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    (Original post by LRLR1998)
    I got t equalled twelve too but everyone was telling me i was wrong and the answer was 5. Can you remember all the numbers in the question?
    I got that too! used 3 pages of working and took 20 mins to do the whole question though!
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    (Original post by LRLR1998)
    I got t equalled twelve too but everyone was telling me i was wrong and the answer was 5. Can you remember all the numbers in the question?
    You aren't wrong!
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    q8 for people that haven't done C4 yet!!Parallel velocities means that the gradient of the magnitude of the velocities are the same, so j/i=j/i for A and B's velocities. Do some rearranging, t^2 cancels, get t=12, use this in position vector to get 106
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    (Original post by LiesandAlibis)
    Quite a few people did, so we're all in the same boat here. I have a feeling it's right though!
    I got that too! Such a relief! was a very challenging question!
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    Can someone explain the Q7 seven marker, the Q6 tension part and the Q3 angle part to me? Thanks
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    Yep me got t=5
    Was pure chance i took it as 5
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    (Original post by QuantumSylar)
    yh i tried finding horizontal and got error sqrt(1.5^2 - 9.8^2)
    got error using x= sin-1(9.8/1.5) too
    yeah probly cus usually the hypotenuse is bigger than than the opposite
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    unofficial mark scheme

    1a. 0.4ms^-1 (3)
    b. 15KG (3)

    2a. p-4 i q+12 j (1)
    b. p=14 q=-17 (4)
    c. distance 17.9m (3)

    3a. accel= 1.5ms^-2 (3)
    b. velocty= 1.8ms^-1 (2)
    c. angle =9 (4)

    4a. V=12ms^1 (2)
    b. resultant= 90.8ms^-1 (2)
    c. bearing =008 (3)

    5a. accel= 3.92 (5)
    b. velcoity =2.50 (3)
    c. max height = 1.12m (4)

    6a. equation (2)
    b. Tension= 43.8m (6)

    7a. time= 1.76 (5)
    b. distance =13.6m (2)
    c. velocity to hit back of box =13.2 (7)

    8. time=35 seconds (10)
    distance =227m
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    (Original post by AQA-Disgrace)
    I got that too! used 3 pages of working and took 20 mins to do the whole question though!


    Yay I got this so haps
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    Was T 12 or 5
    If so how many marks would be lost using rong value of T but doing the rest right eg modulus
    I think A1 mark M2 marks would be lost
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    (Original post by Busted838)
    unofficial mark scheme

    1a. 0.4ms^-1 (3)
    b. 15KG (3)

    2a. p-4 i q+12 j (1)
    b. p=14 q=-17 (4)
    c. distance 17.9m (3)

    3a. accel= 1.5ms^-2 (3)
    b. velocty= 1.8ms^-1 (2)
    c. angle =9 (4)

    4a. V=12ms^1 (2)
    b. resultant= 90.8ms^-1 (2)
    c. bearing =008 (3)

    5a. accel= 3.92 (5)
    b. velcoity =2.50 (3)
    c. max height = 1.12m (4)

    6a. equation (2)
    b. Tension= 43.8m (6)

    7a. time= 1.76 (5)
    b. distance =13.6m (2)
    c. velocity to hit back of box =13.2 (7)

    8. time=35 seconds (10)
    distance =227m
    I think t=12s for the last one but the rest looks fine
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    (Original post by wil_is_he)
    q8 for people that haven't done C4 yet!!Parallel velocities means that the gradient of the magnitude of the velocities are the same, so j/i=j/i for A and B's velocities. Do some rearranging, t^2 cancels, get t=12, use this in position vector to get 106
    This is exactly what I did. Over the moon tbh as all the other aqa a2 exams I have done have gone shocking and I was dreading mechanics but I think its gone okay so hopefully that can boost my overallgrade
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    (Original post by QuantumSylar)
    yh i tried finding horizontal and got error sqrt(1.5^2 - 9.8^2)
    got error using x= sin-1(9.8/1.5) too
    oh i dont know though as i said i didn't know what to do but someone else doing the exam did it and got an answer
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    (Original post by LiesandAlibis)
    Quite a few people did, so we're all in the same boat here. I have a feeling it's right though!
    How many marks would you say getting time = 12 using that method would be?
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    Can anyone remember this 10 mark question? I'll have a go at it
    • Welcome Squad
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    Welcome Squad
    q8)
    particle a:
    v=u+at for i component
    v=4+0.2t
    particle b:
    v=u+at for i component
    v=6-0.2t
    these velocities will be the same because travelling parallel to each other horizontally
    4+0.2t=6-0.2t
    0.4t=2
    t=5 then sub it into the r=ut+1/2at^2 for each particle
    i can't remember any other numbers but there was 32.5 and 46.25 something like that.
    not sure if my answer is right but i got 80.6 in the end
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    i would've said it was a good paper if it wasn't for that last question LOL... general consensus for the last answer was 106m tho
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    (Original post by Busted838)
    unofficial mark scheme

    1a. 0.4ms^-1 (3)
    b. 15KG (3)

    2a. p-4 i q+12 j (1)
    b. p=14 q=-17 (4)
    c. distance 17.9m (3)

    3a. accel= 1.5ms^-2 (3)
    b. velocty= 1.8ms^-1 (2)
    c. angle =9 (4)

    4a. V=12ms^1 (2)
    b. resultant= 90.8ms^-1 (2)
    c. bearing =008 (3)

    5a. accel= 3.92 (5)
    b. velcoity =2.50 (3)
    c. max height = 1.12m (4)

    6a. equation (2)
    b. Tension= 43.8m (6)

    7a. time= 1.76 (5)
    b. distance =13.6m (2)
    c. velocity to hit back of box =13.2 (7)

    8. time=35 seconds (10)
    distance =227m
    2c0 was a displacement so wouldnt you also have to put an angle if you didnt leave it as I and J vectors
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    (Original post by Busted838)
    unofficial mark scheme

    1a. 0.4ms^-1 (3)
    b. 15KG (3)

    2a. p-4 i q+12 j (1)
    b. p=14 q=-17 (4)
    c. distance 17.9m (3)

    3a. accel= 1.5ms^-2 (3)
    b. velocty= 1.8ms^-1 (2)
    c. angle =9 (4)

    4a. V=12ms^1 (2)
    b. resultant= 90.8ms^-1 (2)
    c. bearing =008 (3)

    5a. accel= 3.92 (5)
    b. velcoity =2.50 (3)
    c. max height = 1.12m (4)

    6a. equation (2)
    b. Tension= 43.8m (6)

    7a. time= 1.76 (5)
    b. distance =13.6m (2)
    c. velocity to hit back of box =13.2 (7)

    8. time=35 seconds (10)
    distance =227m
    I'm sure 2c was displacement, not distance and 5c was 1.19 (to 3sf)
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    (Original post by postexamtalk)
    How many marks would you say getting time = 12 using that method would be?
    Full as long as you got the distance to be 106 m.
 
 
 
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