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KE puzzle

  1. Oct 28, 2004 #1
    A man in space fires a CO2 pellet rifle, the pellet accelerates to 500 fps. The KE is calculated to be 4.44 ft-lbs. His buddy happens to fly by his position at 500 fps with perfect timing and trajectory so that he is flying parallel to the pellet. He reaches out, grabs the pellet, places it in his CO2 pellet rifle and fires the pellet at 500 fps along exactly the same path. From his perspective he has also added 4.44 ft-lbs of energy to the pellet. From his buddy's perspective however the pellet is now traveling 1000 fps with an energy of 17.76 ft-lbs. From where did this mysterious 8.88 ft-lbs of energy appear and in what form to we find it? :grumpy:
  2. jcsd
  3. Oct 28, 2004 #2
    I'm not sure but it sounds like the "mysterious 8.88 ft-Ibs" comes from the frame transformation...(2^2 * 4.44 ft-Ibs = 17.76) idk think about that..
  4. Oct 28, 2004 #3


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    There's nothing mysterious about it. If you double the speed you quadruple the kinetic energy. The pellet ends up with eactly the same amount of kinetic energy as it would had the first man fired it at 1,000 fps.
  5. Oct 28, 2004 #4
    I'm aware of the KE formula,

    that's how I got the numbers. What's not clear is why the pellet has 4 times the energy content from a different point of reference. Intuitively since each pellet rifle added 4.44 ft-lbs of energy, you would think you could sum up the amounts. The KE formula yields a different result because it squares the velocity. So did the second pellet gun add an additional 8.88 ft-lbs of energy over and above the perception of the second individual?

    I guess what I'm pointing out is that squaring the velocity doesn't make much sense. I do understand how the formulas are derived, just not comfortable with the results.
  6. Oct 28, 2004 #5
    Energy = force x distance. From the buddy's perspective, the second force was applied over a larger distance, since the second guy was moving relative to the first.

    If you do the math, you'll find that the first observer thinks the bullet goes 3 times farther than the guy on the ship does. So the bullet gets 3 times the energy. You can see this simply because the average velocity of the bullet the first time, assuming uniform acceleration, is 250 fps, whereas the second time its average velocity is 750 fps (as viewed by the first guy), so it goes 3 times as far in the same time period.

  7. Oct 28, 2004 #6
    Slinkie is asking if a horse pulls a carriage at 10 miles per hour, why should it take 4 horses to pull the carriage to 20 miles per hour and not just two so it kind of misses the point to answer : because you can attach more horses to the carriage given a longer time. Why isn't the carriage moving at 40 miles per hour after you've added 4 times as many horses?

    One possibility would be to say that the energy added to move a mass has to also move the energy already attached to the mass. It's as if the horse pulling the original carriage becomes a carriage also so the addition of horses end up pulling more carriages (mv^2 # of carriages) than the previous.
    Last edited: Oct 28, 2004
  8. Oct 28, 2004 #7


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    Think of it this way. Suppose those horses apply a constant force to change the speed of an object. They would have to apply that force over 4 times the distance and do 4 times as much work in order to double the speed of the 'payload.'
  9. Oct 29, 2004 #8

    Again, this analysis provides little insight for me as to why the velocity of the "payload" only doubles. Assuming 1 unit of energy is required to move the mass from "rest" to 2 meters per second, why should it take 3 more units of energy to increase the velocity of the mass only 2 more meters per second? And why isn't the velocity increased to 8 meters per second since 2 X 4 = 8?
    Why isn't kinetic energy just mv?

    One explanation would be that energy is massive so that to increase the velocity of the mass-energy hybrid would require an exponential addition (mv^2) of energy instead of just v if energy had no mass. A simpler explanation would be that the expression is due to experimental error. To measure the kinetic energy of an mass, you can either add energy to it and relate it with the final velocity or collide it to something else. In adding energy to it, it may not be efficient enough to get to the hypothesized true value (like Carnot's engine), and in colliding it, it's hard to attribute which energies were from which mass, leading to miscalculation.
    Last edited: Oct 29, 2004
  10. Oct 29, 2004 #9
    There may be a simpler explanation; perhaps we have mis-defined Energy. As long as KE = Force x Distance we do indeed have to square Velocity. If instead we use KE = Force x Time we would arrive at more intuitive results for adding Energy to an object in motion. We would then have Energy = Mass x Velocity (and not Velocity^2).

    Another non-intuitive example of KE = MV^2 is a rocket pushing an object in space consumes 1 gallon of fuel per minute to exert a force F on an object. Let's ignore the mass of the rocket (assume whatever fuel needs to be consumed to accelerate the rockets mass is consumed) and assume that all of the energy of the 1 gallon is used to generate the force F for one minute. After 1 minute a certain amount of energy has be transferred to the object. After 2 minutes, 4 times the amount of energy has been transferred to the object!!! Wow, something for nothing in that second gallon of fuel. This is what suggests we might have a problem in our definition of KE.

