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OCR A2 Physics A: G484 & G485 - 15 & 21 Jun 2017 [Exam Discussion] Watch

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    (Original post by sadasdasdasd)
    M proportional to pressure and L is proportional to volume therefore graph is a straight line through the original.
    I thought L cant be 0 cause other PV=nrt doesnt apply because they'd be no volume so I put mine positive constant gradient starting further down the x axis?
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    (Original post by A sad Man)
    I thought L cant be 0 cause other PV=nrt doesnt apply because they'd be no volume so I put mine positive constant gradient starting further down the x axis?
    Ahh yeah that sounds better, I honestly was just thrown by the exam so i panicked on every question pretty much
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    Anyone get 1.2 for the last question?
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    (Original post by emmabrindley1)
    what do you reckon the grade boundaries will be ?

    I reckon b 37-41 41-42a, 46-47 a*
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    (Original post by Jack077)
    Anyone get 1.2 for the last question?
    i got 240/290 which is 0.83
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    (Original post by sadasdasdasd)
    dunno exactly what the questions are in order, or the numbers, but can remember the method probably

    1) strontium and alpha nucleus
    a) velocity of alpha nucleus
    energy and mass given therefore 1/2*m*v^2
    b) newtons laws to explain motion of strontium
    strontium experiences a force as alpha particle experiences a change in momentum?
    Repulsion between the two are equal and opposite, also act on each others bodies?
    c) calculate v of strontium
    M(initial)*v(initial) = M(final)^v(final)

    2) ?

    3)

    4)

    5)Thermal physics
    a) work out initial n of baloon
    use 4/3*pi*r^3 and PV=nRT using initial T, P
    b) work out final pressure of baloon before it pops
    use 4/3*pi*r^3 for new V and PV=nRT using final T, initial n, final V
    c) ratio question
    E=3/2KT therefore ratio is T(final)/T(initial)

    Help.. pls
    I'll see how much I can recall. Unfortunately, I can't quite remember all the numbers, feel free to chime in if anyone else does.

    2) SHM
    - Straight line through the origin suggests SHM
    - Why the mass attached to spring went towards the outside when rotated
    - State the vertical axis. =v^2 (the graph was in the form v^2=(k/m)R^2. [Derived from F=kR, F=mv2/R.])
    - Calculate units for k =kg s^-2
    - Calculate k/m (i.e. the gradient.) (they gave units of s^-2, confirming previous answer.)
    - How the mass could be determined (gradient=k/m [k was known], m=(k)/gradient.)
    - Experiment for R, F and v using video camera and ruler.

    3) Gravity
    Mass of Mars is 9.3 times less that of Earth. Mars radius=3400km, Earth radius=6400km
    - Calculate g on Mars (I did a lot of working these 2 marks. I certainly over complicated things.)
    - Comparison of a probe taking off from the surface of Mars compared to Earth (Resultant force difference due to gravity air resistance.)
    - 3 Properties for the satellite to remain in the same point above Mars (Equatorial orbit, period = Marian time, same direction as Mars rotation.)
    - Calculate the radius of satellite (Using, T^2=(4pi^2/GM)r^3. With the mass of Mars calculated before [6.48x10^23 kg])
    - The other satellite had a polar orbit with a particular distance above Mars. Calculate the number of pictures taken per second. (They gave the number of total pictures. [Around 3000.]) (Determine period using T^2 equation [r=radius of Mars+distance above surface] then divide total number by the period.) (answer = 527?).

    4) Photon energy, E=mc(theta).
    45mA of 130keV electrons for 1.6s.
    - Calculate [confirm] the number of electrons [equals 5x10^17]. (=It/e=(45x10-3)(1.6)/(1.6x10-19)=4.5x10^17 electrons)
    - Calculate wavelength. (E=130keV=2.08x10-14J. E=hc/wavelength->wavelength=9.56x10-12 m)
    Water cooled thing down by 90%. c=4200, initial temp=17c, final temp=27c, density of the water=1000kgm^-3[?]
    - Calculate volume of water per second. (Energy cooled by water=(0.9)(number of electrons)(energy of electron)=(0.9)(5x10^17)(2.08x10-14)=9360 J. E=mc(theta)->m=E/c(theta)=(9360)/(4200)(27-17)=0.22 kg. m per second=(0.22)/(1.6)=0.139 kgs-1. volume per second=(0.139)/1000= 1.39x10-4 m^3 s^-1 [Not entirely sure about this one due to forgetting the density.]
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    For question one did anyone get like 1.6x10^7 which is approx 2x10^7 cus that seemed a bit off
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    (Original post by Super199)
    For question one did anyone get like 1.6x10^7 which is approx 2x10^7 cus that seemed a bit off
    I did, thought it was some terrible rounding on their part
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    (Original post by adsy9)
    I did, thought it was some terrible rounding on their part
    It makes sense but i was expecting 1.9 something lol
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    (Original post by headyb)
    I'll see how much I can recall. Unfortunately, I can't quite remember all the numbers, feel free to chime in if anyone else does.

