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Cheeky little velocity puzzle

  1. May 7, 2008 #1
    Oh no. My old mechanics teacher would probably disown me if he reads this. But today I was thinking about this puzzle, and I couldn't figure it.

    A space vehicle of mass 2kg accelerates from 0 to 100m/s from it's rest frame. The energy required to do so is given by E = 0.5mv^2 = 10000 J. Then the space vehicle increases it's velocity to 101m/s.
    From the initial rest frame, the total energy required to get a 2kg mass from rest to 101m/s is
    E = 0.5*mv^2 = 0.5*2*101^2 = 10201 J
    Taken with the fact that it takes 10000 J for a 2kg mass to reach 100m/s from rest, this implies it takes 201 J more for the mass to reach 101m/s than 100m/s
    E(101)-E(100) = 0.5*2*(101^2 - 100^2) = 201 J

    However, taking 100m/s as the rest frame, it only takes 1 J to acel from 100 to 101m/s.

    :confused:

    How much energy does the space vehicle actually require to acel that last 1m/s? Why is one frame correct and the other wrong? This isn't an SR question is it? What is wrong with my understanding?

    help me, please. I am feeling foolish. Right now I'm thinking it's an SR situation, but if so, it's the 1st time I've objectively realised SR effects are central to macro mechanics at v<<c.
     
    Last edited: May 7, 2008
  2. jcsd
  3. May 7, 2008 #2
    Cheeky indeed!

    I think the answer is that the energy increases with the square of the velocity.

    So to increase speed by an additional m/s at the beginning is easier than to do so later on. That's why it takes less energy to go from 0 to 1 m/s than to go from 100 to 101 m/s.

    It's kind of like trying to lose weight. The first few pounds are the easiest to drop, but the more you lose, the harder it is to lose more!
     
  4. May 7, 2008 #3

    rcgldr

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    In the case of a rocket, the total amount of work done includes both the work done on accelerating the mass of the spent fuel backwards from the rocket engine, and accelerating the mass of rocket and the remaining unspent fuel forwards.

    Since the point of application of this force is at the rocket engine, the rocket could be used as a frame of reference for calculating work done and power, as if the rocket weren't moving at all, but only accelerating spent fuel from it's engine. The power would be equal to the 1/2 times the (mass of fuel spent/unit of time) x terminal velocity^2 of the spent fuel from the rocket engine. The numbers for work done on fuel and rocket, and power from the engine should be independent of the frame of reference, so although the math would be more complicated, the results should be the same using any normal (non-accelerating) frame of reference (the rocket frame of reference is just easier to calculate).
     
    Last edited: May 7, 2008
  5. May 7, 2008 #4

    Redbelly98

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    Recall that
    Work = Force x distance = change in energy

    The distance (over which the force acts) will be different for different reference frames.

    Thus the work, and energy change, is frame dependent. In the frame going at 100 m/s, the acceleration to 101 m/s appears to act over a shorter distance, hence less kinetic energy gain.
     
    Last edited: May 7, 2008
  6. May 11, 2008 #5
    No need. Just take a little time to realise that kinetic energy is calculated by ½mv² , not by
    [tex]1/2 m(\Delta v)^2[/tex]

    Then, as v depends on your frame of reference, kinetic energy unavoidably also depends on your frame of reference. No point in changing your frame of reference halfway through an experiment. That tends to confuse things :smile:
     
  7. May 12, 2008 #6

    Andy Resnick

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    You can't change reference frames in the middle! In the new reference frame, the rocket is only going 1 m/s (from rest), not 101 m/s.

    Cheeky problem, indeed :)
     
  8. May 12, 2008 #7

    rcgldr

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    Note the power from a rocket engine is independent of the frame of reference. Assuming the power is constant, then the rate of work done on spent fuel and rocket is constant, also independent of the frame of reference. The change in energy of the spent fuel expelled by the rocket engine and the change in energy of the rocket and it's remaining unspent fuel per unit time is also independent of the frame of reference. You can't ignore the work done on the spent fuel and expect to get reasonable results; this is the issue with some of the earlier posts in this thread.