    Just shrugging our shoulders and saying it's that way because the formula says so doesn't cut it for me. This isn't religion and blind faith should have no place here. Mathematical abstractions are not reality and are only useful to the degree that they model and/or predict reality.

    So I put it to you all: Why is energy defined as Force x Distance?
  11. Oct 29, 2004 #10


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    Read up on impulse, Slinkie.
    In addition, learn the difference between vectors and scalars.
    Energy is not misdefined.
  12. Oct 29, 2004 #11
    Thank you for that condescending remark. That certainly clears things up a bit. Anyone else care to say "Nuh Uh".
  13. Oct 29, 2004 #12
    Well, gravity is a force. Let's say a 40 lb rock falls for a second. Then its velocity rises to 32 ft/sec, and it falls 16 feet.

    If you stop it after one second, and you try and put the rock back up where it started, you will need to lift it 16 feet.

    But, suppose you let it fall for one additional second. Its velocity rises to 64 ft/sec, and it travels 64 feet. So if you want to put it back, you will need to lift it 64 feet.

    So, if something falls for 2 seconds instead of 1 second, you will need to do 4 times the work to haul it back up to its initial position, and trust me, you will expend 4 times the energy and be 4 times as tired when you are done!

  14. Oct 30, 2004 #13
    here's the real reason

    For every action, there is an opposite but equal reaction (Newton's Third Law).

    So the energy that propels the bullet also propels something in the opposite direction.

    To simplify your example, assume there are two bricks with an explosive charge between them. You ignite the charge, and one brick goes to the left with velocity v, and the other goes to the right with velocity v. Each has energy 1/2 mv**2, so the total energy from the charge is mv**2.

    Now, assume there's a train going to the left with velocity v, and there's a guy on the train with another brick. So, he grabs the left brick from the first explosion, slaps his brick next to it, puts a charge in between, and ignites it.

    Now, that left brick has a velocity of 2v as observed from the ground, so its energy is 2mv**2. However, the right hand brick from the train now has velocity 0, so its energy is 0. And the charge had energy mv**2.

    So, that left brick had energy of 1/2mv**2, the explosion released energy of mv**2, and the right brick gave up energy of 1/2mv**2, and so the left brick's energy is 1/2mv**2 + mv**2 + 1/2mv**2, or 2mv**2.

    So the answer to your riddle, which was quite interesting, is that the charge has twice the energy that you thought, and that some object on the train must slow down and also give up energy, and that's why two charges appear to give you four times the energy.

    Another way to say it is that when you fire a gun on the ground, the recoil uses up kinetic energy, but if you do it on a moving train, the recoil actually slows the gun (and train) down and releases kinetic energy, and that's why it looks like a 2 for 1.

    Last edited: Oct 30, 2004
  15. Oct 31, 2004 #14
    After the second explosion, the left brick has velocity 2v as viewed from the ground. So, its energy as viewed from the ground is:
    1/2m * ((2v)^2 )= 1/2m * 4(v^2 )or 2mv^2, . The symbol 'v' here is the velocity of the left brick after the first explosion as viewed from the ground, and 2v is its velocity after the second explosion, as viewed from the ground.

    Everything I said is from the viewpoint of an observer on the ground. Before the second explosion, the observer sees two bricks on the train, both moving to the left with velocity v. After the second explosion, the observer sees one brick going to the left with velocity 2v, and and the other brick is now sitting on the ground with velocity 0. I'm assuming that the bricks are free to move without friction to keep it simple.

    So the observer on the ground sees a brick lying there with velocity 0 and energy 0. Before the second explosion, it had velocity v and energy 1/2mv^2. So, the question is, where did its energy go?

    The answer is, it went into the other brick! Now, you're probably asking "how did it get transferred to the other brick?" That's a good question, and you need to model the explosion in order to get the answer. If you model the explosion as a coiled spring between the two bricks which is suddenly released, then it should be easy to see. Basically, as the right brick slows down, an observer on the ground sees it as giving its energy to the spring, which then gives its energy to the left brick.

    You might object to the spring analogy, but the fact is that in order for the explosion to increase the left brick's velocity to 2v (as viewed from the ground), the explosion must also push against something on the right side. An observer on the train thinks that the explosion gives energy to the object on the right; an observer on the ground thinks that the object on the right adds its energy to the explosion's as it slows down.

    To put it another way, an observer on the train thinks that the spring pushes against both bricks and accelerates both of them simultaneously. However, an observer on the ground thinks that the right brick pushes the expanding spring against the left brick, and thereby transfers its energy to the left brick.

  16. Oct 31, 2004 #15
    The charge doesn't move inside the ship, the bricks do so it's the charge that imparted kinetic energy to the bricks, either from the ground point of view or from the ship's point of view. I think your example shows more the confusion surrounding Newton's third law than it goes to explain why kinetic energy is not one to one with velocity increase.
    Last edited: Oct 31, 2004
  17. Oct 31, 2004 #16
    No, from the ground's view, the charge makes the right hand brick slow down, so the energy of the right hand brick decreases.

