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Question with Momentum and KE

  1. May 5, 2003 #1

    I have a question that has been really been bothering me, even after my teacher spent awhile trying to discuss it. The question was:

    A 5.0e4 kg space probe is traveling at a speed of 11,000 m/s through space. Rockets are fired along the line of motion in order to reduce the probe's speed. The rockets generate a force of 4.0e5 N over a distance of 2500km. What is the final speed of the probe?

    My answer (incorrect according to teacher) was this:

    using W=Fdx (W=KE) KE=(4e5 N)(2500000m)=1e12 J (amount of Joules used to change rocket speed)
    KE=.5mv^2 so 1e12 J=(.5)(5e4 kg)(v^2) v=6325 m/s (change in velocity)
    SO: 11,000 m/s - 6,325 m/s = 4675 m/s Final Speed

    My teacher's explanation was this:

    1e12 J (same as above -- amount of Joules used to change rocket speed)
    For ship itself: KE=(.5)(5e4 kg)(11,000^2) 3.025e12 J
    3.025 J - 1e12 J = 2.025e12 J
    2.025e12 J = (.5)(5e4 kg)(v^2) V = 9000 m/s Final Speed

    After looking over this answer, it has made me VERY confused about why mine didn't work, and I'll support my explanation:

    When moving in space, velocity can only be measured in comparison to other objects that are considered to have no velocity. Because of this, the amount of kinetic energy put into it, and its corresponding velocity, could be the only thing used to measure a change in velocity.

    For example: Say you have 2 ships seemingly moving at eachother. In reality, you are moving towards the ship at 11,000 m/s, while the other ship is stationary. However, it appears from your point of view that you are actually stationary, instead, and that the other ship is moving towards you. You know that the ship can only withstand a collision at 6,000 m/s, so you must compensate by propelling yourself up to 5,000 m/s away from the ship, with KE=(.5)(5e4)(5,000^2) = 6.25e11 J. However, according to my teacher's theory, this would not work, because you are actually moving at 11,000 m/s, or (.5)(5e4)(11,000^2) = 3.025e12 .
    Using the same amount of KE found by a "still" ship, you would do 3.025e12 - 6.25e11 = 2.375e12 J SO 2.375e12=(.5)(5e4)(v^2) v=9,747 m/s, or a change of 1,253 m/s.

    As you saw above, the numbers come out differently once again, depending on who's actually moving. However, when your in space, it should not matter whether you are moving at 11,000 and the other ship is still, or vice versa, the velocity change should still come out the same, correct?

    Quickly, here's another example of my teacher's theory that I'm having trouble with:
    Say you are in a stationary car with a mass of 1kg (to keep it simple), and want to move 1 m/s with respect to the ground. You would need (.5)(1)(1^2) = .5 J of energy in the car to get it moving at that speed. HOWEVER, take into consideration the fact that the Earth is spinning. We will say, spinning at 500 m/s. This means you actually want to go 501m/s to move 1 m/s with respect to the ground. because of this, you would ACTUALLY need (.5)(1)(501^2)= 125500.5 m/s MINUS the kinetic energy you already have when you were traveling with the Earth: (.5)(1)(500^2) = 125000. 125500.5-125000 = 500.5 J you ACTUALLY need to move 1 m/s, not .5 m/s.

    Thanks for your help!
  2. jcsd
  3. May 5, 2003 #2

    Tom Mattson

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    Hi, and welcome to PF.

    Your teacher is correct. The work-energy theorem states (assuming no changes in PE):


    which is what your teacher has.

    What you have is:


    which is not the same.
  4. May 5, 2003 #3
    Hey again--
    I dont exactly see where your pulling my (Vf -Vi )^2
    But look at it more through the technical aspect of it rather than mathematically: Why would it take more energy to change the velocity of an object from 5 m/s to 10 m/s, rather than 0 m/s to 5 m/s? According to this theory, it does.

    And even so, pls check back to my original post, in the last paragraph where I dealth with the car and movement of the Earth.

    Thanks again..
  5. May 5, 2003 #4
    Because the energy isn't proportional to the velocity, it's proportional to the SQUARE of the velocity.
  6. May 5, 2003 #5

    Tom Mattson

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    Look at what you wrote as your solution:

    The red part is where you made the error. You set the work equal to (1/2)m(Δv)2.

    Gnome has adequately handled your other question, so I'll leave it at that.
  7. May 5, 2003 #6
    As to the other question, conservation of energy only applies within a particular inertial frame of reference. You can't expect energy changes to appear the same to two observers who are in frames that are moving relative to each other. [In fact, for the moment forget about the CHANGE in kinetic energy, think about the amount kinetic energy. Let v = velocity of car relative to earth. Let w = velocity of earth relative to sun. Let x = velocity of sun relative to the center of the galaxy.

    Is the kinetic energy of the car equal to (1/2)mv^2 = 0?
    Or (1/2)mw^2?
    Or (1/2)mx^2?

    The answer is all of those. It just depends on which frame of reference you choose to use.
  8. May 5, 2003 #7
    Alright, but theres no 0 velocity. It is only relative to other objects. Earth is whizzing through to solar system at incredible speeds, and the solar system is going through the milky way, too. If you take the velocity of US relative to the center if the milky way, we'd be going VERYYY fast, and a very small velocity change yields a LOT of energy change. Hence, it would be just about impossible to move.

    If there's a person standing on a distant, watching Earth go by, they could see us as going at, say, 100,000 m/s. At that speed, in order for us to have a 1kg car drive just 10m/s away from that distant planet, it would need 5e9J, and I doubt it takes that much energy for a 1kg car to drive 10 m/s
  9. May 5, 2003 #8

    Tom Mattson

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    That's right: and different observers in different reference frames will not agree on the change in KE of a body under the action of a force. They will only agree that the energy is conserved.
  10. May 5, 2003 #9
    Ah ok, so if the AMOUNT kinetic energy is observed as different from every point of reference, how are you supposed to measure KE lost in a 2-body collision in deep space?? From there, it may not be that big a deal. However, from an Earth observer it may seem like an incredible event of energy release....
  11. May 5, 2003 #10

    Tom Mattson

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    You would need to know the same things that you would need to know when measuring it in the lab: the masses, initial velocities, and final velocities.

    Again, "incredible event of energy release" is a relative term. No two observers in different reference frames will agree on it.
  12. May 5, 2003 #11
    Exactly. First they see it moving at 100000m/s, with 5e9j of energy, and then they would see it accelerate to 100010m/s, and they would deduce that 1e6j of energy was added to it. So?

    You're right, it doesn't take 1e6j of energy to accelerate form 0 to 1m/s, but it does take that much energy to accelerate from 100000 to 100010 m/s, and in their frame of reference, that's what they see.

    The hurdle is to free yourself from thinking of energy as if it is like a gallon of gas, some constant quantity of physical stuff that looks the same to every observer, and realize that instead, it is a quantity that is only meaningful in relation to a specific frame of reference.

    [Even more perplexing (at least to me) is that the same applies to force. I'm still working on it myself.]
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