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

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    FOr the EM induction q which asked you to work out the current- for the area, you had to use Pi x r^2 , right?
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    11b) i said current down for a) and then using resistive force (opposite to motion) due to lenz law, the B field in out of the page, is this correct? I had seen this in a textbook and this is why i thought it was 2 marks
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    (Original post by BP_Tranquility)
    FOr the EM induction q which asked you to work out the current- for the area, you had to use Pi x r^2 , right?
    Yep and you should get a maximum current of 4.1A i think.
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    (Original post by Edac)
    2.48*10^-5 , think it was joules ?
    2.49 ×10-28kg?
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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
    Wasnt Hz 18,000 rounded more?


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    Id the capacitor time dichRge in a very unusual mathematical way which i think is right. I will post it later on. My tecaher said it is right but where the examiner understands it or not would be different.


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    (Original post by jbrown2)
    11b) i said current down for a) and then using resistive force (opposite to motion) due to lenz law, the B field in out of the page, is this correct? I had seen this in a textbook and this is why i thought it was 2 marks
    I did exactly the same, for the same reason as you. But it's pestering me as I don't think the question was linked to electromagnetic induction at all.
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    (Original post by daniel666545)
    So for question 17 (a):

    Can anybody tell me what they got for the constant E0? So what is the Permittivity of free space in base units? Since in the formula sheet is says: 8.85 x 10^-12 F m^-1, i said: F = m a = kg x m/s^2, therefore kg x m^-1 x m/s^2 which is kg/s^2.

    Would be great if you can tell me if that is correct!
    Stupidly thid is what I did. It was completely wring.


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    Guys what would 65-66 marks be in ums? Need an A for uni
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    (Original post by Bishoy)
    Guys what would 65-66 marks be in ums? Need an A for uni
    Probably a high A or low A*- so about 105-110 ums
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    (Original post by Bishoy)
    Guys what would 65-66 marks be in ums? Need an A for uni
    scroll down to bottom and download the edexcel mark converter, see if you can find an exam with similar difficulty and look how many ums 66 marks would be
    http://qualifications.pearson.com/en...nd-grades.html
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    (Original post by Bishoy)
    Guys what would 65-66 marks be in ums? Need an A for uni
    Easily an A*. Last year was 62 for A* and this years was harder imo.


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    Is there an unofficial mark scheme by any chance?
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    Ugh need a B for uni that was such a rubbish paper. Failed.
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    is there a list of the questions and the marks they were worth?

    Or at least the list of Section A's multiple choice questions?
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    Did anyone else only just finish in time? I was working right to the end, so I didn't have time to check my answers. If I had I might have noticed that I used electron mass instead of proton mass in the frequency question, and made a mistake working out the circular motion answer. I got the method right, I just did something wrong somewhere :/
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    So many papers were easier than this and still had quite low boundaries for an A grade, for example jan 2011 had a boundary of 57 for an A*, so even if they rise, don't think it will be a significant increase.
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    (Original post by VibraGenesis)
    is there a list of the questions and the marks they were worth?

    Or at least the list of Section A's multiple choice questions?
    Edexcel Physics Unit 4 (11th June 2015)
    Unofficial Mark Scheme
    UNIDENTIFIED QUESTIONS:
    Stuff from MC Questions:

    Acceleration of an electron: 3.5x10^15 ms^-2 (+1)

    Which question was the one about the difference between a uniform electric field and an electric field due to a point charge?
    you had to draw a diagram of the electric field between 2 parallel plates and of a point charge. and then compare the electric field strength of them. I.e. Point Charge: E (proportional) 1/r^2. And E = constant for the parallel plates. I drew the arrows pointing towards the positive charge! am i wrong? Yes will i get any marks? Unlikely. If it was 2 marks for the diagram, you may get 1 mark for having the lines equally spaced. But like the other person said… unlikely.
    Diagram:
    Question 1.
    Number of neutrons in a Uranium atom (Z = 92, A= 238) 238-92=146
    +

    Question 2.

    Question 3.
    The conclusion you can’t make from Alpha Particle Experiment = That Nucleus contains Protons and Neutrons.+5+1

    Question 4.
    Emission of electrons from heating a metal: thermionic emission (+2)+1
    Question 5.

