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PHYA2 2016 resit unofficial mark scheme (in progress) pls add Watch

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    1) resultant of rock = 196N

    moments taken from rock

    1.2m*32g>0.8m*46g hence plank wont tilt

    2) 11.83ms for speed

    High point was 1.3m above starting point using suvat

    Horizontal distance before hitting floor was 13.9M done by using pythagorus using 11.83^2-5^2 then square root to find horizontal velocity then times by 1.3 seconds which was time of flight

    A was at bottom of arc and B was at the highest point

    3) This was the spring question these were the given properties;
    extension =0.2m
    distance travelled 1.2m
    mass=0.4kg

    A) Velocity of brick after leaving spring = 3.3ms (used KE=mv^2/2)

    B)friction force was 1.8N

    C)elastic potential.

    KE from elastic potential gave brick 2.2j.....but you need to add the energy lost from resistive force

    force was 1.8 and distance was 0.2 (extension of spring) so elastic potential = 1.8*0.2+2.2=2.56j

    D) Use E to find force produced by the spring

    2E/extension = F = 23.8N (no1 else put this so probs wrong)

    4) the rocket question....

    A) sum total of rockets mass, then f=ma to find acceleration = A

    B)then the newtons laws part,

    the difference was the direction
    the similarity was the magnitude of forces

    Find if the rocket reduced speed to safe speed
    RF=670N
    Mass= 890?

    hence 670/890=A

    U=5.5
    safe V=3

    time =3s

    TA= speed decrease = 2.25

    5.5-2.25=3.25 so not enough? I think my mass was wrong or the other values please add

    5) if this was the 6 marker?

    Graph showed acceleration where the driving force was greater than the air resistance and frictional forces, where the graph straightens is where the truck hits terminal velocity, 0 resultant force, driving force= air resistance + friction

    Your graph showing STEEPER gradient and HIGHER end velocity. Greater acceleration as F=ma so F/m=a so the lower the mass of the truck the more acceleration it has.

    Higher end velocity as although air resistance stays the same, the frictional force doesnt. F=μR and as Reaction is lower then friction will be lower (provided μ stays the same)

    6) Define a progressive wave- Not actually sure on this, I put
    A wave that transfers energy along the path of the waves trajectory, without transferring mass (probably wrong)

    Tick 2 boxes, I ticked longitudinal and can interfere

    then you needed frequency which i think was time = 44ms so 1/t =f f=22.7 round down to 22hz

    wavelength I think was 0.08m

    amplitude was 0.6

    then velocity which was lambda*f= 1.82ms

    7) A.) What colour would you see first as you turn to first order? (Violet)

    as violet smallest wavelength so least diffraction and microscope looked in to out

    B.) What angle is the first order at? (dsin(theta)=lambda)

    C.) How many maxima will there be? (d=n(lambda) gave 3.39 so 3)

    8) refractive index diamond (a) 24.4 (b) 18.5 (c) draw the path (d) IR speed in core = 4.65x108 (e) wavelength in core 2x10-6 mm

    Question regarding material with higher n than diamond, it does reflect better because sinc=n2/n1 hence angle is smaller so more waves will reflect and less will refract through due to smaller angle available



    I realise there are lots of gaps...anyone who has better versions of the later questions post them up......Its hard to recreate paper when you panicked for the last 30 minutes
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    Heres help with the 6marker i think.

    Graph showed acceleration where the driving force was greater than the air resistance and frictional forces, where the graph straightens is where the truck hits terminal velocity, 0 resultant force, driving force= air resistance + friction

    Your graph showing STEEPER gradient and HIGHER end velocity. Greater acceleration as F=ma so F/m=a so the lower the mass of the truck the more acceleration it has.

    Higher end velocity as although air resistance stays the same, the frictional force doesnt. F=μR and as Reaction is lower then friction will be lower (provided μ stays the same)

    obvs my qwc isnt great here but i think thats the correct info
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    My numerical answers in the order they were in the exam without any x10^ (and whether they agree with your answers):

    196 (agrees with you)
    11 (kind of agrees...)
    1.3 (agrees)
    14 (agrees)
    0.206 (agrees)
    2.86
    3.32 (agrees)
    0.18 (likely incorrect?)
    1.8 (agrees)
    0.6 (agrees)
    0.058 (dont know what this was)
    23 (dont know what it is, the frequency?)
    1.32 (dont know what this was)
    Sound waves longitudinal and interfere
    17 (dont know)
    3 (agrees)
    24.4 (agrees)
    18.5
    19
    8.4
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    (Original post by WillRose)
    Heres help with the 6marker i think.

    Graph showed acceleration where the driving force was greater than the air resistance and frictional forces, where the graph straightens is where the truck hits terminal velocity, 0 resultant force, driving force= air resistance + friction

    Your graph showing STEEPER gradient and HIGHER end velocity. Greater acceleration as F=ma so F/m=a so the lower the mass of the truck the more acceleration it has.

