textbooks. Did you have physics in high school?
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  32. #32  
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    Quote Originally Posted by cincirob View Post
    xOx: The velocity changes from -v to +v so the total impulse change is 2mv and the total speed change is 2v.

    cinci: Impulse and energy are two different animals. Impulse is linear with speed, energy is proportional to the square of velocity.

    For instance, using the classical equation for energy, if you have two particles of equal mass and speed on a collision course the energy of each is

    E = .5mv^2

    and the energy release by their collision is

    E = 2 X (.5mv^2) = mv^2.


    Their total impulse before AND after collision in the above exercise is 0.

    It is not is not

    E = .5m(2v)^2 = 2mv^2

    as your analysis suggests.
    The total impulse is in THIS exercise. You are comparing two very different experiments, other wise known as comparing apples and oranges. Carry on.
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    Quote Originally Posted by pikpobedy View Post
    It's a figure of speech. A metaphor for motivation.
    It is taken from a 9-th or 10-th grade high school physics textbook. The OP (or the motivational speaker) misquoted the exercise a little.
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    Quote Originally Posted by x0x View Post
    It is an exercise often found in high school textbooks. Did you have physics in high school?
    Yes, but I never came across anything in those text books that failed to conserve energy.
    For your equations to have validity the person being motivated would need to be responsible for getting it up to speed in the first place, slowing it down and then getting it back up to speed again. Do you think that is what the motivational speaker had in mind?
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    Quote Originally Posted by Jilan View Post
    Yes, but I never came across anything in those text books that failed to conserve energy.
    So, you took physics in high school but failed it. Figures.

    For your equations to have validity the person being motivated would need to be responsible for getting it up to speed in the first place, slowing it down and then getting it back up to speed again. Do you think that is what the motivational speaker had in mind?
    Have no idea what he had in mind but I know where he borrowed it from (modulo his misquoting the ACTUAL exercise). From a high school physics exercise.
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    xOx: Their total impulse before AND after collision in the above exercise is 0.

    cinci: I agree. But hat you're calling impulse is momentum, not energy. So it doesn't apply to your 3X issue at all. Energy is the issue.

    xOx: It is not is not

    E = .5m(2v)^2 = 2mv^2

    cinci: I'm glad you agree. Using 2v in either this equation or m(2v)^2 as you di is wrong.

    xOx: The total impulse is in THIS exercise. You are comparing two very different experiments, other wise known as comparing apples and oranges. Carry on.

    cinci: The issue as stated is this: "Regaining momentum takes three times as much energy as sustaining momentum".

    Momentum and energy are different but related, like apples and oranges. Regaining momentum, which is a function of v, requires energy which is a function of v^2. The classical form of the energy equation is adequate to show that the 3X value of energy to recover lost momentum is wrong.

    Conservation of energy requires that it takes the same amount of energy to slow an object down from v as it does to speed it up to v. According to you, I could shoot a jet of water into the air and when it came back down to me it would have 3 times the energy that it took to get it up. That's even better than perpetual motion.

    Carry on.
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    Quote Originally Posted by cincirob View Post
    xOx: Their total impulse before AND after collision in the above exercise is 0.

    [B]cinci: I agree. But hat you're calling impulse is momentum, not energy.
    Momentum and impulse are one and the same thing, and yes, impulse is different from energy, all these are correct and also irrelevant to the discussion.
    I was simply pointing out that your example has no relevance to the original exercise.
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    xOx: Momentum and impulse are one and the same thing, and yes, impulse is different from energy, all these are correct and also irrelevant to the discussion.

    I was simply pointing out that your example has no relevance to the original exercise.

    cinci: The original "exercise" is "Regaining momentum takes three times as much energy as sustaining momentum". So it refers to both momentum and energy but the assertion that it takes 3 times the energy to restore momentum is unequivocally wrong.
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    My concern is for the young minds on the forum who might actually be swallowing the **** that X0X is coming up with.
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    Jilan: My concern is for the young minds on the forum who might actually be swallowing the **** that X0X is coming up with.

    cinci: Just keep telling him the truth. maybe it will set him free. :-)
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    Impulse and momentum have the same base units. Are they the same thing? hmmm
    Torque and energy have the same base units. Are they the same thing? hmmm
    Interesting when using dimensional analysis to try to develop an empirical relationship in an unfamiiar process (speaking in general in the world at large)
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    The impulse is generally the change in momentum.
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    Yes impulse means change in momentum - Jilan ain't as dumb as he or she looks! Just kidding Jilan, you do make sense you me the few times I think I know what the heck you are talking about.

    Momentum and Impulse Connection
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    Quote Originally Posted by cincirob View Post
    xOx: Momentum and impulse are one and the same thing, and yes, impulse is different from energy, all these are correct and also irrelevant to the discussion.

