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    What does getting 2.4*10^12 and 7.2*10^12 score you out of 3 i think it was? And, for example, if I forgot to square root my suvat answer to get 3.5, will i get the mark for the height if i worked it out correctly using my previous answer? Also, what do you think 46/60 is in ums?
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    For Q5)b) I just basically showed that m1 is proportional to 1/r1 and same for m2, therefore m1/m2=(1/r1)/(1/r2) = r2/r1.... do I still get the marks?
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    (Original post by ComputerMaths97)
    For Q5)b) I just basically showed that m1 is proportional to 1/r1 and same for m2, therefore m1/m2=(1/r1)/(1/r2) = r2/r1.... do I still get the marks?
    you need to start your derivation from F1 equation, that's what the questions asks.
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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!
    If you got through the derivation of GMT^2 = 4 pi^2 R^3

    you'll find that you're assuming that R = distance between objects is the same as R = radius of orbit. That doesnt work here so you cant use that equation.
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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
    Why is it G not A?
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    Guess I got between 47-52 out of 60 If this is right of course!
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    Why would kepler's 3rd law not apply for 5e?

    http://www.astro.cornell.edu/academi...ler_binary.htm

    http://csep10.phys.utk.edu/astr162/l...ries/mass.html
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    (Original post by speed1✈️✈️)
    Guess I got between 47-52 out of 90 If this is right of course!
    full marks 60
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    (Original post by teachercol)
    If you got through the derivation of GMT^2 = 4 pi^2 R^3

    you'll find that you're assuming that R = distance between objects is the same as R = radius of orbit. That doesnt work here so you cant use that equation.
    could you adapt the equation and just say that R2=R-R1,
    or something? anything like that ahah
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    can you have A for 3a) ii)
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    (Original post by Jm098)
    can you have A for 3a) ii)
    Also what id like to know


    Posted from TSR Mobile
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    (Original post by lai812matthew)
    full marks 60
    Whoooops yeah 60
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    (Original post by Kevinwong)
    Also what id like to know


    Posted from TSR Mobile
    i wrote A as well
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    Looks like i got 45... Pushing for an A*? lol
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    (Original post by teachercol)
    If you got through the derivation of GMT^2 = 4 pi^2 R^3

    you'll find that you're assuming that R = distance between objects is the same as R = radius of orbit. That doesnt work here so you cant use that equation.
    I dont get it because I used R2 in my equation, It's only about the orbit of one object, and it orbits around a fixed point, of which the radius is R2(3.6x10^12). We know the radius then, we know the period, we know G why can't we use Kepler's third? Am I missing something?
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    Any grade boundaries predictions TeacherCol?
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    also @teachercol could you have for an assumption in the lead question that there is no time taken/lost between inversions of the tube? thanks
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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
    Please check 5e again, I'm fairly confident the answer is 1.0x10^33 (One of us must have made an error since we've used the same working)
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    (Original post by Spectral)
    Please check 5e again, I'm fairly confident the answer is 1.0x10^33 (One of us must have made an error since we've used the same working)
    yh im pretty certain it is this too
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    teachercol a few really quick questions if you have 1 min spare!

    1)Would I get a mark for saying "they occur at the same time" as one of the ways that paired forces are the same?
    2)For the KE and EPE graph, will I get any of the 2 marks for doing the inverted hyperbole (completely inverted from EPE, like on normal graphs) but forgetting that is must not start from 0?
    3)Due to getting the graph wrong, I then obviously got the max v wrong, and then got the f wrong, but still used all the correct methods. Would I get any of the 3 marks here (ecf) from the earlier max KE error? I feel I must get 1 from the f calculation for the method, but will it have the ecf? I doubt the calculation for the max v will have an ecf though. Any thoughts? Referring to question Q3)b) in it's entirety.
    4) For question 5 part b I think I just used the (correct) equation from above to show that M1 is proportional to 1/R1. I don't think I showed, but I implied, that this was the same for M2 (just seemed obvious in my head, I must've not thought it was necessary to specify something so obvious in my head). Therefore M1/M2 = (1/R1)/(1/R2) = R2/R1 as required - any of the two marks here?
    5) Question 6)d I got that the temp doesn't change but I think I just mentioned that the masses are cancelled in the specific calculation - do I get 1/2 or 2/2?

    Also how you feeling about grade boundaries?
 
 
 
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