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    (Original post by MathsNerd1)
    Thanks and to be honest I can't really get anywhere with DJ's question as I'm not too confident in proving things in Mechanics :-/
    Well, first of all think back to basics. If you have a constant force acting on the body, what does this imply? Does this provide any hints on other things you could potentially use?

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    F=ma

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    constant acceleration

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    SUVAT is appropriate for constant acceleration

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    End of the line, sorry. :teehee:




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    (Original post by DJMayes)
    Prove that, if a constant force acts on a body, then the work done by the force is equal to the change in kinetic energy of the body.
    No peeking MathsNerd
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    Assuming not travelling close to speed of light.
    F=ma \Rightarrow a =\frac{F}{m}
    a = constant
    Consider:
    v^2 = u^2 +2as ,



\Rightarrow \frac{1}{2}(v^2 - u^2) = as



\Rightarrow \frac{1}{2}(v^2 - u^2) = \frac{F}{m}s



\Rightarrow \frac{1}{2}m(v^2 - u^2) = Fs
    Is that sufficiently rigorous?
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    (Original post by joostan)
    No peeking MathsNerd
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    Assuming not travelling close to speed of light.
    F=ma \Rightarrow a =\frac{F}{m}
    a = constant
    Consider:
    v^2 = u^2 +2as , \ \ let \ u=0



\Rightarrow \frac{1}{2}v^2 = as



\Rightarrow \frac{1}{2}v^2 = \frac{F}{m}s



\Rightarrow \frac{1}{2}mv^2 = Fs
    Is that sufficiently rigorous?
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    "Assuming not travelling close to speed of light." :rofl:

    We can derive a more general equation for change in kinetic energy if we do not assume  u=0 .
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    (Original post by DJMayes)
    Well, first of all think back to basics. If you have a constant force acting on the body, what does this imply? Does this provide any hints on other things you could potentially use?

    More explicit hint:

    Spoiler:
    Show


    F=ma

    Even more explicit hint:

    Spoiler:
    Show


    constant acceleration

    Even more explicit hint:

    Spoiler:
    Show


    SUVAT is appropriate for constant acceleration

    even more explicit hint:

    Spoiler:
    Show


    End of the line, sorry. :teehee:




    Well I know what the equation will be for change in KE is 1/2*m(V^2-U^2) and if you have a constant force acting upon something then you'll have your resultant force equation with a constant acceleration. I'm guessing some rearrangement of this will get me the answer?
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    (Original post by MathsNerd1)
    Hmm, I'm not sure with that as I'm not too strong in the Mechanics section but I'd probably start by drawing a diagram and going from there?
    I haven't really done much mechanics, but I'm pretty sure no diagrams are involved.

    I tried doing a calculus proof, but got nowhere so I gave up and cheaply used SUVAT :teehee:
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    (Original post by joostan)
    No peeking MathsNerd
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    Assuming not travelling close to speed of light.
    F=ma \Rightarrow a =\frac{F}{m}
    a = constant
    Consider:
    v^2 = u^2 +2as , \ \ let \ u=0



\Rightarrow \frac{1}{2}v^2 = as



\Rightarrow \frac{1}{2}v^2 = \frac{F}{m}s



\Rightarrow \frac{1}{2}mv^2 = Fs
    Is that sufficiently rigorous?
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    You don't need to assume u=0, change in K.E. is \frac{1}{2}m(v^2-u^2)
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    (Original post by Scientific Eye)
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    "Assuming not travelling close to speed of light." :rofl:

    We can derive a more general equation for change in kinetic energy if we do not assume  u=0 .
    (Original post by justinawe)
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    You don't need to assume u=0, change in K.E. is \frac{1}{2}m(v^2-u^2)

