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Kinetic energy

  1. Apr 4, 2005 #1

    tony873004

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    A system consists of 10 0.50-kg balls, each moving radially outward from a common center in a symmetrical pattern, all at a speed of 12 m/s. What is the kinetic energy of the system?

    My instinct is to say 0, since it it symmetrical and every ball moving one direction will be cancelled by a ball of the same mass and speed moving the opposite direction.

    But energy is a scalar, right? And scalars only have magnitude, not direction, right?

    So the answer could be

    [tex]e_k=10*\frac{1}{2}mv^2[/tex]

    [tex]e_k=10*\frac{1}{2}0.50kg *(12m/s)^2[/tex]

    [tex]e_k=360 J[/tex]
     
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  3. Apr 4, 2005 #2
    My guess is 0, but I'm not sure why.
    I think if they all are radiating outward from a common center, then they must have been motionless at some point during which the potential enegry and kinetic energy is zero.
    Energy is a scalar though and they don't cancel each other out, I'd still like to know why its not 0 in a definitive fashion though.
     
  4. Apr 4, 2005 #3

    HallsofIvy

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    No, your "second thought" was correct. E= (1/2)mv2 where you can think of v2 as meaning either the dot product or the square of the magnitude of the v vector.

    In any case, kinetic energy, unlike momentum, IS a scalar, not a vector, and is always positive. The kinetic energys of the two objects do not cancel, they add.

    No, if they are radiating outward, they are moving. If they were "motionless at some point" there must have been some force acting to cause them to move. That force does work and contributes energy.
     
  5. Apr 4, 2005 #4

    tony873004

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    I'm guessing that the velocities cancel. Since they're vectors and not scalars, I can cancel them before plugging them into the K equation. Then one of my factors is 0 which makes the whole answer 0.

    I think.
     
  6. Apr 4, 2005 #5
    Well take the balls to be physical objects. If they are all moving outwards at t=t1, then isnt it feasible to say at some t<t1 they were at the same point? If so, motionless?

    Its kind of like the big bang.
     
  7. Apr 4, 2005 #6

    tony873004

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    I posted before I saw your post HallsofIvy. Can't I cancel the v's first?
     
  8. Apr 4, 2005 #7

    tony873004

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    Thinking about it, I guess it couldn't be 0. I could always spin down this system with a generator and light a light bulb for a moment.
     
  9. Apr 4, 2005 #8
    Cancelling out the velocities would impose a direction requirement on the kinetic energy, which we already know it doesn't have.
     
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