    I was simply pointing out that your example has no relevance to the original exercise.

    cinci: The original "exercise" is "Regaining momentum takes three times as much energy as sustaining momentum". So it refers to both momentum and energy but the assertion that it takes 3 times the energy to restore momentum is unequivocally wrong.
    The simple high school physics says that regaining the momentum takes 3x the energy for acquiring said momentum. I corrected the problem statement in my very first post. The fact that you joined the group of mayflow and Jilan is your problem.
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    Quote Originally Posted by Jilan View Post
    My concern is for the young minds on the forum who might actually be swallowing the **** that X0X is coming up with.
    What you swallow is not coming from me.
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    Quote Originally Posted by cincirob View Post
    xOx: The velocity changes from -v to +v so the total impulse change is 2mv and the total speed change is 2v.

    cinci: Impulse and energy are two different animals. Impulse is linear with speed, energy is proportional to the square of velocity.

    For instance, using the classical equation for energy, if you have two particles of equal mass and speed on a collision course the energy of each is

    E = .5mv^2

    and the energy release by their collision is

    E = 2 X (.5mv^2) = mv^2.

    It is not is not

    E = .5m(2v)^2 = 2mv^2

    as your analysis suggests.
    In the OP, the momentum change is , in your example the momentum change is ZERO. This makes your example IRRELEVANT. Get this? I am quite sure that you will make this thread as long as the thread of misconceptions that you posted about the "relativistic wheel". The trouble with your posts is not (only) that they are patently wrong but that you continue after being shown wrong by multiple posters.
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    X0X, however fast you back peddle this is not going to turn out well. The OP had nothing to do with the maths you offered, under any circumstances.

    (As a side issue I did get interested though in how to solve the DE where the drag is proportional to the square of the velocity. It involves logarithms.)
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    Synopsis: Well, the OP was really about some quote from someone wanting to make money by motivating people by feeding them BS. I give xOX credit in the discussion for not asking for money.
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  49. #49  
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    Quote Originally Posted by Jilan View Post
    X0X, however fast you back peddle this is not going to turn out well. The OP had nothing to do with the maths you offered, under any circumstances.
    You need to learn English, you keep repeating the same mis-spelling. And, you are wrong, this is indeed a problem taken out from the high school books, it is becoming clear that you flunked physics.

    (As a side issue I did get interested though in how to solve the DE where the drag is proportional to the square of the velocity. It involves logarithms.)
    Why would you get into advanced subjects when you can't grok elementary mechanics? What does the ODE have to do with your inability to understand a much simpler exercise? You need to solve :



    Based on your posts on this forum, you do not even know where to start.
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    What mis-spelling are you referring to? I will correct it.
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    Quote Originally Posted by Jilan View Post
    What mis-spelling are you referring to? I will correct it.
    Is English your second language?
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    Quote Originally Posted by x0x View Post
    Is English your second language?
    Sorry to remove one of your poison arrows for the wounding of the vulnerable. I live in England.
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    Quote Originally Posted by Jilan View Post
    Sorry to remove one of your poison arrows for the wounding of the vulnerable. I live in England.
    I know you live in England but you aren't English. And your spelling is atrocious. Were you born in Iran?
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    No, i was born in England of English parents, grandparents, great grandparents etc. What spelling mistake do you have an issue with!? i can correct it.
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    Quote Originally Posted by x0x View Post
    I know you live in England but you aren't English. And your spelling is atrocious. Were you born in Iran?
    Ds tis men dat yo is to std to undstn I am saing this were a rdiclus pst t mk on a phys ferm?

    Incidentally, I didn't really notice anything wrong with Jilan's English, but I suppose it could be because I am American and not English, and do not (thankfully) speak xOxian.

    ps: It is not cool to say were you born in Iran as an insult. Lrn sm mnnrs and repct.
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    Quote Originally Posted by Jilan View Post
    No, i was born in England of English parents, grandparents, great grandparents etc. What spelling mistake do you have an issue with!? i can correct it.
    "back peddle".

    Correct your spelling and give your trolling a rest. Your spelling is as bad as your physics, if you trolled less, you could be learning more.
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    Peddle is usually used to sell something. Pedal is usually referring to riding a bicycle. In Jilan's post about back stepping either one seems correct to me.
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    Quote Originally Posted by mayflow View Post
    Peddle is usually used to sell something. Pedal is usually referring to riding a bicycle.
    yep


    In Jilan's post about back stepping either one seems correct to me.
    nope.
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    You pedal a bike but you peddle a myth. I got top grades in English.
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  60. #60  
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    Quote Originally Posted by Jilan View Post
    You pedal a bike but you peddle a myth. I got top grades in English.
    In this case, your mis-spelling applies to you perfectly
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    Quote Originally Posted by Jilan View Post
    You pedal a bike but you peddle a myth. I got top grades in English.
    Looking at it this way, pedal seems more about physical activities and peddle more about mental ones. Still, it seems that just the word or phrase is less important than the context it is used in. For instance "you are a pig" could mean all kinds of different things in different contexts and spoken to different beings in different situations and surroundings.
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    Funny as it is this is a physics forum. Does anyone still think there is merit in this 3:1 rule? I am worried that some people may still take it seriously!
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    Let's get back to the physics and stop all the petty bickering please, folks.
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