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    Definitely No Arsey?
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    (Original post by samma1)
    hey, can you just remind me what question 2 was? i cant remember....
    A lift accelerating with a mass inside. Had to find tension for part A and Normal reaction of the mass for part B
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    Im thinking i got low 60s, hopefully an A
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    was μ meant to be given to 2sf or is 3sf ok?
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    If anyone knows the marks for each question could you please post them

    thanks
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    (Original post by KloppOClock)
    Message me if anything is wrong.
    Also, if anyone remembers how many marks each question was worth, please comment so I can add them on too.

    Question 1:
    Spoiler:
    Show
    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
    Question 2:
    Spoiler:
    Show
    2(a) T = 20.6N
    2(b) 15.45 N => 15.5 N (3sf)
    Question 3:
    Spoiler:
    Show
    3 Ns
    Question 4:
    Spoiler:
    Show
    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
    Question 5: (10 Marks)
    Spoiler:
    Show
    μ=0.73 to 2 significant figures
    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
    M = 42kg
    Question 7:
    Spoiler:
    Show
    7(a) F2 = 2.5i + 2.5j (this required simultaneous equations)
    (b) V = 12i +5j, thus speed = 13 ms-1
    Question 8:
    Spoiler:
    Show
    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

    For question 2) (b)

    I wrote the reaction force is equal to 15.45N.
    Buth then i wrote the force exerted on the steel pan is 15.45 - 1.5g, so i took away the weight of the brick, and got 0.75N.

    Will i stil get full marks because i wrote dwon that R=15.45N, though this was not my final answer?
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    can anyone remember question 8
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    (Original post by VA123)
    was μ meant to be given to 2sf or is 3sf ok?
    I give it to 2 sf because I used the value for g as 9.8 in my calculations when finding the weights of the mass which is also to 2 sf
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    For the magnitude of the force on the pulley I didn't do root T^2 + T^2 because i got confused when it said work out the resultant magnitude and it messed me up. I thought the resultant force was the root of ((T-friction)^2 + (weight - T)^2)

    Was this stupid of me? I thought 11.47^2 with an angle of 45 from the horizontal was too simple for 4 marks.
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    So on the last question i completely missed out the fact it said 'and direction' so i didn't say southwest or 225degs or anything but I did just out of chance draw the arrow in the right direction on the diagram that they provided. Does this get you any marks or no?
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    (Original post by KloppOClock)
    Message me if anything is wrong.
    Also, if anyone remembers how many marks each question was worth, please comment so I can add them on too.

    Question 1:
    Spoiler:
    Show
    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
    Question 2:
    Spoiler:
    Show
    2(a) T = 20.6N
    2(b) 15.45 N => 15.5 N (3sf)
    Question 3:
    Spoiler:
    Show
    3 Ns
    Question 4:
    Spoiler:
    Show
    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
    Question 5: (10 Marks)
    Spoiler:
    Show
    μ=0.73 to 2 significant figures
    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
    M = 42kg
    Question 7:
    Spoiler:
    Show
    7(a) F2 = 2.5i + 2.5j (this required simultaneous equations)
    (b) V = 12i +5j, thus speed = 13 ms-1
    Question 8:
    Spoiler:
    Show
    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
    Q1 10 marks
    2 is 6
    3 is 7
    4 is 12
    7 is 11
    8 is 12
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    (Original post by KloppOClock)
    Message me if anything is wrong.
    Also, if anyone remembers how many marks each question was worth, please comment so I can add them on too.

    Question 1:
    Spoiler:
    Show
    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
    Question 2:
    Spoiler:
    Show
    2(a) T = 20.6N
    2(b) 15.45 N => 15.5 N (3sf)
    Question 3:
    Spoiler:
    Show
    3 Ns
    Question 4:
    Spoiler:
    Show
    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
    Question 5: (10 Marks)
    Spoiler:
    Show
    μ=0.73 to 2 significant figures
    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
    M = 42kg
    Question 7:
    Spoiler:
    Show
    7(a) F2 = 2.5i + 2.5j (this required simultaneous equations)
    (b) V = 12i +5j, thus speed = 13 ms-1
    Question 8:
    Spoiler:
    Show
    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
    I said North East for 8b get the marks ?
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    (Original post by KloppOClock)
    Message me if anything is wrong.
    Also, if anyone remembers how many marks each question was worth, please comment so I can add them on too.

    Question 1:
    Spoiler:
    Show
    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
    Question 2:
    Spoiler:
    Show
    2(a) T = 20.6N
    2(b) 15.45 N => 15.5 N (3sf)
    Question 3:
    Spoiler:
    Show
    3 Ns
    Question 4:
    Spoiler:
    Show
    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
    Question 5: (10 Marks)
    Spoiler:
    Show
    μ=0.73 to 2 significant figures
    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
    M = 42kg
    Question 7:
    Spoiler:
    Show
    7(a) F2 = 2.5i + 2.5j (this required simultaneous equations)
    (b) V = 12i +5j, thus speed = 13 ms-1
    Question 8:
    Spoiler:
    Show
    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
    Yo I think the very last question was not a bearing. So it would be 45 degrees
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    (Original post by kprime2)
    I'll be posting model answers tomorrow (edexcel)
    Thanks kprime2. Good man.
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    (Original post by marshmellow:))
    I give it to 2 sf because I used the value for g as 9.8 in my calculations when finding the weights of the mass which is also to 2 sf
    Ah ok, hopefully they'll allow it!
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    I gave μ=0.7, will this lack of significant figures make me lose marks?
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    could someone show the working for 5 bc i got 1.07 for μ and obviously im wrong but i have no idea what i did lmao ;;A;;
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    I completely guessed 7a) F=2.5i and forgot to write j component. Showed no working. How many marks could i get??
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    (Original post by Jackhawkins21)
    My teacher reckons it will be
    A=60
    A*=66
    70=100ums
    Its an AS module, you cant get an A*?
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    Guys,
    How did u do q 2b?

    It said find the force on the pan exerted by the brick, not the other way around.
    i got the answer to be 13.95 (or 14). Some other people also got this.
 
 
 
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