    Force was F/2 of initial value. +9+1


    Question 6.
    Pendulum: T = mg + mv2/x +2+1
    Question 7.
    mv2=p2/m +1+1
    (Times through by m such that M2V2=P2)
    (Root Entire Thing)
    Question 8.
    not sure which q question but rate of change of momentum =kgms^-2 (+1)+1
    Question 9.
    The one about what you can use E =mc^2 +
    data in the back for:

    C Work out energy released in a proton anti porton anihalation +2+1
    Question 10.
    High energies required to
    make particles with large mass +1
    OR
    make proton collide +3 +1 -1 definitely not this!(E=mc^2, large energy ->large mass


    Question 11.
    (a) Draw the direction of current in the copper rod.
    Down. +7
    The diagram showed the current going anti clockwise, so at the copper rod, it would’ve been going down.

    (b) State the direction of magnetic field that will cause the rod to move to the right.
    Into the page. +6
    Fleming's Left hand rule current down, force right, hence mag field into page +1+2


    Question 12.
    Particle physics - identify pion minus quark composition
    (a) Arrangement of Pion
    (http://en.wikipedia.org/wiki/Pion)+1 Down anti-up +8

    (b) Convert 140MeV/c2 to kg
    2.49 ×10-28kg +10

    Question 13.
    T was about the boy in the merry-go-round thing. Find the minimum time taken for the boy to complete one revolution given that the boy's mass is 20 kg and the size of the frictional force is 0.35*(weight of the boy). The second part was: Explain how the time changes if a child of greater mass sits on the merry-go-round.
    (a)
    time = 3.03 +2
    Fcentripetal=Ffriction
    mr2=0.35mg
    2=0.35gr
    //ω = 2π/T
    4π2T2=0.35g0.8
    T=4π20.80.35g=3.03s

    (b)
    independent to mass +8
    F = μmg = mrω² therefore μg = rω² so ω independent of mass so T independent of mass

    you probably disagree - I read this one as to ask how the minimum time changes when a fat kid sits down. if so, would it not be true that the merry go round can withstand greater velocities, due to the greater frictional force, before throwing the boy off? or nah (despite the time period for any particular spin being independent of mass.)

    I don’t really understand what you mean..
    Do you mean: “the question is asking what will happen if an extra fat kid sit on the roundabout” ?

    Question 14.
    This was the 6 marks question about the probe in space. Discuss whether conservation of momentum and energy applies and why its speed increased.
    Space probe momentum question
    *Answer from previous MS - http://puu.sh/ikFJk/a2e6029c79.png

    +1
    TE = KE + PE. When KE increases, PE decreases. Idk. TE constant. Chemical energy transferred to heat and light?


    Question 15.
    (a)
    Permitivity of Free space in base units = A2·s4·kg−1·m−3 in SI base units, +1

    (b)Charge was 4x10^-7C Mass = 2.7g Distance between the two plates: 25cm
    θ = 41 +5
    , T=0.035? +5
    Tension = √[mg^2 + F^2]. Where mg = weight of one of the balls. And F = the electrostatic force between the balls (from coulombs law).
    Then θ = tan-1(o/a) = tan-1(mg/F) = 41°.


    Question 16.
    A question about capacitors. A circuit diagram of a battery and capacitor connected in series was given. The capacitance was 220 nF and the charge on each plate was 3.3 uC, with one having a plus sign and the other a minus sign.
    a) Calculate the voltage of the battery and the energy stored by the capacitor.
    b) Calculate the time taken for the capacitor to discharge 80% of its charge.
    c) Use and equation to determine whether the time taken for the energy stored in the capacitor is greater than or less than the time taken for the charge on the capacitor to half
    d) State two advantages of using a data logger to measure the voltage on the capacitor instead of a voltmeter and a stopwatch.
    a) V= Q/C = (3.3 x10-6)/(220 x10^-9) = 15v energy=2.475*10^-4 J +4
    R = 20MΩ?

