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Edexcel Unit 4: Physics on the Move 6PH04 (11th June 2015) watch

  • View Poll Results: What do you think the grade boundary for an A will be?
    GCE-65+
    3
    3.57%
    GCE-60-64
    28
    33.33%
    GCE-55-59
    31
    36.90%
    GCE-50-54
    8
    9.52%
    GCE-45-49
    1
    1.19%
    GCE- Lower than 44
    0
    0%
    IAL-65+
    5
    5.95%
    IAL-60-64
    6
    7.14%
    IAL-55-59
    2
    2.38%
    IAL-50-54
    0
    0%
    IAL-45-49
    0
    0%
    IAL- Lower than 44
    0
    0%

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    So what would it be for an A roughly?
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    Isn't time period for the roundabout question 3.03, not 3.08? :

    mw^2r=0.35mg
    w^2r=0.35g
    w=sqrt(0.35g/r)
    2pi/T=sqrt(0.35g/r)
    T=2pi*sqrt(r/0.35g)
    T=3.03 (3.s.f.)
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    (Original post by PhysicsFiend)
    Isn't time period for the roundabout question 3.03, not 3.08? :

    mw^2r=0.35mg
    w^2r=0.35g
    w=sqrt(0.35g/r)
    2pi/T=sqrt(0.35g/r)
    T=2pi*sqrt(r/0.35g)
    T=3.03 (3.s.f.)
    I got 3.03s as well
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    Do we know how much each question was worth? I've got a couple of them down, not sure about Qs 11,12 and 18 though! Help Please!
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    (Original post by Dancatpro)
    wait what, how do you know your coursework mark?!

    our school didnt tell us ours ?!
    Lol. xD Our teacher told us the moderated mark

    (Original post by cerlohee)
    A B or a C. But the actual grade you'll get depends on what you got last year as well as how you do in unit 5!!
    i need around 100 UMS for an A in unit 5 :/ I will be happy with a B as uni wants a B. But A will be good
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    (Original post by Freddy-Francis)
    Lol. xD Our teacher told us the moderated mark



    i need around 100 UMS for an A in unit 5 :/ I will be happy with a B as uni wants a B. But A will be good
    You can do it bro!
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    11a=1 mark, 11b = 2 marks
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    (Original post by Dancatpro)
    You can do it bro!
    Thanks buddy
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    (Original post by BP_Tranquility)
    For the capacitor q, I said that since W=½QV , if the charge halves then the energy stored is divided by 4. So the energy must halve before the charge halves . is that correct?

    Posted from TSR Mobile
    nope sorry mate, voltage aint constant, you have to substitute voltage for Q/C to get W=1/2 Q^2/c
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    (Original post by PhysicsFiend)
    11a=1 mark, 11b = 2 marks
    for people who said current is upwards. will get 1 mark in that section if the magnetic field is in the right direction. As in field must be out of the page.
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    (Original post by Freddy-Francis)
    for people who said current is upwards. will get 1 mark in that section if the magnetic field is in the right direction. As in field must be out of the page.
    I sadly did this exact thing, silly error :/ I'm assuming at least 1 ecf mark. Also, since 11b was 2 marks, I have a feeling they wanted something about it being perpendicular (to the current), perhaps?
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    My numerical answers:

    3.03s (roundabout question)
    15V
    2.48e-5
    7.08s for charge or something to decrease to 0.2
    4.1A (emf induced)
    0.035N (tension)
    41 degrees
    18000Hz
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    (Original post by Edac)
    My numerical answers:

    3.03s (roundabout question)
    15V
    2.48e-5
    7.08s for charge or something to decrease to 0.2
    4.1A (emf induced)
    0.035N (tension)
    41 degrees
    18000Hz
    Agreeing with all of them!

    (Apart from I got 1.5V but miscalculated so the 15V is right)
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    hey guys for the kid on the roundabout question, i did mv^2/r= 20x0.35. so 20v^2/0.8=7. then rearrange to get 20 v^2=5.6 so v=0.53.
    since V=d/t and d=2pi r shouldnt t=d/v work to give 9.48seconds??!?!?!?!?!?
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    (Original post by horatiohorn)
    hey guys for the kid on the roundabout question, i did mv^2/r= 20x0.35. so 20v^2/0.8=7. then rearrange to get 20 v^2=5.6 so v=0.53.
    since V=d/t and d=2pi r shouldnt t=d/v work to give 9.48seconds??!?!?!?!?!?
    The frictional force was given by the weight of the child, so the frictional force was 0.35x20x9.81
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    (Original post by LukeBarnett)
    The frictional force was given by the weight of the child, so the frictional force was 0.35x20x9.81
    oh....fml im an idiot. goodbye uni. cheers mate
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    (Original post by horatiohorn)
    oh....fml im an idiot. goodbye uni.
    I did that too at first had to double take, sure you'll still get method marks, don't worry about it too much
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    Grade boundary predictions?
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    (Original post by Edac)
    My numerical answers:

    3.03s (roundabout question)
    15V
    2.48e-5
    7.08s for charge or something to decrease to 0.2
    4.1A (emf induced)
    0.035N (tension)
    41 degrees
    18000Hz
    2.48e-5?
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    (Original post by h.h.h.h.h)
    2.48e-5?
    2.48*10^-5 , think it was joules ?
 
 
 
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