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Edexcel Physics IGCSE 1P Unofficial Mark Scheme 25th May 2016 Watch

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    (Original post by Hsevras)
    Yes youd get all 3 marks for that

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    Yes I know that, I made the mark scheme, haha. The guy just wasn't sure so I was just explaining it to him.
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    (Original post by conradliebers)
    Yes I know that, I made the mark scheme, haha. The guy just wasn't sure so I was just explaining it to him.
    apoligies

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    (Original post by Hsevras)
    Diagonally to right, horizontally left, diagonally right

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    Thank you
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    Snow is a reflector of UV radiation - that's why you wear sun cream when you go skiing. Therefore the eyes of the polar bear were reflectors and the mouth and body of bear as well as sky were absorbers.
    The reason that refractive index has no units is because it is a ratio between the refractiveness of certain objects, however alone it is simply an arbitrary unit.
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    What about the question on which stage is the comet closest to the earth during its orbit? I got the first one (I think the options where stage 7, 8, 9 or 10)
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    (Original post by amkay)
    What about the question on which stage is the comet closest to the earth during its orbit? I got the first one (I think the options where stage 7, 8, 9 or 10)
    Thanks a lot! It was stage 9.
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    What if I put thermal energy for fission (instead of kinetic) because it heats the water in the reactor to steam that turns the turbine
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    (Original post by BOBQ)
    What if I put thermal energy for fission (instead of kinetic) because it heats the water in the reactor to steam that turns the turbine
    Ur point's completely valid but I think you'd lose a mark cos they're anal like that


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    [QUOTE=pk789;65248669]Ur point's completely valid but I think you'd lose a mark cos they're anal like that

    ah ok thx
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    Some please help with the last question…..what is the answer please…..is it same density or the fact that UV doesnt penetrate through?
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    I thought there was a lot in the exam for the time limit - I did all the past papers from 2011 and always had 20-30 mins at the end to go over my work. In this exam, I was still 2 questions short with 15 minutes to go, so I had to rush and panicked a little. Only had 1 couple of minutes at the end to go over my work. So not a good performance and expecting a lower grade than on all the past papers.
    Just hope the paper 2 is a little kinder as I need to do really well to rescue some of the errors from the first paper
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    Not seen any thoughts about what may be in paper 2.
    So will it cover areas not in paper 1
    Moments, Momentum, Static electricity, Sankey diagrams and efficiency, Transformers, C-K conversions, marsden and giegar experiment

    or can we expect to see questions like - A hunter shoots a polar bear, if the bullet in the gun has a mass of 25g and moves at a speed of 300 m/s what speed will the shooter recoil ? :-)
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    (Original post by pro567)
    Some please help with the last question…..what is the answer please…..is it same density or the fact that UV doesnt penetrate through?
    The hairs weren't dense enough to slow down the uv enough for total internal reflection to occur

    (I think)
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    Refractive index is wavelength dependent, so can only suggest that the hollow hair fibre does not have sufficient change in refractive index from the fibre to air for UV light to undergo TIR. Some truly terrible questions on this paper asking what is the examiner thinking.
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    (Original post by coedybrenin)
    Not seen any thoughts about what may be in paper 2.
    So will it cover areas not in paper 1
    Moments, Momentum, Static electricity, Sankey diagrams and efficiency, Transformers, C-K conversions, marsden and giegar experiment

    or can we expect to see questions like - A hunter shoots a polar bear, if the bullet in the gun has a mass of 25g and moves at a speed of 300 m/s what speed will the shooter recoil ? :-)
    DEFINITELY about the polar bears, or possibly some bees.
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    (Original post by Laith Selman)
    I'm pretty sure there was a question about variables. Something about light shining through a hole onto an LDR I think it was. Can anyone remember what they put down as the variables?
    Something like a student measuring the amount of light affecting the resistence on a light dependent resistor at a fixed length from the resistor, Used card with different sized holes and recorded results - what was the dependent and independent variables .
    Why is it important to keep the distance between the hole and the resistor constant
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    Will 90/120 A*
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    (Original post by Hsevras)
    Will 90/120 A*
    Borderline maybe. If you think you got that you should study hard for paper 2.
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    (Original post by escg)
    the exam was brutal, my god
    I did 1pr was literally the best test I've ever did. This mark scheme looks wayyyyy harder than ours
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    (Original post by conradliebers)
    Tell me the questions which I'm missing out - I'll add them in. Please try your best to remember the rough question number. Thanks for all the contributions, including Advait10.

    Do any of you remember which question they gave you a figure of 250000km or something like that. I'm trying to think of it but I can't remember. Tag me if you can remember it.

    Do the poll!

    1.a)
    Constant velocity with horizontal line. (1)

    1.b)
    Distance travelled is area under line. (1)

    1.c)
    Average velocity is distance moved / time taken. (1)

    2.a)
    Coolant to reduce energy.Shielding to absorb radiation.Fuel rods for Uranium fuel. (2)

    2.b)
    Kinetic energy. (1)

    2.c)
    The purpose of the moderator is to slow down fast moving neutrons to become slow moving thermal neutrons. This makes it easier for neighbouring uranium nuclei to absorb the neutrons and hence go under fission. (2)

    2.d)
    A controlled chain reaction where only 1 out of 3 (on average) of the neutrons emitted through fission are absorbed by neighbouring uranium nuclei. This is helped by using boron control rods to absorb these remaining neutrons. This prevents an uncontrolled chain reaction. The neutron which is absorbed by a uranium nucleus becomes unstable hence splits into 2 daughter nuclei and 2-3 fast moving neutrons. This process keeps repeating. (3)

    2.e)
    Draw week 6 position of complete. (1)

    2.f)
    X between comet and Earth. (1)

    2.g)
    The distance from 1-3 is is roughly the same. 3-4 is increase in displacement. 4-5 distance decreases. Time is the same so change in displacement = change in speed as s = d/t.

