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    (Original post by teachercol)
    OCR Physics A Newtonian World 20/6/16

    iii) 5pi/4 = 3.93 rad (might be SF penalties here if left in pi) (1)

    Good Luck Col
    Thanks teachercol apperiate it.

    With 3a)iii) Would I get a mark for writing down 0.75π or does the phase difference depend on the direction of the wave?

    In order words, is 5/4pi = 3/4pi ? in terms of the phase difference.

    Regardless, should have just wrote down 5/4pi and not try to over complicate it.
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    (Original post by lai812matthew)
    you need to start your derivation from F1 equation, that's what the questions asks.
    I used the equation from part ii (that I got correct) to show that relationship yes, so you saying I got the marks then?
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    right could someone literally anyone respond please, i wrote as an assumption that there is no time between the inversions of the tube in the lead experiment? reckon that's okay?
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    (Original post by Jm098)
    Looks like i got 45... Pushing for an A*? lol
    How were you doing in Past Papers?

    I was getting A's/ Low A*s and I got 45, I expect it to be very close to the A* boundary regardless.

    The A* boundary will not be under 42 but it won't be like 49.
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    (Original post by Asurat)
    Looking like sky high grade boundaries.
    Way to scare everyone
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    (Original post by teachercol)
    Quite right - big typo - I got 0.59m.
    I'll edit it.
    You edited it but you put 0.69 instead of 0.59.
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    (Original post by Jm098)
    right could someone literally anyone respond please, i wrote as an assumption that there is no time between the inversions of the tube in the lead experiment? reckon that's okay?
    I don't know I have to say, I said that we assume that all the weight of the pellets stays in the top and therefore all of it falls through the entire length of the tube, and that no thermal energy was lost to the surroundings.
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    (Original post by ComputerMaths97)
    How were you doing in Past Papers?

    I was getting A's/ Low A*s and I got 45, I expect it to be very close to the A* boundary regardless.

    The A* boundary will not be under 42 but it won't be like 49.
    Well to be honest, got full UMS in the 2015 as a mock, fingers crossed ay! you'd think it would be an A though surely?
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    TeacherCol do you have any predictions for grade boundaries? Usually you're quite spot on with them
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    (Original post by teachercol)
    OCR Physics A Newtonian World 20/6/16 Unofficial mark scheme

    Usual disclaimers. These are just my answers worked through quickly this morning.
    They may contain errors, typos and omissions.


    Didn't get chance to see my guys when they came out but the word is a bit mixed. Some liked it; others didn't.
    Seems very mathematical to me - not much heat. hardly any 'explain' questions.
    Heavily loaded to mechanics.

    Q1 a) i) Gradient = acceleration of free-fall so same for both (1)
    ii) Area under graph = distance travelled.
    Ball loses energy in bounce so speed after is less than speed before
    so doesn't go so high (2)
    I'm not sure whether you can just use the graph for i)ii) and iii) by reading off v and calculating the area. I suspect not.
    b) i) suvat v^2=u^2+2as so v = (-) 5.8 ms-1 (1)
    ii) Impulse = FT = 1.2Ns = change in momentum
    so new momentum = 0.13 x 5.8 - 1.2 = 0.754Ns (up) so v = 3.4 ms-1 (2) (Doesn't match graph)
    iii) suvat so V^=U^2 +2as again so s=0.69m (1)
    Total: 7

    Q2 a) Omnia corpore ... OK - in English Every object continues in a state of rest or uniform motion in a straight line unless acted upon by a resultant force. (1)
    b) i) same magnitude; same type (2)
    ii) opposite directions; act on different objects (2)
    c) Nice diagram ...
    i) Mass per s = vol per s x density = 3.3E-4 x 25 x 1.0E3 = 8.25 kg per s (1)
    ii) R = mg + Fsin55 = 92 x 9.81 + (8.25 x 25)sin 55
    = 902.5+169 = 1071N (3)
    Total: 9

    Q3 a) i) CF (1)
    ii) G (1)
    iii) 5pi/4 = 3.93 rad (might be SF penalties here if left in pi) (1)
    b) i) Inverted parabola - max KE = 50 mJ (2)
    ii) 1/2 x 0.45 x v^2 = 50E-3 so v = 0.471ms-1
    iii) max v = 2 pi f A so F = 0.471 /( 2 x pi x 5.0E-2) = 1.50Hz
    so T = 0.67s (2)
    Total: 8

    Q4 a) i) g = GM/R^2 so M = 3.7 x (3.4E6)^2 / 6.67E-11 = 6.41E23 kg (2)
    ii) If double r, then g is 4x less so g= 0.925 Nkg-1 (1)
    b) i) Period squared is prop to radius of orbit cubed (1)
    ii) (7.7/30)^2 = (9.4E3/R)^3 so R^3 = 15.18 x 9.4E3^3
    so R = 2.33E4km (2)
    c) mv^2/r = GMm/r^2 so v^2 = GM/r if lower orbit
    if r decreases v will increase. (1)
    Total: 7