    As I mentioned earlier, you can use the accelerating rocket itself as a frame of reference, which in effect is constantly changing the frame of reference, yet the power output and work done on spent fuel and rocket (zero for the rocket if it's the frame of reference) is the same as any other frame of reference.
     
    Last edited: May 12, 2008
  9. May 12, 2008 #8
    As Redbelly98 already suggested, it's really quite logical (what isn't in physics once you see it) and has everything to do with the work done.

    Let's suppose equal accelerations, whatever the reference frame.
    [tex]a=\frac{\Delta v}{\Delta t}[/tex]
    That means that for the same change in speed we need the same change in time.
    [tex] a=\frac{F}{m}[/tex]
    So, for the same mass and the same acceleration we need the same force.

    Take whatever acceleration you like, but let's have here (for simplicity's sake) 1m/s². So the whole acceleration needs tot take one second. (F= m·a ==> F= 2 N )

    In the one reference frame we accelerate from 0 to 1 m/s , thus average speed during acceleration 0,5 m/s, during 1 s means a path of 0,5 m.

    Work is force times path, W= F·s = 2 x 0.5 = 1 J
    So additional kinetic energy will be 1 J.

    In the other reference frame we accelerate from 100 to 101 m/s , average speed during acceleration 100,5 m/s, during 1 s means a path of 100,5 m.

    Work is force times path, W= F·s = 2 x 100.5 = 201 J
    So additional kinetic energy will be 201 J.

    So in a way, it's your force that has a longer way tot travel to do it's bit....
     
    Last edited: May 12, 2008
  10. May 14, 2008 #9
    thanks for the feedback. Interesting responses. But I'm not sure I'm totally clear. Let's round up whats been said.

    1. KE \ Work done is frame dependant.
    2. Not allowed to change frame during KE analysis without tricky math
    3. Analysis can be done correctly from vehicle's accelerating frame.

    All important points, which are likely part of the answer, but I don't think they describe the answer in a complete and intuitive way.

    In trying to find the best way to understand the problem consider this.
    KE is frame dependant, but fuel energy is frame independant. (i.e. your velocity wrt to a gallon of petrol won't alter your perspective of how much chemical energy is in the petrol, it's still a gallon a petrol's worth )
    So back to the space vehicle ( assuming linear accel ), wrt to any rest frame, the vehicles KE will change with the square of time. For the fuel energy, since it is frame independant, it can only change linearly with time. i.e. for any velocity v, set v as rest frame v=0, the fuel change for any v to v+1 is same as v=0 to 1.
    How can KE be non linear, while fuel useage be linear wrt to time and velocity?

    Considering the KE of the burning fuel expulsion is important. It may be at the root of the puzzle. Can someone do the maths? It's tricky. i.e.
    v is velocity of vehicle
    it's true to say
    for v=1 to 2, KE of vehicle goes from 1 to 4J
    but it's not true to say
    by newton's 2nd law,
    for v=1 to 2, KE of fuel emission goes from 1 to 0J ( equivalent to a vehicle mass deccelerating in the opposite direction)
    It would be ok if the fuel spent was a discrete mass emitted at an instance, but in fact the fuel emission is not discrete ( at the classical macro scale ), but a continuous distribution of energies. The maths will involve the integral of the energy distribution of the fuel emission.
     
  11. May 14, 2008 #10

    rcgldr

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    1. Total work done and total change in KE is frame independent, if you include the work done on both spent fuel and rocket.

    2. KE of each component, fuel and rocket, is frame dependent, but their sum is frame independent, and reflect the work done by the rockets engine.

    3. Correct, it's easiest to calculate the power and work done by simply considering the rocket fixed, with only the spent fuel moving.

    Momemtum is preserved, so the sum of momentum of the spent fuel and the rocket (along with the rocket's unspent fuel) remains the same (zero if the initial frame of reference is the initial velocity of the rocket).

    The fuel's KE increases by much more than the rocket because it's effective terminal velocity relative to the rocket engine is very high.
     
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