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OCR PHYSICS B G495~ 18th June 2015 AM ~ A2 Physics Watch

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    (Original post by Cowy97)
    it felt like it was there as a troll
    I thought that was the end, until i accidentally flicked through, with under 5 minutes left. Think i got the 4 marker right, but i didn't even read the last two questions.
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    A friend and I have managed to remember these and *think* these are the answers. If there's anything we're missing, please do post!



    SECTION A
    1 Particle Physics [2]
    a. (neutrino is) Lepton and Neutral
    b. (proton is) Nucleon and Hadron

    2 Scattering [3]
    Particle deflected more
    Because particle is moving slower and so force has more time to act on it

    3 Graph of V vs r [3]
    Field is uniform electric field
    Field strength is the gradient
    Gradient is constant, so E is constant
    E = dV/dr = 3000/0.07 = 4.3 x 10^4 Vm-1

    4 Alpha Safety [1, 2]
    a. Mass is NOT conserved.
    b. Agree (they are dangerous inside the body)
    Alpha particles have high ionising ability, because they have a 2+ charge and can take electrons away from atoms easily.
    They also have a low penetrating ability, so they can’t escape the body if ingested.
    (ORA: Alpha particles highly ionising, but are stopped by a few cm of air/skin, so cannot get inside the body unless ingested/swallowed)

    5 Transformers [1]
    C (6 V and 0.36A)

    6 Transformers (again) [2, 3]
    a. flux is the same everywhere
    flux density is double at point Z than at the sides/top of the core [must say double/twice/etc]
    b. Crack introduces air gap
    Air has much lower permeability compared to iron (about 1/8000th)
    So overall permeance of core decreases
    Φ is proportional to Λ, so flux increases
    B is proportional to Φ, so flux density increases

    7 Energy Levels [2]
    ΔE = -1.5 eV + 3.4 eV = 1.9 eV = 3.04x10^-19 J
    f = 4.6 x 10^14 Hz
    λ = 6.5 x 10^-7 m
    Lamda or frequency?

    8 ???


    SECTION B
    9 Electric Fields [1, 1, 1, 1, 4, 2]
    a i. field strength greater at X since field lines are closer together.
    a ii.draw circle around the charge passing through X.
    b i. Divide by 4, so (+)405 V
    b ii. Divide by 16 so 6062.5 N
    c. For Fig 9.3a, resultant force is 0 N, so E = F/q = 0 NC-1
    (OR calculations and statement: Eleft = Eright, so no resultant field strength)
    For Fig 9.3b, resultant force is 2 x kQq/r2 = 6.3 x 10^7 x q N
    So E = F/q = 6.3 x 10^7 NC-1
    d. For +ve charge, Velec = +kQ/r
    For -ve charge, Velec = -kQ/r
    So total V = +kQ/r - kQ/r = 0

    10 Motors
    a. F = BIL = ???
    b. If L is halved, F is halved (since F is proportional to L), so the acceleration halves, and the motor moves more slowly. --QUERY: could be as F halves, but m halves, acceleration is the same, so no change in rate.
    c. max NΦ = NBA = 0.0243 Wb-turns
    d. average e.m.f = d(NΦ)/dt = 0.12(15) V
    e. When moving, the motor is experiencing a change of flux linkage, and so an emf is induced.
    Current is induced too, and by Lenz's law the induced current must flow in such a way that it opposes the change creating it, so it acts in the opposite direction to the current already there.
    So the resultant current decreases while the motor turns.
    When the motor is jammed (not moving) there is no change in flux linkage, so no emf is induced, and so no opposing current is induced. This means the current remains as it is (higher than when rotating).

    11 Particles in Magnetic Fields [1?, 1, 2, 4]
    a. Force acts perpendicular to direction of motion, so motion is circular
    b. r = mv/Bqc. Alpha particle is much heavier than beta particle, so you would expect the radius of the circular path it takes to be much larger, as predicted in the diagram.
    Gamma straight because it is uncharged, so is unaffected by the magnetic field.
    d. rα = 0.20625/B
    rβ = 0.0011375/B
    rβ is 180 times smaller than rα, so diagram incorrect in the sense that it only shows slight difference in radius.

    12 Radiation [1, 2, 3, 2, 3]
    a. LHS: Lepton number = 0 [nuclei have no lepton number]
    RHS: Lepton number = 0 + 1 - 1 = 0 [nuclei have no lepton number, electron = +1, anti-neutrino = -1]
    So lepton number conserved.
    b. λ = 1.0 x 10-6 s-1A0 = 5.49 x 10^5 Bq
    c. A12 weeks = 394 Bq [345ish if you use their values)
    d. Absorbed dose = 0.89 Gy
    Assumption is all the beta particles are absorbed by the gland
    e. γ rays have high penetrating ability, so can escape from body
    Radiation can therefore be detected by external sensors to track movement through body.
    γ rays have low ionising ability, so do not cause much damage (as long as not too much is given)
    Iodine-131 not used anymore, because the half-life is too long.
    The total absorbed dose by the thyroid gland due to 131-I was too high (so increases risk of cancer by unacceptable amounts).