    2.h)
    Orbits are closet to each other at 9. (1)

    2.g)
    Orbital speed = 150,000,000 * 2 * pi = 94247796.
    1km365 * 24 = 8760so divide the first by the second to get 107588.7895km/hroughly equal to 108000km/h. (3)

    3.a)
    Pressure is P1V1 = P2V2 so: (101*1700)/12 = 14308.3kPa. (3)

    3.b)
    Pressure with depth = height x density x g. (1)

    3.c)
    113.08kPa. (3)

    3.d)
    113.08 + 101 = 214.08kPa. (1)

    3.e)
    Bubbles get larger as they rise due to the decrease in pressure with the decrease in depth as pressure = hpg, so air expands into larger volume. Also, smaller bubbles join together to form larger bubbles. (2)

    4.a)
    Gamma for tracing.X-rays for internal objects.Microwaves for food. (3)

    4.b)
    Particles carrying the energy (vibrations) travel perpendicular to the direction of the wave in motion. (1)

    4.c)
    You have to half the time because the time measured was for the wave travelling to the point of reflection and back as the sound wave travelled double the distance (to what you are measuring) so to get the depth from the point of emission to reflection, the time is divided by 2. (2)

    4.d)
    As v = frequency x wavelength, upon entering a denser medium the speed of sound increases as the wavelength increases. (2)

    5.a)
    Draw voltmeter parallel to LDR. (1)

    6.a)
    DC current is current that flows in one direction. (1)

    6.b)
    You can measure resistance by taking voltmeter reading (ensure it is constant) and ammeter reading and then do R = V/I (Ohms). (2)

    6.c)
    Graph. (4)

    6.b)
    The purpose for the split ring commutator and carbon brushes is so that once the coil goes past the half turn, when the coil retouches the commutator the direction of the current reverses hence magnetic field hence the effect of the force on the coil, ensuring it travels unidirectionally. (3)

    6.c)
    The coil continues to rotate clockwise. (1)

    7.a)
    Refractive index in glass is higher than air. YES
    If i = 0, the ray doesn't deviate. YES
    All rays that enter are totally internally reflected. NO (4)

    7.b)
    Draw point on graph. (1)

    7.c)
    It fits with correlation and gradient of graph as point is plotted above point following. (1)

    7.d)
    Refractive index can't be 0, only <1. (1)
    Critical angle can't be 0. (1)

    7.e)
    Critical angle was 41.82 so rounds to 42. (3)

    7.f)
    Refractive index is the simplest ratio between speed of light in air: speed of light in material medium. (2)

    7.g)
    Draw rays and arrows. (4)

    8.a)
    Ruler. (1)

    8.b)
    Hooke's law: Measure length of spring using ruler. Add 1N mass, measure new length, subtract original length from this to get extension. Repeat this process for a range of masses. Repeat whole experiment x3 and average for increased reliability. Plot graph of extension vs load, if follows Hooke's law, the gradient line between points will be contrast as F = ke (extension is proportional to the load). (5)

    9.a)
    5.3 cm. (2)

    9.b)
    5.12cm (1)

    9.c)
    Differing results.
    String measurements are to 0.1 cm.
    Digital measurements are to 2 decimal places.
    String measurements are averaged.
    Digital only takes 1 measurement.
    String measurement subject to parallax error due to human judgement.
    Calliper more accurate.
    Calliper measures diameter.
    String measures circumference. (4)

    10.a)
    P = IV (1)

    10.b)
    Max power = (4000*600)/1000000 = 2.4 MJ. (2)

    10.c)
    Work done = Force x distance

    10.d)
    76000000J of energy. (3)

    10.d)
    35.02 seconds. (2)

    10.e)
    Decreasing power increases time taken for the same amount of work done. (1)

    11.a)
    KE = 1/2 x m x v(squared). (1)

    11.b)
    Velocity = 12m/s. (3)

    11.c)
    GPE = mxgxh. (1)

    11.d)
    GPE is gained as the lift moves up from platform to surface hence increase in height hence increase in GPE. (2)

    11.e)
    Slope up causes transfer of KE to GPE hence decrease in velocity hence deceleration hence less work required by brakes. Slope down converts GPE into KE hence increase in speed hence acceleration hence less work required by motor. Energy is conserved. (4)


    12.a)
    Sky has lowest temperature and snow and bear have same temperature which is warmer. (2)

    12.b)
    Hair acts as an insulator by trapping in a pocket of air, hence preventing heat loss via convection as convection requires for the movement of gas particles. (2)

    12.c)
    Snow was best at reflecting, worst at absorbing. Sky and polar bear good at absorbing, bad at reflecting. (2)

    12.d)
    UV is not reflected by the sky, it is absorbed. (1)

    12.e)
    Black surfaces = good emitter, good absorbers, bad reflectors of radiation.
    White surfaces = good reflectors, bad absorbers, bad emitters of radiation.
    Hence with white fur on top, it increases the surface area of the fur exposed to the air, minimising loss of thermal energy via radiation. By having the black skin underneath, it limits heat loss via radiation and any heat that is radiated is reflected back by fur hence retaining heat. (4)

    12.f)
    UV waves don't reach skin as they are good absorbers, bad reflectors hence do not undergo total internal reflection hence doesn't reach skin. (3)
    as far as i remember .. in the last question they stated that the polar bear hair is like a tube with air in it.. so why tir dont take place..so i wrote that the medium is same for both .. for TIR to happen light must go from more dense to less dense medium... hair tubes were filled with air....

    hope i m right
 
 
 
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