    Q5 Been a while since we had a binary question.
    a) i) F = GM1M2/(R1+R2)^2 (1)
    ii) F1 = M1v1^2/R1 = M1 (2 pi R1/T)^2 / R1 = 4 pi^2 R1 / T^2 (1)
    b) Centripetal force = grav force so same on each object
    M1 x (4pi^2) x R1 / T^2 = M2 x (4pi^2) x R2 / T^2
    so M1/M2 = R2/R1 (2)
    c) M1/M2 = R2/R1 = 3 R1+R2 = 4.8E12
    so divide 4.8E12 in ratio 1:3
    R1 = 1.2E12 and R2=3.6E12 (3)
    d) V1 = 2 pi R1 / T = 6.08E4 ms-1 (2)
    e) v1^2 = GM2/R1) so M2 = 6.65E31 kg so M1 = 3M2 = 2.0E32kg (3)
    Kepler's law of planetary motion isn't going to apply here to a binary system.
    Total: 12

    Q6 On to something a lot easier!
    a) As falls, grav PE -> KE
    when hits bottom, KE -> heat (2)
    b) 50 x mgh = mcdT
    50 x 0.025 x 9.81 x 1.2 = 0.025 x c x 4.5 so c=131 J kg-1 K-1 (4)
    c) assume - no air resistance so all PE ->KE
    assume - all KE - heating le3ad shot and no heat transferred to container (2)
    d) If double mass, double input energy but heat shared among 2x mas so same change in temp (2)
    Total: 10

    Q7 a) Ideal gas so all internal energy is translational KE
    KE is prop to Kelvin temp so internal energy is prop to Kelvin temp (2)
    b) i) PV = nRT
    1.0E5 V = (80/0.004) x 8,81 x (21+273) so V = 489 m^3 (3)
    ii) n = PV/RT = 1.3E3 x 1.4E4 / (*.81 = (273-40) = 8677
    so need to lose 20000-8677 = 11300 moles (2)
    Total: 7

    So there we have it. Good for maths experts; not so good for the medicos.

    Good Luck Col
    For 2cii) isn't the force of water upwards? So mg=R+169 therefore R=mg-169? Or have I missed something?
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    (Original post by ComputerMaths97)
    Way to scare everyone
    Yeah probs going to delete the post as it's not well timed... Personally I found it to be more difficult than previous papers but I'm just judging by the reaction on twitter tbh.
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    (Original post by teachercol)
    OCR Physics A Newtonian World 20/6/16 Unofficial mark scheme

    Usual disclaimers. These are just my answers worked through quickly this morning.
    They may contain errors, typos and omissions.

    Didn't get chance to see my guys when they came out but the word is a bit mixed. Some liked it; others didn't.
    Seems very mathematical to me - not much heat. hardly any 'explain' questions.
    Heavily loaded to mechanics.

    Q1 a) i) Gradient = acceleration of free-fall so same for both (1)
    ii) Area under graph = distance travelled.
    Ball loses energy in bounce so speed after is less than speed before
    so doesn't go so high (2)
    I'm not sure whether you can just use the graph for i)ii) and iii) by reading off v and calculating the area. I suspect not.
    b) i) suvat v^2=u^2+2as so v = (-) 5.8 ms-1 (1)
    ii) Impulse = FT = 1.2Ns = change in momentum
    so new momentum = 0.13 x 5.8 - 1.2 = 0.754Ns (up) so v = 3.4 ms-1 (2) (Doesn't match graph)
    iii) suvat so V^=U^2 +2as again so s=0.69m (1)
    Total: 7

    Q2 a) Omnia corpore ... OK - in English Every object continues in a state of rest or uniform motion in a straight line unless acted upon by a resultant force. (1)
    b) i) same magnitude; same type (2)
    ii) opposite directions; act on different objects (2)
    c) Nice diagram ...
    i) Mass per s = vol per s x density = 3.3E-4 x 25 x 1.0E3 = 8.25 kg per s (1)
    ii) R = mg + Fsin55 = 92 x 9.81 + (8.25 x 25)sin 55
    = 902.5+169 = 1071N (3)
    Total: 9

    Q3 a) i) CF (1)
    ii) G (1)
    iii) 5pi/4 = 3.93 rad (might be SF penalties here if left in pi) (1)
    b) i) Inverted parabola - max KE = 50 mJ (2)
    ii) 1/2 x 0.45 x v^2 = 50E-3 so v = 0.471ms-1
    iii) max v = 2 pi f A so F = 0.471 /( 2 x pi x 5.0E-2) = 1.50Hz
    so T = 0.67s (2)
    Total: 8

    Q4 a) i) g = GM/R^2 so M = 3.7 x (3.4E6)^2 / 6.67E-11 = 6.41E23 kg (2)
    ii) If double r, then g is 4x less so g= 0.925 Nkg-1 (1)
    b) i) Period squared is prop to radius of orbit cubed (1)
    ii) (7.7/30)^2 = (9.4E3/R)^3 so R^3 = 15.18 x 9.4E3^3
    so R = 2.33E4km (2)
    c) mv^2/r = GMm/r^2 so v^2 = GM/r if lower orbit
    if r decreases v will increase. (1)
    Total: 7