    Higher end velocity as although air resistance stays the same, the frictional force doesnt. F=μR and as Reaction is lower then friction will be lower (provided μ stays the same)

    obvs my qwc isnt great here but i think thats the correct info
    I will add it bare with, do you know if you actually had to draw on the graph? I know it said you can but dont think it was required (he says hoping)
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    For the newtons question, what was the object they were asking about,,,,is my answer wrong was it not even in equilibrium hence 3rd law wont apply? damn i dropped more marks i think
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    (Original post by JamieOH)
    My numerical answers in the order they were in the exam without any x10^ (and whether they agree with your answers):

    196 (agrees with you)
    11 (kind of agrees...)
    1.3 (agrees)
    14 (agrees)
    0.206 (agrees)
    2.86
    3.32 (agrees)
    0.18 (likely incorrect?)
    1.8 (agrees)
    0.6 (agrees)
    0.058 (dont know what this was)
    23 (dont know what it is, the frequency?)
    1.32 (dont know what this was)
    Sound waves longitudinal and interfere
    17 (dont know)
    3 (agrees)
    24.4 (agrees)
    18.5
    19
    8.4
    Thanks man, yeah the 0.058 sounds very familiar but cannot remember what it was for!!
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    (Original post by philo-jitsu)
    I will add it bare with, do you know if you actually had to draw on the graph? I know it said you can but dont think it was required (he says hoping)
    no wasnt required i dont think
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    (Original post by philo-jitsu)
    For the newtons question, what was the object they were asking about,,,,is my answer wrong was it not even in equilibrium hence 3rd law wont apply? damn i dropped more marks i think
    You didnt have to draw on graph for 6 marker. As for rocket it was slowing down so there was a resultant force therefore the forces were not equal -> N3L does not apply.
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    Pretty sure one question was the wavelength of light in the core is 1300nm, what is the frequency of light in the core or something? Refractive index of core = 1.57?
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    (Original post by WillRose)
    Pretty sure one question was the wavelength of light in the core is 1300nm, what is the frequency of light in the core or something? Refractive index of core = 1.57?
    If that was a question (can anyone confirm) the answer would be 1.47x10^14 Hz i think
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    also sound isnt transverse its longitudinal, just spotted that in your MS, other ones correct
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    (Original post by WillRose)
    also sound isnt transverse its longitudinal, just spotted that in your MS, other ones correct
    cheers...thankfully i did it right in the exam!

    and yeah I remember the nanometer one but vaguely.....will have a think then add it
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    I've just blitzed the paper in 30 mins and done all the numerical stuff, will double check later

    Q1 - Moments (a) centre of mass (b) 196N (c) boy balances

    Q2 - Projectiles (a i) 11.84 ms is about 12 ms-1 (a ii) 10.73 ms-1 (b i) 1.3 m height (b ii) 13.9 horiz distance (b at top, a at bottom)

    Q3 - Rocket questions (a) 0.21 ms-2 (b) newtons law Q's (d) rocket will slow down, 670 force slows down enough

    Q4 springs (b i) velocity = 3.3 ms-1 (b ii) 1.83 yup (b iii) 0.183J

    Q5 6 marker - graph goes higher I think?

    Q6 waves - (b) (I) 0.6mm (ii) 0.08m (iii) 22.7 Hz (iv) 1.81 ms-1

    (c) higher frequency = less diffraction so lower amplitude and shorter spread

    Q7 difraction grating (a) tick violet (b) constructive interence question (c) 17.1 (d) n = 3rd order

    Q8 refractive index diamond (a) 24.4 (b) 18.5 (c) draw the path (d) IR speed in core = 4.65x108 (e) wavelength in core 2x10-6 mm
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    (Original post by MiniMan64)
    I've just blitzed the paper in 30 mins and done all the numerical stuff, will double check later

    Q1 - Moments (a) centre of mass (b) 196N (c) boy balances

    Q2 - Projectiles (a i) 11.84 ms is about 12 ms-1 (a ii) 10.73 ms-1 (b i) 1.3 m height (b ii) 13.9 horiz distance (b at top, a at bottom)

    Q3 - Rocket questions (a) 0.21 ms-2 (b) newtons law Q's (d) rocket will slow down, 670 force slows down enough

    Q4 springs (b i) velocity = 3.3 ms-1 (b ii) 1.83 yup (b iii) 0.183J

    Q5 6 marker - graph goes higher I think?

    Q6 waves - (b) (I) 0.6mm (ii) 0.08m (iii) 22.7 Hz (iv) 1.81 ms-1

    (c) higher frequency = less diffraction so lower amplitude and shorter spread

    Q7 difraction grating (a) tick violet (b) constructive interence question (c) 17.1 (d) n = 3rd order

    Q8 refractive index diamond (a) 24.4 (b) 18.5 (c) draw the path (d) IR speed in core = 4.65x108 (e) wavelength in core 2x10-6 mm
    Cheers dude very helpful, you by any chance have the points for each question?

    Also I thought my velocity looked wrong, I forgot about the extra zero in wavelength...has been corrected
 
 
 
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