    2) SHM
    - Straight line through the origin suggests SHM
    - Why the mass attached to spring went towards the outside when rotated
    - State the vertical axis. =v^2 (the graph was in the form v^2=(k/m)R^2. [Derived from F=kR, F=mv2/R.])
    - Calculate units for k =kg s^-2
    - Calculate k/m (i.e. the gradient.) (they gave units of s^-2, confirming previous answer.)
    - How the mass could be determined (gradient=k/m [k was known], m=(k)/gradient.)
    - Experiment for R, F and v using video camera and ruler.

    3) Gravity
    Mass of Mars is 9.3 times less that of Earth. Mars radius=3400km, Earth radius=6400km
    - Calculate g on Mars (I did a lot of working these 2 marks. I certainly over complicated things.)
    - Comparison of a probe taking off from the surface of Mars compared to Earth (Resultant force difference due to gravity air resistance.)
    - 3 Properties for the satellite to remain in the same point above Mars (Equatorial orbit, period = Marian time, same direction as Mars rotation.)
    - Calculate the radius of satellite (Using, T^2=(4pi^2/GM)r^3. With the mass of Mars calculated before [6.48x10^23 kg])
    - The other satellite had a polar orbit with a particular distance above Mars. Calculate the number of pictures taken per second. (They gave the number of total pictures. [Around 3000.]) (Determine period using T^2 equation [r=radius of Mars+distance above surface] then divide total number by the period.) (answer = 527?).

    4) Photon energy, E=mc(theta).
    45mA of 130keV electrons for 1.6s.
    - Calculate [confirm] the number of electrons [equals 5x10^17]. (=It/e=(45x10-3)(1.6)/(1.6x10-19)=4.5x10^17 electrons)
    - Calculate wavelength. (E=130keV=2.08x10-14J. E=hc/wavelength->wavelength=9.56x10-12 m)
    Water cooled thing down by 90%. c=4200, initial temp=17c, final temp=27c, density of the water=1000kgm^-3[?]
    - Calculate volume of water per second. (Energy cooled by water=(0.9)(number of electrons)(energy of electron)=(0.9)(5x10^17)(2.08x10-14)=9360 J. E=mc(theta)->m=E/c(theta)=(9360)/(4200)(27-17)=0.22 kg. m per second=(0.22)/(1.6)=0.139 kgs-1. volume per second=(0.139)/1000= 1.39x10-4 m^3 s^-1 [Not entirely sure about this one due to forgetting the density.]
    Slightly worried as I don't have any recollection of your Q3!
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    (Original post by yelash)
    Slightly worried as I don't have any recollection of your Q3!
    Lol I couldn't work out mass of earth because I was in a rush so I just wacked on 6X10^24 because i remembered it from practise questions, So unless they accept that, I've pretty much lost the entire question apart from explanations
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    (Original post by Jack077)
    Anyone get 1.2 for the last question?
    I got that but it might've been 0.83 because apparently it was the other way around so we both must've rushed it :/
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    (Original post by A sad Man)
    Lol I couldn't work out mass of earth because I was in a rush so I just wacked on 6X10^24 because i remembered it from practise questions, So unless they accept that, I've pretty much lost the entire question apart from explanations
    Same, that's what I did, I even put a little note lol.
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    (Original post by headyb)
    I'll see how much I can recall. Unfortunately, I can't quite remember all the numbers, feel free to chime in if anyone else does.