    Q5 Been a while since we had a binary question.
    a) i) F = GM1M2/(R1+R2)^2 (1)
    ii) F1 = M1v1^2/R1 = M1 (2 pi R1/T)^2 / R1 = 4 pi^2 R1 / T^2 (1)
    b) Centripetal force = grav force so same on each object
    M1 x (4pi^2) x R1 / T^2 = M2 x (4pi^2) x R2 / T^2
    so M1/M2 = R2/R1 (2)
    c) M1/M2 = R2/R1 = 3 R1+R2 = 4.8E12
    so divide 4.8E12 in ratio 1:3
    R1 = 1.2E12 and R2=3.6E12 (3)
    d) V1 = 2 pi R1 / T = 6.08E4 ms-1 (2)
    e) v1^2 = GM2/R1) so M2 = 6.65E31 kg so M1 = 3M2 = 2.0E32kg (3)
    Kepler's law of planetary motion isn't going to apply here to a binary system.
    Total: 12

    Q6 On to something a lot easier!
    a) As falls, grav PE -> KE
    when hits bottom, KE -> heat (2)
    b) 50 x mgh = mcdT
    50 x 0.025 x 9.81 x 1.2 = 0.025 x c x 4.5 so c=131 J kg-1 K-1 (4)
    c) assume - no air resistance so all PE ->KE
    assume - all KE - heating le3ad shot and no heat transferred to container (2)
    d) If double mass, double input energy but heat shared among 2x mas so same change in temp (2)
    Total: 10

    Q7 a) Ideal gas so all internal energy is translational KE
    KE is prop to Kelvin temp so internal energy is prop to Kelvin temp (2)
    b) i) PV = nRT
    1.0E5 V = (80/0.004) x 8,81 x (21+273) so V = 489 m^3 (3)
    ii) n = PV/RT = 1.3E3 x 1.4E4 / (*.81 = (273-40) = 8677
    so need to lose 20000-8677 = 11300 moles (2)
    Total: 7

    So there we have it. Good for maths experts; not so good for the medicos.

    Good Luck Col
    Thank you. As with most students I have a few questions to ask so if you have the time I'd greatly appreciate your opinion.

    2) Would I lose 1 or 2 marks by saying R = mg - Fsin55 having got the right value for Fsin55.

    3) If I left it as 5/4pi will this lose the mark?

    6) For the uncertainties in question 6, will by stating something on the lines of "not all the pebbles behave the same way ie not all pebbles hit both sides of the tube at the same time, therefore are not all subjected to the same energy transfers" gain any credit for the second mark besides the obvious reasons?

    Lastly, grade boundary predictions please, sir
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    If this is right then I actually did better than I thought I did. Hopefully I can still get an A.

    Did anyone else think that was harder than the June 2014 paper?
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    (Original post by Asurat)
    Yeah probs going to delete the post as it's not well timed... Personally I found it to be more difficult than previous papers but I'm just judging by the reaction on twitter tbh.
    The most intelligent aren't on twitter. Let's be honest. They're revising for the next exam, or they're on TSR. People expecting a c or b (or even a low A) will be happy with todays paper as there was lots of free marks.

    I just feel that there was more easily-lost marks than usual.
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    (Original post by tombrunt45)
    In what way does Kepler's third not apply? I have not seen that answer with anyone I've asked (Q.5) I can't say I agree with your answer here, but thanks for the quick first efforts!
    it cant apply to a binary system as one star doesnt orbit another they both orbit a third object, not each other
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    (Original post by NickLCFC)
    If this is right then I actually did better than I thought I did. Hopefully I can still get an A.

    Did anyone else think that was harder than the June 2014 paper?
    In the June 2014 paper, done 2 days before the actual exam, I got 46/60.

    Considering I got 45-51 in this paper, I'd conclude it was considerably easier, considering my revising up to doing the June 2014 paper included notes from my teachers - which must've included stuff from that paper in some form.
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    (Original post by Tizzydag)
    it cant apply to a binary system as one star doesnt orbit another they both orbit a third object, not each other
    no there's no third object in this case they orbit around a fixed point (their CoM) that was shown in the question.
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    5/4pi difference is same as 3/4pi difference so should get the mark.
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    (Original post by TheFarmerLad)
    Thank you. As with most students I have a few questions to ask so if you have the time I'd greatly appreciate your opinion.

    2) Would I lose 1 or 2 marks by saying R = mg - Fsin55 having got the right value for Fsin55.

    3) If I left it as 5/4pi will this lose the mark?

    6) For the uncertainties in question 6, will by stating something on the lines of "not all the pebbles behave the same way ie not all pebbles hit both sides of the tube at the same time, therefore are not all subjected to the same energy transfers" gain any credit for the second mark besides the obvious reasons?

    Lastly, grade boundary predictions please, sir
    Probably one mark
    You might lose a mark for not calculating it out - but I expect not.

    Reasonable to say not all pebbles fall the same distance but what you said seems too vague to me and I wouldn't allow it.

    Sorry - no idea for this one.
 
 
 
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