    SECTION C
    13 Lenses [1, 2]
    a. magnification = v/u = 89/-5.3 = (-)16.79… = (-)17x
    b. power = 200 D

    14 Imaging [2, 2]
    a. 5.2 x 10-7 m
    b. Approx. 270 lenses in picture, so 20,000 lenses in total, but they will probably accept values in a large range
    c. A part c????

    15 Gases and Boltzmann Factor [3, 4]
    a. ??? something here ???
    b. Number of particles per cubic cm was 2.4 x 10^9
    c. I is proportional to eE/kT... Boltzmann factor can be used because...
    BF is proportion of electrons with enough energy to escape.
    Rate of electron release is proportional to BF
    Current is proportional to number of electrons flowing (since current is rate of flow of charge)
    So current is proportional to BF
    d. T = 2310 K

    16 B Fields [2, 3]
    a. N = 3800 turns [show that]
    b. 380-390 V (allow range as ft from a)

    17 Wavelength of Electrons
    a. i. E = 6.4 x 10^-15 J
    ii. p = 1.1 x 10^-22 kgms-1
    b. λ = 6.1 x 10^-12 m
    Resolution = 3.1 x 10^-12 m
    c. number of electrons needed to change temperature by 1 K = 1.8 x 10^6

    18 Deflecting Electrons [1, 4, 2]
    a. E = 1.1 x 10^5 Vm-1
    b. L = 0.027m
    c. E is constant everywhere (it is a uniform field) so force is the same everywhere / process in (b) is the same everywhere.

    19 Relativity [3]
    γ = 1.39
    v = 2.1 x 10^8 ms-1
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    (Original post by Greating)
    A friend and I have managed to remember these and *think* these are the answers. If there's anything we're missing, please do post!



    SECTION A
    1 Particle Physics [2]
    a. (neutrino is) Lepton and Neutral
    b. (proton is) Nucleon and Hadron

    2 Scattering [3]
    Particle deflected more
    Because particle is moving slower and so force has more time to act on it

    3 Graph of V vs r [3]
    Field is uniform electric field
    Field strength is the gradient
    Gradient is constant, so E is constant
    E = dV/dr = 3000/0.07 = 4.3 x 104 Vm-1

    4 Alpha Safety [1, 2]
    a. Mass is NOT conserved.
    b. Agree (they are dangerous inside the body)
    Alpha particles have high ionising ability, because they have a 2+ charge and can take electrons away from atoms easily.
    They also have a low penetrating ability, so they can’t escape the body if ingested.
    (ORA: Alpha particles highly ionising, but are stopped by a few cm of air/skin, so cannot get inside the body unless ingested/swallowed)

    5 Transformers [1]
    C (6 V and 0.36A)

    6 Transformers (again) [2, 3]
    a. flux is the same everywhere
    flux density is double at point Z than at the sides/top of the core [must say double/twice/etc]
    b. Crack introduces air gap
    Air has much lower permeability compared to iron (about 1/8000th)
    So overall permeance of core decreases
    Φ is proportional to Λ, so flux increases
    B is proportional to Φ, so flux density increases

    7 Energy Levels [2]
    ΔE = -1.5 eV + 3.4 eV = 1.9 eV = 3.04x10-19 J
    f = 4.6 x 1014 Hz
    λ = 6.5 x 10-7 m
    Lamda or frequency?

    8 ???


    SECTION B
    9 Electric Fields [1, 1, 4, 2]
    a i. Divide by 4, so (+)405 V
    a ii. Divide by 16 so 6062.5 N
    b. For Fig 9.3a, resultant force is 0 N, so E = F/q = 0 NC-1
    (OR calculations and statement: Eleft = Eright, so no resultant field strength)
    For Fig 9.3b, resultant force is 2 x kQq/r2 = 6.3 x 107 x q N
    So E = F/q = 6.3 x 10^7 NC-1
    c. For +ve charge, Velec = +kQ/r
    For -ve charge, Velec = -kQ/r
    So total V = +kQ/r - kQ/r = 0

    10 Motors
    a. F = BIL = ???
    b. If L is halved, F is halved (since F is proportional to L), so the acceleration halves, and the motor moves more slowly.
    c. max NΦ = NBA = 0.0243 Wb-turns
    d. average e.m.f = d(NΦ)/dt = 0.12(15) V
    e. When moving, the motor is experiencing a change of flux linkage, and so an emf is induced.
    Current is induced too, and by Lenz's law the induced current must flow in such a way that it opposes the change creating it, so it acts in the opposite direction to the current already there.
    So the resultant current decreases while the motor turns.
    When the motor is jammed (not moving) there is no change in flux linkage, so no emf is induced, and so no opposing current is induced. This means the current remains as it is (higher than when rotating).