    Basically, I'm trying to illustrate that energy is conserved in the ground reference frame. Before the second explosion, from the ground frame's view, each brick has an energy of 1/2mv^2, for a total energy of mv^2. The charge generates energy of mv^2, and after the charge the left brick has energy 2mv^2 and the right brick has energy 0.

    Lastly, I'd like to point out again that if a brick falls for a second, it has velocity 32 ft/sec and it falls 16 ft. If it falls for two seconds, then it has velocity 64 fps and it falls 64 ft. So if you want your brick to have a velocity of 32 ft/sec, you have to carry it up 16 ft. And, if you want it to hit the ground with a velocity of 64 ft/sec (double the velocity), you have to carry it up 64 ft (4 x the height, and 4x the work). These are physical facts.

    I think that historically the definition of energy developed from observations like these. Basically, the height that you need to lift something up to in order to attain a given velocity goes up with the square of the velocity (the exact formula in feet and seconds is: height = ((desired velocity)^2)/(2g)): if you can give a better definition for energy I'd sure like to hear it.

  18. Nov 1, 2004 #17


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    I'm going back to the original question because no one has given a satisfactory answer yet. And Slinkie has a really good question here. But it has come up many times before. The thing that is always forgotten is that the platform from which the pellet is fired the second time suffers a change in momentum that exactly accounts for the energy change.

    The real question is that even though each CO2 firing expends the same energy, the total pellet energy after 2 firings is not 2 times this energy, but 4 times.

    Let m be the mass of the pellet, and v be its speed after being fired from a gun at rest. Each firing releases an amount of energy equal to [itex]mv^2/2[/itex]. Let M be the mass of the buddy plus his pellet gun plus whatever else he is riding in. By conservation of momentum, this assembly changes velocity by [itex]\Delta v[/itex], where [itex]M\Delta v=-mv[/itex]. So the second firing changes the pellet energy from [itex]mv^2/2[/itex] to [itex]m(2v)^2/2=2mv^2[/itex] for a change of [itex](3/2)mv^2[/itex]. But we know the energy release on firing is only [itex](1/2)mv^2[/itex], so there is an extra energy of [itex]mv^2[/itex]. Now calculate the energy change of the "buddy platform". It is [itex](1/2)M\Delta(v^2)[/itex]. Let [itex]\Delta v=v_2-v[/itex], then [itex](1/2)M\Delta(v^2)=(1/2)M(v_2-v)(v_2+v)=(1/2)M\Delta v(2v)=-mv^2[/itex]. There it is; your missing energy.
  19. Nov 1, 2004 #18

    Actually, if slinkie was interested in the conservation of momentum, a collison would've served as a better example. I think what you have said above is that if K.E. is defined to be 1/2mv^2 then the pellet will have 2mv^2 k.e. after the second firing, that it got this 1.5mv^2 more energy from the second firing, and that because of Newton's third law, momentum gained by the pellet is equal to momentum lost by the platform and if pure conversion to and transfer of kinetic energies are assumed, the 1.5mv^2 kinetic energy gained by the pellet is accounted for by the lost of 1.5mv^2 of kinetic energy by the platform. That's a lot of assumptions to make for one. Another is that even if they are all correct, all you and bruceg2 (whose calculations
    were wrong even if his solution was similar to yours) appear to be saying is that if the platform loses 1.5mv^2 of kinetic energy to the pellet, the pellet gains 1.5mv^2 of kinetic energy. I don't consider that a great revelation. [all considerations with respect to the "stationary" observer]

    What slinkie wanted to know was why, without defining it first with the equation 1/2mv^2, you need 4 times as much horsepower to move a vehicle
    only twice as much distance in the same amount of time? An obvious answer is that an 8 cylinder engine would be more massive than a 2 cylinder so that some of that horsepower is used to move the extra mass. Or more relevant to the case, energy probably has mass.
    Last edited: Nov 1, 2004
  20. Nov 1, 2004 #19


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    Uh Eyesaw, I presented no assumptions. What I gave was a proof based on the conservation of momentum plus Slinkie's original gedanken. You may need to read it again. Or maybe you do not realize that conservation of momentum has been considered a principle rather than an assumption for over 300 years now.
  21. Nov 1, 2004 #20
    I probably read more into Slinkie's gedunken than
    you guys did because if slinkie was just asking where the "extra" energy
    came from, the answer would have been trivial- the switching of inertial frames to calculate the kinetic energy of the pellet. Or more precisely, the
    "extra" energy is a consequence of K.E. = 1/2mv^2. Or more facetiously, it came from the firing of the gun.

    No, I read into the gedanken as questioning why the expression for kinetic energy is such and not something like K.E. = mv. My guess was confirmed by a later post by slinkie that asked exactly that.
    Last edited: Nov 1, 2004
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