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    for question 2 part b, i got 5.15N because the question was about the force on the scale pan, so i thought it would be:R-0.5g = 0.5 X 0.5which would give you 5.15 im not sure though im slightly confused :/
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    can i express answers in terms of g because i don't border to multiply it.
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    does anyone remember what the question 7 part b was? ive done part a and need to upload part b
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    (Original post by Jackhawkins21)
    My teacher reckons it will be
    A=60
    A*=66
    70=100ums
    I think thats quite low, considering the paper was a lot easier than last year, maybe 62-63 for an A?
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    (Original post by KloppOClock)
    does anyone remember what the question 7 part b was? ive done part a and need to upload part b
    resultant force on a pulley
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    (Original post by lai812matthew)
    can i express answers in terms of g because i don't border to multiply it.
    yes
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    Guys how many marks do you think I would have lost for the following:

    1c. I found t=32 but got something like 640i 320j as the final answer

    7a. I got the value of 3/2 for lambda but again got the wrong answer overall due to a careless error
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    (Original post by lai812matthew)
    resultant force on a pulley
    that was question 8, im on about the details of the question about 7b where you had to work out velocity of a vector at a time
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    How did u get 225 for bearing, isnt it 45 as tension was acting east and north so resultant would be in north east which is 45 as bearing.

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    (Original post by Major-fury)
    yeah that's fine you'll get full marks but could you help me with question 2 part b please? stating the method ty.
    Thanks

    For 2 part b what you had to do - it was worded very funny but you had to work out the R force from the scale pan upwards which was,
    Resolving upwards : R-1.5g = 1.5a
    R= 1.5g + 1.5a
    = 1.5*9.8 + 1.5 * 0.5
    = 14.7 + 0.75
    = 15.45N
    = 15.5N - 3 sig fig.
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    (Original post by Assmaster2)
    Thanks

    For 2 part b what you had to do - it was worded very funny but you had to work out the R force from the scale pan upwards which was,
    Resolving upwards : R-1.5g = 1.5a
    R= 1.5g + 1.5a
    = 1.5*9.8 + 1.5 * 0.5
    = 14.7 + 0.75
    = 15.45N
    = 15.5N - 3 sig fig.
    yeh i see what i did wrong by accident fk ... should ahve changed the mass
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    (Original post by KloppOClock)
    that was question 8, im on about the details of the question about 7b where you had to work out velocity of a vector at a time
    Yes, here it is (according to what you have labelled as Q7 on the first page):

    If the initial velocity is 3i - 22j and the acceleration is 3i + 9j, find the speed when t = 3s.


    For the section where we are asked the velocities of P and Q as a function of time, I remember it was something like this:
    P moves with velocity 15i + 20j and Q moves with velocity 20i-5j. At t = 0, P has position vector 400i and Q has position vector 800i. Find an expression for the position vectors of P and Q at time t.
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    (Original post by Maim56)
    How many marks was 8b

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    4 marks
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    (Original post by hadtosignup)
    Yes, here it is:

    If the initial velocity is 3i - 22j and the acceleration is 3i + 9j, find the speed when t = 3s.
    thanks
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    (Original post by neek101)
    for question 2 part b, i got 5.15N because the question was about the force on the scale pan, so i thought it would be:R-0.5g = 0.5 X 0.5which would give you 5.15 im not sure though im slightly confused :/
    You had to use the mass of the brick, not the steel thing. Or maybe you just read the masses the wrong way around?
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    did everything correct for the moments except i supidly did 6-d not 4-d....how many marks will i have lost you think?.....also anyone got the final question....I cant remember my answer
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    (Original post by suhaylpatel786)
    These are my answers:

    1(a) 104, (90+14)

    1(b) p = (400+15t)i + (20t)j
    q = (20t)i + (800-5t)j

    1(c) The 'j' vectors for both are equal, as it is due west.
    800-5t = 20t
    t = 32
    therefore q = 640i + 640j