    2) SHM
    - Straight line through the origin suggests SHM
    - Why the mass attached to spring went towards the outside when rotated
    - State the vertical axis. =v^2 (the graph was in the form v^2=(k/m)R^2. [Derived from F=kR, F=mv2/R.])
    - Calculate units for k =kg s^-2
    - Calculate k/m (i.e. the gradient.) (they gave units of s^-2, confirming previous answer.)
    - How the mass could be determined (gradient=k/m [k was known], m=(k)/gradient.)
    - Experiment for R, F and v using video camera and ruler.

    3) Gravity
    Mass of Mars is 9.3 times less that of Earth. Mars radius=3400km, Earth radius=6400km
    - Calculate g on Mars (I did a lot of working these 2 marks. I certainly over complicated things.)
    - Comparison of a probe taking off from the surface of Mars compared to Earth (Resultant force difference due to gravity air resistance.)
    - 3 Properties for the satellite to remain in the same point above Mars (Equatorial orbit, period = Marian time, same direction as Mars rotation.)
    - Calculate the radius of satellite (Using, T^2=(4pi^2/GM)r^3. With the mass of Mars calculated before [6.48x10^23 kg])
    - The other satellite had a polar orbit with a particular distance above Mars. Calculate the number of pictures taken per second. (They gave the number of total pictures. [Around 3000.]) (Determine period using T^2 equation [r=radius of Mars+distance above surface] then divide total number by the period.) (answer = 527?).

    4) Photon energy, E=mc(theta).
    45mA of 130keV electrons for 1.6s.
    - Calculate [confirm] the number of electrons [equals 5x10^17]. (=It/e=(45x10-3)(1.6)/(1.6x10-19)=4.5x10^17 electrons)
    - Calculate wavelength. (E=130keV=2.08x10-14J. E=hc/wavelength->wavelength=9.56x10-12 m)
    Water cooled thing down by 90%. c=4200, initial temp=17c, final temp=27c, density of the water=1000kgm^-3[?]
    - Calculate volume of water per second. (Energy cooled by water=(0.9)(number of electrons)(energy of electron)=(0.9)(5x10^17)(2.08x10-14)=9360 J. E=mc(theta)->m=E/c(theta)=(9360)/(4200)(27-17)=0.22 kg. m per second=(0.22)/(1.6)=0.139 kgs-1. volume per second=(0.139)/1000= 1.39x10-4 m^3 s^-1 [Not entirely sure about this one due to forgetting the density.]
    for the first part of 2) the direction of the springs acceleration and extension were oppositely directed, this was the first reason for SHM

    first part of 3) was weird, I just calculated mass of marks first then used that for the next parts

    no. of photos for 3) I got 1670 3.s.f.

    density for 4) for 1000kgm^-3

    everything there I got the same pretty much
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    (Original post by A sad Man)
    Lol I couldn't work out mass of earth because I was in a rush so I just wacked on 6X10^24 because i remembered it from practise questions, So unless they accept that, I've pretty much lost the entire question apart from explanations
    I was joking before but I think I may have actually missed it! Whenn you say its Q3 I pressume thats not the actual Q number as the long answer Questions started at Q15 or something?
    Oh well hopefully I'm just blocking the pain of rememebring wrong answers from my brain
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    (Original post by sadasdasdasd)
    for the first part of 2) the direction of the springs acceleration and extension were oppositely directed, this was the first reason for SHM

    first part of 3) was weird, I just calculated mass of marks first then used that for the next parts

    no. of photos for 3) I got 1670 3.s.f.

    density for 4) for 1000kgm^-3

    everything there I got the same pretty much
    (Original post by A sad Man)
    Lol I couldn't work out mass of earth because I was in a rush so I just wacked on 6X10^24 because i remembered it from practise questions, So unless they accept that, I've pretty much lost the entire question apart from explanations
    I'm sure this was overkill for the 2 marks but it seemed to work out.
    http://i.imgur.com/J1R0RPB.jpg
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    Does anyone have the 2016 papers?
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    (Original post by clear_water)
    Does anyone have the 2016 papers?
    this
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    Have they increased the time for the G485 paper. Because I did it last year and I'm retaking this year and every paper so far has beem 1h 45mins but the place where I am doing it says 120mins (2h). Is it possible the college just gave extra time?
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    (Original post by k_nikolov)
    Have they increased the time for the G485 paper. Because I did it last year and I'm retaking this year and every paper so far has beem 1h 45mins but the place where I am doing it says 120mins (2h). Is it possible the college just gave extra time?
    they've been 2 hours for a few years now I'm pretty sure.
 
 
 
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