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About conservation of energy

  1. Dec 14, 2008 #1
    Dear All,

    Suppose we have a movable inclined plane (initially at rest) which has height h (if we attach wheels, then it should be movable) and assume that there is no friction anywhere. If a block travels with speed V towards the plane and goes up to the distance h, I wonder ;

    Is mechanical energy of the block alone is conserved ? (not kinetic energy!)

    First of all, I would like to know whether it is a collision or not. If the block is going through the plane, are they colliding ?

    And if they are colliding, then it certainly is not an elastic collision. So total kinetic energy must not be conserved. But would equation

    mblock * g * hblock = 1/2 mblock * Vinitial^2 - 1/2 mblock * Vfinal^2

    hold ? (I thought I could use conservation of momentum to calculate Vfinal )

    PS: I didn't want to post it as homework question to the other forum because it isn't a homework, only a conceptual exercise that I have imagined but have been unable to come up with a reasoning.
  2. jcsd
  3. Dec 14, 2008 #2


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    Hi ximath! :smile:

    Yes, it is a collision, and it's "sharp" (as opposed to gradual), so energy won't be conserved.
    It's still "homework or coursework" … do use that forum in future. :smile:
  4. Dec 14, 2008 #3
    Hi again!

    What if we consider block and wedge together: then could we write U1 + K1 = U2 + K2 to relate V with h, g and masses ?

    If we can't, what would be the correct approach to relate V with those quantities ?

    I'll use that forum next time :)
  5. Dec 14, 2008 #4
  6. Dec 14, 2008 #5


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    Most collisions are not elastic collisions and so energy is not conserved.

    Good rule of thumb … if it makes a noise, then energy isn't conserved!!!! :biggrin:
    Hi ximath! :smile:

    Let's examine this systematically, since it's obviously worrying you …

    as you've worked out, you need two equations to solve a collision problem like this …

    and you're thinking one of them is always conservation of momentum, but if the other one isn't conservation of energy, where do I get my second equation from? :cry:

    The answer is that it will usually be a geometrical constraint …

    for example, if a bullet embeds itself in a block, then v1 = v2, which is the second equation you need.

    In this case, the geometrical constraint is that the block is obliged to stay on the slope … that gives you your one extra (slightly complicated! :rolleyes:) equation!

    That gives you the "initial final speed" of the block, just as it starts up the slope. Obviously, you can use conservation of energy after that. :smile:
  7. Dec 14, 2008 #6

    I am not sure I understand... Could you give me an idea of how to derive that equation in that case ? And how can I use conservation of energy after that, if energy is not conserved ?
  8. Dec 14, 2008 #7
    Well, that's like your perception, man...

    If you view it in this way, then I bet you that the total energy of the accelerated air molecules, emitted light, heated surface, and kinetic energy of molecules inside the blocks is conserved.

    Next time, instead of using "energy is not conserved", use "kinetic energy of the blocks is not conserved" :P
  9. Dec 14, 2008 #8


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    Like most people, I'll continue to say "conservation of energy" when mechanical or kinetic energy is obviously implied.
    It's like a mid-air collision … you have to split the experiment into two parts … the collision itself, in which energy is not conserved … and then the aftermath, in which energy is conserved.
    Nooo! Have a go yourself! :smile:
    Last edited: Dec 14, 2008
  10. Dec 14, 2008 #9

    I guess I solved it.

    My approach:

    Suppose X is the distance wedge travels and n is the normal force acted by wedge to the
    block, w is the weight of the block and M is the mass of the weight.


    n sin(theta) * ( X + h/tan(theta) ) = Ki - Kf

    and since w = n cos (theta) ;

    w*tan(theta)*(X + h/than(theta)) = Ki - Kf

    Vi^2 - Vf^2 = 2g(tan(theta)X + h)

    and if we substitute Vi = (m + M) * Vf / m (conservation of momentum) to find an equation.

    The other equation I have found was from the work-energy theorem for the wedge.

    nsin(theta)*X = 1/2 M Vf^2 and thus Xtan(theta) = MVf^2 / 2mg

    And solving these two equations give the result I expected.

    Thanks a lot for that!
  11. Dec 15, 2008 #10

    I’m not sure about “most people” but this is a physics forum and saying that “energy is not conserved” is at least very confusing and at most very wrong. Energy is ALWAYS conserved, and that is one of the most fundamental statements in all of physics. Maybe kinetic energy was “implied” and maybe it wasn’t, but why not take the extra precaution of specifically stating the type of energy involved to avoid any confusion? It just seems a reasonable approach.
  12. Dec 15, 2008 #11


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    so sue me …

    Sorry, but I am!

    Virtually everyone says "conservation of momentum" and "conservation of energy" (… "does or does not apply" in a particular case), without qualifying the word "energy" in any way.

    Further or alternatively, m'luds, the context is almost always "in" something (usually "in a collision"), and so the conservation or non-conservation clearly falls to be interpreted as applying in that context.
    … and not a legal one! :biggrin:
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