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    (Original post by Terminatoring)
    Why does the graph of potential energy in 2014 show a vertically oscillating Spring mass system as having 0 potential energy at some points? Surely at the equilibrium it has gravitational potential energy and at the bottom elastic?
    Depends where you pick your zero level for GPE - as long as you're consistent you're fine.
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    (Original post by kingaaran)
    Depends where you pick your zero level for GPE - as long as you're consistent you're fine.
    Yeah but how would it ever have 0 potential energy? There's always either grav or elastic and the mark scheme is strict
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    Guys can I just check this with you (can't find my notes on this topic)
    Elastic collisions- momentum conserved, kinetic energy conserved
    Inelastic collisions- momentum conserved, kinetic energy not conserved
    Explosions- momentum conserved, kinetic energy not conserved
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    (Original post by kingaaran)
    Depends where you pick your zero level for GPE - as long as you're consistent you're fine.
    AND if you take the equilibrium as 0 pe than that introduces NEGATIVE GRAV POTNETIAL ENRGY,!
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    (Original post by thatcooldude2.0)
    Guys can I just check this with you (can't find my notes on this topic)
    Elastic collisions- momentum conserved, kinetic energy conserved
    Inelastic collisions- momentum conserved, kinetic energy not conserved
    Explosions- momentum conserved, kinetic energy not conserved
    Sounds good to me
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    (Original post by thatcooldude2.0)
    Guys can I just check this with you (can't find my notes on this topic)
    Elastic collisions- momentum conserved, kinetic energy conserved
    Inelastic collisions- momentum conserved, kinetic energy not conserved
    Explosions- momentum conserved, kinetic energy not conserved
    All right. Momentum is conserved unless an outside force is acting.
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    (Original post by Terminatoring)
    Yeah but how would it ever have 0 potential energy? There's always either grav or elastic and the mark scheme is strict
    I think the 0 level is just relative, eg if you pick initial position as the 0 level, the graph will then show whether it has gained or lost GPE relative to that position, as well as the magnitude of the energy
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    (Original post by Terminatoring)
    AND if you take the equilibrium as 0 pe than that introduces NEGATIVE GRAV POTNETIAL ENRGY,!
    And? As long as you're consistent, it's fine.

    Link me to the question and I'll have a look
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    (Original post by Terminatoring)
    Yeah but how would it ever have 0 potential energy? There's always either grav or elastic and the mark scheme is strict
    at the position of equilibrium all of the potential energy (GPE + Elastic)
    is in the form of kinetic energy for a mass spring system
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    (Original post by rxns_00)
    I think the 0 level is just relative, eg if you pick initial position as the 0 level, the graph will then show whether it has gained or lost GPE relative to that position, as well as the magnitude of the energy
    Absolutely
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    (Original post by Terminatoring)
    AND if you take the equilibrium as 0 pe than that introduces NEGATIVE GRAV POTNETIAL ENRGY,!
    Energy is a scalar quantity so you cant get negative energy
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    (Original post by kingaaran)
    And? As long as you're consistent, it's fine.

    Link me to the question and I'll have a look
    I don't understand how you can sit there and tell me negative energy isn't a problem here. Let's dispel the notion that negative energy isn't a problem - it certainly is. Now tell me how we solve it.
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    (Original post by d14m)
    at the position of equilibrium all of the potential energy (GPE + Elastic)b
    is in the form of kinetic energy for a mass spring system
    Only if you take the equilibrium as 0 potential energy , but the the bottom has NEGATIVE GRSVITSTIONAL POTENTIAL - problem!!
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    (Original post by thatcooldude2.0)
    Guys can I just check this with you (can't find my notes on this topic)
    Elastic collisions- momentum conserved, kinetic energy conserved
    Inelastic collisions- momentum conserved, kinetic energy not conserved
    Explosions- momentum conserved, kinetic energy not conserved
    Yeah this is correct, in the explosion potential energy is converted to kinetic
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    (Original post by Brailey)
    Energy is a scalar quantity so you cant get negative energy
    Except for all of the gravitational potential that is negative.
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    (Original post by Brailey)
    Energy is a scalar quantity so you cant get negative energy
    Isn't all gravitational potential negative?

    Cause it's max is zero....I'm partially kidding....partially
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    How do you do this question?

    Water of density 1000 kg m–3 flows out of a garden hose of cross-sectional area7.2 × 10–4m2 at a rate of 2.0 × 10–4m3 per second. How much momentum is carriedby the water leaving the hose per second?

    A 5.6 × 10–5N s
    B 5.6 × 10–2N s
    C 0.20 N s
    D 0.72 N s
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    (Original post by Terminatoring)
    I don't understand how you can sit there and tell me negative energy isn't a problem here. Let's dispel the notion that negative energy isn't a problem - it certainly is. Now tell me how we solve it.
    No, I am talking about when you are doing energy calculations. The negative sign simply indicates that you have lost energy and it has been expelled into another form.

    Take a particle performing vertical circular motion and your zero level to be at the highest point the particle reaches into its motion. Then when you go beneath that zero level, you have lost potential energy, which will give you a negative sign. Doesn't mean energy is negative - just that it has been lost to another form.

    I'm not going to sit here and claim you can have negative energy, because that of course makes no sense. I would have thought it'd have been obvious that I am talking about energy losses.
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    (Original post by Music With Rocks)
    How do you do this question?

    Water of density 1000 kg m–3 flows out of a garden hose of cross-sectional area7.2 × 10–4m2 at a rate of 2.0 × 10–4m3 per second. How much momentum is carriedby the water leaving the hose per second?

    A 5.6 × 10–5N s
    B 5.6 × 10–2N s
    C 0.20 N s
    D 0.72 N s
    Momentum = mv
    m = density * volume
    v = volume per second divided by area

    Does this help?
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    (Original post by Terminatoring)
    I don't understand how you can sit there and tell me negative energy isn't a problem here. Let's dispel the notion that negative energy isn't a problem - it certainly is. Now tell me how we solve it.
    So you cannot get negative energy, it is simply a graph of energy on the y axis and displacement/amplitude on the x axis. This therefore give your max gravitational energy at max displacement on the negative x axis but still positive energy.

    Look at the attached graph
    Attached Images
     
 
 
 
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