    11 Particles in Magnetic Fields [1?, 1, 2, 4]
    a. Force acts perpendicular to direction of motion, so motion is circular
    b. r = mv/Bqc. Alpha particle is much heavier than beta particle, so you would expect the radius of the circular path it takes to be much larger, as predicted in the diagram.
    Gamma straight because it is uncharged, so is unaffected by the magnetic field.
    d. rα = 0.20625/B
    rβ = 0.0011375/B
    rβ is 180 times smaller than rα, so diagram incorrect in the sense that it only shows slight difference in radius.

    12 Radiation [1, 2, 3, 2, 3]
    a. LHS: Lepton number = 0 [nuclei have no lepton number]
    RHS: Lepton number = 0 + 1 - 1 = 0 [nuclei have no lepton number, electron = +1, anti-neutrino = -1]
    So lepton number conserved.
    b. λ = 1.0 x 10-6 s-1A0 = 5.49 x 105 Bq
    c. A12 weeks = 394 Bq
    d. Absorbed dose = 0.89 Gy
    Assumption is all the beta particles are absorbed by the gland
    e. γ rays have high penetrating ability, so can escape from body
    Radiation can therefore be detected by external sensors to track movement through body.
    γ rays have low ionising ability, so do not cause much damage (as long as not too much is given)
    Iodine-131 not used anymore, because the half-life is too long.
    The total absorbed dose by the thyroid gland due to 131-I was too high (so increases risk of cancer by unacceptable amounts).

    SECTION C
    13 Lenses [1, 2]
    a. magnification = v/u = 89/-5.3 = (-)16.79… = (-)17x
    b. power = 200 D

    14 Imaging [2, 2]a. 5.2 x 10-7 mb. Approx. 270 lenses in picture, so 20,000 lenses in total, but they will probably accept values in a large range
    I feel like there was a question here15 Gases and Boltzmann Factor [3, 4]a. ??? something here ???b. Number of particles per cubic cm was 2.4 x 109c. I is proportional to eE/kT... Boltzmann factor can be used because...BF is proportion of electrons with enough energy to escape.Rate of electron release is proportional to BFCurrent is proportional to number of electrons flowing (since current is rate of flow of charge)So current is proportional to BFd. T = 2310 K
    16 B Fields [2, 3]a. N = 3800 turns [show that]b. 380-390 V (allow range as ft from a)
    17 Wavelength of Electronsa. i. E = 6.4 x 10-15 Jii. p = 1.1 x 10-22 kgms-1b. λ = 6.1 x 10-12 mResolution = 3.1 x 10-12 mc. number of electrons needed to change temperature by 1 K = 1.8 x 106
    18 Deflecting Electrons [1, 4, 2]a. E = 1.1 x 105 Vm-1b. L = 0.027mc. E is constant everywhere (it is a uniform field) so force is the same everywhere / process in (b) is the same everywhere.
    19 Relativity [3]γ = 1.39v = 2.1 x 108 ms-1
    for 10 i said when length halfed the force halfed but so did the mass so f=ma had the same acceleration
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    (Original post by Cowy97)
    for 10 i said when length halfed the force halfed but so did the mass so f=ma had the same acceleration
    Hmm... I was tempted by that argument in the exam, but there are a few points to note:

    As the side length of the coil halves, the mass of the total coil would quarter, not halve (so if F halves, but m quarters, a would double, so rate of rotation would increase).

    However, since the rate of rotation is determined by some hideously complicated moment of inertia (which we definitely aren't required to do in physics A Level), I just took the reasoning that let us assume the mass is still the same, so force halves and rate decreases.

    You could be right, I don't know...


    EDIT: this is incorrect. See QuantumHover's response below.
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    (Original post by HennersPD)
    I hope so too becauae although i liked section C, section A and B were very tough in places :/ whatd you think 80/100 on this will be in UMS and 45/60 on G494?
    Personally my guess is about 48/60 for 81 ums on g494 and 82/100 for 135 ums in this paper but that may be optimistic... 😜
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    (Original post by Greating)
    Hmm... I was tempted by that argument in the exam, but there are a few points to note:

    As the side length of the coil halves, the mass of the total coil would quarter, not halve (so if F halves, but m quarters, a would double, so rate of rotation would increase).