    Isn’t Q 6.6x10-6? no that was the difference in charge, but you do not use this, so it is 3.3x10-6 C
    How many marks would I lose for using 6.6?


    b) Find the time for capacitor to discharge to 0.2 (0.8 through the capacitor)
    RC = τ = 20*106 * 220*109 = 4.4s
    Q/Q0= 5 (. Sorry, put Q and Q0 the wrong way around. It’s correct now.
    Therefore t = 4.4 ln(5) ≡ -4.4 ln(0.2) = 7.08 s. +4




    c) Because and C is constant
    Hence E = kQ2
    for E become 1/2 E
    Q need to become 1/√2 Q
    1/√2 ≈ 0.7 (≈0.689)
    Q reaches 0.7Q0 before reaching 0.5Q0
    Q reaches 0.7Q0 = Energy reaches half
    TEnergy reaches half < TCharge reaches half

    So the time is less for the energy to decrease by half
    +2

    d)
    Reduced/ gets rid of human error in measuring time (+2
    Means that many measurements can be taken - as voltage changes very quickly for a discharging capacitor. +1
    The data logger can plot a graph automatically when connected to a computer, and this graph is likely to be more accurate than the graph drawn be the student +3

    Question 17.

    Part 1: State what is meant by Faraday's law.
    Emf induced is directly proportional to the rate of change of flux linkage in a system
    ε= ᐃΦ/ᐃtI., so EMF is directly proportional to the change in flux linkage and indirectly proportional to change in time.
    Part 2: A diagram of a magnet hung vertically (with N at the bottom and S at the top) on a spring with it placed inside a copper ring was given. The question was: The magnet is displaced and then let go. It oscillates. Why does this produce an alternating current in the ring?
    Change in B field

    alternates as magnet oscillates up and down
    Rate of change of flux greatest when speed of magnet greatest
    meaning a maximum current


    Part 3: The given value was 0.035 Ts^-1 Which is (n(BA))/Time Therefore A = GIVEN 0.05^2*PI So EMF = 0.035*pi*0.05^2
    EMF = I(r +R) but as there is no internal resistance, r=0 so EMF = IR
    Therefore divive the emf/given resistance
    Gives Currrent = 4.01A +7
    I = 4.10 A ?
    Is it just me that thinks its 4.10 A? It could be.
    I can’t remember.. It’s something like that +1
    Fair enough
    I’m around 70% sure its 4.01
    its 41.0A because there were 10 turns in the wire
    I don't think we are supposed to count the number of turns from the graph..
    Anyone remember the Resistance?

    Question 18.
    (a) Diagram of a cyclotron- draw the the path that a proton would take
    (b) What direction is the magnetic field for the path y(u have drawn


    b)Mine spiralled outwards anticlockwise so said the magnetic field is into the page +2
    =Show that B = 2fme.
    From formula sheet:
    r = mV/BQ and v = ωr and ω = 2π/T = 2πf

    B = mV/Qr = mω/e = m2πf/el

    calculate the frequency of the alternating potential difference in the cyclotron
    when B = 1.2 mT-
    Answer = 18.3KHz (to 3 SF) (18,298 Hz) +3

    (c) Describe why the kinetic energy increases or something +1
    KE Increases between the plates because of alternating electric field. Electric field applies a force on proton, therefore, there is an acceleration, so velocity increases. So KE increases.(is this correct?) The magnetic field doesn’t do any work on the particle because the force acts at a right angle

    Question 19.
    Question: Describe and explain the events shown in Photograph?
    -Charged particle collision with proton created two particles, one charged and one neutral
    -Charged particle goes along path of X
    -Neutral particle travels from X to Y
    -Neutral particle leaves no track as no charge (only charged particles show in bubble chamber)
    -Neutral decays into matter/anti matter pair at Y as tracks are equal (radius of curvature) but in opposite directions (momentum conserved) so mass is same for both but opposite charges - Is it correct to say pair production here? I don’t think so. You can’t tell if there are particles and anti-particles. You can say they could be anti-particles but not definitely
    Well they have the same mass because momentum is the same no? Idk I explained how just two particles of the opposite charge are produced so hopefully that will be ignored.
    Also, pair production is two photo goes to two particles with mass? NOT TRUE lol what Oh okay fair enough hopefully i’ll still get the marks
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    In case you didn't see it:

    HERE is the link for the UNOFFICIAL MARK SCHEME which many have helped and collaborated on.

    Please help and the completed final version will be posted back here!
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    (Original post by MangleKuo)
    In case you didn't see it:

    HERE is the link for theUNOFFICIAL MARK SCHEMEwhich many have helped and collaborated.

    Please help and the completed final version will be posted back here!
    Great Idea !
 
 
 
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