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    I like to cover all my bases.
    Yeah you get a \Delta E_k But I was going for the \frac{1}{2}mv^2
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    (Original post by joostan)
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    I like to cover all my bases.
    Yeah you get a \Delta E_k But I was going for the \frac{1}{2}mv^2
    u=0 was not a condition though and it's supposed to be change in K.E., so I'm not sure if your proof covers all the bases :sly:
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    (Original post by justinawe)
    u=0 was not a condition though and it's supposed to be change in K.E., so I'm not sure if your proof covers all the bases :sly:
    Are you happy now? I edited it in
    I've not done M2 I was going for the physics KE equation
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    (Original post by joostan)
    Are you happy now? I edited it in
    I've not done M2 I was going for the physics KE equation
    I haven't even done M1, so that's not an excuse! :noway:
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    (Original post by justinawe)
    I haven't even done M1, so that's not an excuse! :noway:
    (Original post by DJMayes)
    .
    Hehe:
    OK . . .Calculus proof:
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    W = \displaystyle\int F \ dx = \displaystyle\int ma \ dx = \displaystyle\int m \frac{dv}{dt} \ dx

\Rightarrow W = \displaystyle\int m \frac{dx}{dt} \ dv

\Rightarrow W = \displaystyle\int mv\ dv = \frac{1}{2}mv^2
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    Some limits'll give you the delta
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    (Original post by joostan)
    Hehe:
    OK . . .Calculus proof:
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    W = \displaystyle\int F \ dx = \displaystyle\int ma \ dx = \displaystyle\int m \frac{dv}{dt} \ dx

\Rightarrow W = \displaystyle\int m \frac{dx}{dt} \ dv

\Rightarrow W = \displaystyle\int mv\ dv = \frac{1}{2}mv^2
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    Some limits'll give you the delta
    Yeah, did that too I was trying something else out of curiosity to see if it worked
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    (Original post by justinawe)
    Yeah, did that too I was trying something else out of curiosity to see if it worked
    Yeah I played around with:
    a = \dfrac{d^2x}{dt^2} but nothing seemed to come of it
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    (Original post by joostan)
    Yeah I played around with:
    a = \dfrac{d^2x}{dt^2} but nothing seemed to come of it
    This is probably utter nonsense, but:

    W = \displaystyle \int F \ dx

    \Rightarrow W = \displaystyle \int \frac{dp}{dt} \ dx

    \Rightarrow W = \displaystyle \int \frac{dx}{dt} \ dp

    \Rightarrow W = \displaystyle \int v \ dp

    integrating p wrt v apparently gives K.E., but "wrt v" is pretty meaningless I think... but anyway, I got something slightly different
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    For C2, a lot of things need so much 'mathematical imagination'. How do people deal with these situations? Like when sometimes you just have to 'know' to use SOHCAHTOA, sometimes you have to know to use a circle theorem etc etc
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    (Original post by justinawe)
    This is probably utter nonsense, but:

    W = \displaystyle \int F \ dx

    \Rightarrow W = \displaystyle \int \frac{dp}{dt} \ dx

    \Rightarrow W = \displaystyle \int \frac{dx}{dt} \ dp

    \Rightarrow W = \displaystyle \int v \ dp

    integrating p wrt v apparently gives K.E., but "wrt v" is pretty meaningless I think... but anyway, I got something slightly different
    That works out quite nicely if you say that:
    v = \frac{1}{m}p

    (Original post by GCSE-help)
    For C2, a lot of things need so much 'mathematical imagination'. How do people deal with these situations? Like when sometimes you just have to 'know' to use SOHCAHTOA, sometimes you have to know to use a circle theorem etc etc
    I'm not sure that knowing SOHCAHTOA is mathematical imagination, it's just expected knowledge I guess.
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    (Original post by joostan)
    That works out quite nicely if you say that:
    v = \frac{1}{m}p
    Oh, right, thanks for that
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    (Original post by GCSE-help)
    For C2, a lot of things need so much 'mathematical imagination'. How do people deal with these situations? Like when sometimes you just have to 'know' to use SOHCAHTOA, sometimes you have to know to use a circle theorem etc etc
    Having experience really helps. So, the more you practise with these types of questions, the more you develop your intuition as to when you need to use those mathematical tools.
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    (Original post by justinawe)
    Oh, right, thanks for that
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    Omg i cant do the soloman papers i can never finish one help
 
 
 
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