    However, since the rate of rotation is determined by some hideously complicated moment of inertia (which we definitely aren't required to do in physics A Level), I just took the reasoning that let us assume the mass is still the same, so force halves and rate decreases.

    You could be right, I don't know...
    *******s, i was tempted to put the first answer
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    (Original post by Greating)

    As the side length of the coil halves, the mass of the total coil would quarter, not halve
    No, because the coil is only on the outside of the square the increase is linear not quadratic.
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    (Original post by QuantumHover)
    No, because the coil is only on the outside of the square the increase is linear not quadratic.
    Yes, of course. Sorry, my misunderstanding.
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    (Original post by STATER)
    Sounds good. As always the main thing holding me back is I misread questions, and do stupid calculation errors. I might have scraped an A* in maths hopefully. G495 went very well as far as I know - only around 6 marks that I lost for sure. G494 was borderline - probably an A but not an A*.
    Ah always the way, the feeling when you walk out and realise the stupid errors. Happens with me a lot, but I've put the effort in this year so I should hope for good results! I think G494 went very well for me with calculations, but think I'll lose more marks than I'd be OK with due to insufficient info in explanations. Talking about explanations, what did you put for the two charges with 2x10^-3 gap and the field strength at their minimum? I have a feeling I didn't put enough, as I calculated the field strengths and said on one they cancel out and didn't even calculate the potential for the next part.
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    I agree with a lot of those answers Cowy97 has given, so I'm pretty confident I've got an A and might be being a little optimistic but maybe an A*.... I seem to mainly get A/A* in homeworks and stuff but physics exams are a fair bit different to those homework questions, and I'd rather study in school with lessons rather than on my own haha. Let's hope all of this work pays off, and everyone elses work pays off for them!
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    (Original post by RealTime)
    I agree with a lot of those answers Cowy97 has given, so I'm pretty confident I've got an A and might be being a little optimistic but maybe an A*.... I seem to mainly get A/A* in homeworks and stuff but physics exams are a fair bit different to those homework questions, and I'd rather study in school with lessons rather than on my own haha. Let's hope all of this work pays off, and everyone elses work pays off for them!
    Yeah too think i may have got an A on this or even A*! Hope so anyway! what'd you draw for the equipotential at X?
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    (Original post by megatovskimaths)
    Personally my guess is about 48/60 for 81 ums on g494 and 82/100 for 135 ums in this paper but that may be optimistic... 😜
    It was 81/100 for 135 ums last year (and the year before). Considering that (through the polls linked to this thread and last year's equivalent) this year's paper seems to have been slightly harder than last year's, I'm hoping for a slightly lower 90 percent mark, although I suppose this could be wishful thinking.
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    (Original post by HennersPD)
    Yeah too think i may have got an A on this or even A*! Hope so anyway! what'd you draw for the equipotential at X?
    I just drawn a circle
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    (Original post by AccountName)
    It was 81/100 for 135 ums last year (and the year before). Considering that (through the polls linked to this thread and last year's equivalent) this year's paper seems to have been slightly harder than last year's, I'm hoping for a slightly lower 90 percent mark, although I suppose this could be wishful thinking.
    I'd assume that it'd be similar, the paper was around the same difficulty as others but that's just my opinion
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    (Original post by Cowy97)
    for 10 i said when length halfed the force halfed but so did the mass so f=ma had the same acceleration
    For 7 don't you do 3.4-1.5 or whatever it was?


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    (Original post by Mutleybm1996)
    For 7 don't you do 3.4-1.5 or whatever it was?


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    Yeah and then E=mc^2, you're on about that one right?
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    (Original post by RealTime)
    I'd assume that it'd be similar, the paper was around the same difficulty as others but that's just my opinion
    There was about 10 marks of truly challenging stuff
    If some people missed the last page....drops lower
    Apart from that I thought it was okayish


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    (Original post by RealTime)
    I just drawn a circle
    I did around the point X, is that correct or was it through X?
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    (Original post by Mutleybm1996)
    There was about 10 marks of truly challenging stuff
    If some people missed the last page....drops lower
    Apart from that I thought it was okayish


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    yeah thats true! did you get the last question with the gamma factor? got v = 2.1*10^8 ms-1 and how did you work the resolution out for the flea image?
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    (Original post by HennersPD)
    I did around the point X, is that correct or was it through X?
    Oh I don't understand why it would be what you done, since equipotentials cut the field lines at 90 degrees and lies on a line of equal electric field strength, which would be a circle around the point charge
 
 
 
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