    2(a) T = 20.6N
    2(b) 15.45 N => 15.5 N (3sf)

    3(a) Acceleration = -g/8
    velocity after rebound = 3.5
    thus Impulse = 3 Ns

    4(a)
    4(b) Area under graph = 975
    Area for slower car for first 25 seconds = 750
    975-750 = 225
    1/2 * b * 30 =225
    b = 15
    total time = 15+25 = 40
    so area under faster car = 975 = 1/2 * (40)(T+40)
    T = 8.75 s

    5) μ = 0.73

    6) For 1st situation, R(T) = 0
    For 2nd situation, R(S) = 0

    M(S) => 0.5M = 30d-15
    M(T) => M = 60 - 15d
    Therefore,
    d = 1.2m
    M = 42kg

    7(a) F2 = 2.5i + 2.5j (this required simultaneous equations)
    (b) V = 12i +5j, thus speed = 13 ms-1

    8(a) Fmax = 0.3g
    acceleration = 0.6g
    Tension = 11.76N = 11.8 N (3sf)

    8(b) RF = √(11.76)^2 + (11.76)^2 = 16.6
    It has a bearing of 225
    Hello, for question 3 I put 1.225 for acceleration but I was aware that it's actually a deceleration of -1.225 and I did get the correct answer. Do you think I'll lose one or two marks for that? All the other answers are the same as urs.
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    For question 1c I equated the i's instead of the j's! stupid mistake I know, but if I did it right do you guys reckon I'd get any marks?
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    If u get below an E (40ums) does that mean u get automatically 0 UMS (U) and it counts as if u've never done the exam?
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    (Original post by KloppOClock)
    IMPORTANT NOTE: After speaking with some maths teachers, they have said that they believe the grade boundaries will be relatively low due to the simultaneous vector question and question three! :banana2:

    *So people stop asking, when using 'g' 2 or 3 significant figures is fine.

    Question 1: (10 marks)
    Spoiler:
    Show
    1(a) 104, (3 marks)

    1(b) p = (400+15t)i + (20t)j (3 marks)
    q = (20t)i + (800-5t)j

    1(c) Q = 640i + 640j (4 marks)

    Model Answer Below
    Question 2: (6 marks)
    Spoiler:
    Show
    2(a) T = 20.6N (3 marks)
    2(b) 15N or 15.5N (3 marks)
    Question 3: (7 marks)
    Spoiler:
    Show
    3 Ns (7 marks)

    Model Answer Below
    Question 4: (12 marks)
    Spoiler:
    Show
    4(a) (4 marks)

    4(b) T = 8.75 s (8 marks)

    Model Answer Below
    Question 5: (10 marks)
    Spoiler:
    Show
    μ=0.73 or μ=0.727 (10 marks)

    Model Answer Below
    Question 6: (7 marks)
    Spoiler:
    Show
    For 1st situation, R(T) = 0
    For 2nd situation, R(S) = 0

    M(S) => 0.5M = 30d-15
    M(T) => M = 60 - 15d
    Therefore,
    d = 1.2m (4 marks)
    M = 42kg (3 marks)
    Question 7: (11 marks)
    Spoiler:
    Show
    7(a) F2 = 2.5i + 2.5j (this required simultaneous equations) (7 marks)
    (b) V = 12i +5j, thus speed = 13 ms-1 (4 marks)
    Question 8: (12 marks)
    Spoiler:
    Show
    8(a) Fmax = 0.3g
    acceleration = 0.6g
    Tension = 11.76N = 11.8 N (3sf) (8 marks)

    8(b) RF = √(11.76)^2 + (11.76)^2 = 16.6
    It has a bearing of 225
    or
    45 degree angle between horizontal plane and vertical rope
    or
    draw a diagram and annotate angles (4 marks)
    Model Answers: (click spoiler to show images).
    Lower than 59 last year? I think I got 61ish, do you think that's enough to get me an A?
 
 
 
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