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    p.s. sorry about my bad handwriting. I was rushing.
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    (Original post by jelmes96)
    Well, g is a constant, as 9.8, so having put 16g, won't matter. 16g is a number, it's just neat On the papers they use g as a constant, so why can't we? Anyway, I've had some problems, but it's uploading now.
    Bro, no joke. It said magnitude and in all past papers they require the exact number. Not trying to take the mick or anything :/
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    (Original post by Axion)
    Bro, no joke. It said magnitude and in all past papers they require the exact number. Not trying to take the mick or anything :/
    My teacher said it will be fine, and if it isn't, oh well. It's just one mark xD
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    Well, here it is guys...

    http://db.tt/STbgP2wY
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    (Original post by jelmes96)
    My teacher said it will be fine, and if it isn't, oh well. It's just one mark xD
    lol why does everyone always try and cite a teacher to back up their claims . Not saying you are lying, but in general when 80% of them are lying :P

    True :P, i've lost lots of one marks all over the paper!

    Thanks for the paper btw bro! you may have lost two marks on question 1 thoguh as you didn't separate the normal reaction force from below and above into 7g and 9p respectively

    Repped you
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    For question 4, xi = yj (i.e 2i = 2j) if the resultant is 045 degrees right? So (16-t^2) = (31-8t) which gives 0=t^2-8t+15. Well if you solve this quadratic, you get to positive values, however the answer is looking for one? Have I done something wrong, or is the question worded incorrectly?

    Couldn't do any of 7 or 8 after part iii.

    LOL

    *cries*

    Also, and this honestly isn't an excuse for my sub-C grade, our teacher for Mechanics has been absolutely awful, and everyone in my class who left that exam was either crying, shouting or just like YOLO. If there is a substantial difference with our class' results and everyone else, will something be done? We're considering making a formal complaint. We weren't taught 3D vectors, tension or F=ma stuff whatsoever.

    According the the unofficial mark scheme, I got less than 50%, that's for sure.
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    (Original post by jelmes96)
    Well, here it is guys...

    http://db.tt/STbgP2wY
    Name:  IMG_0007.jpg
Views: 432
Size:  368.6 KB
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    wasnt one of the forces for part 5 -10newtons?
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    (Original post by ryanm95)
    For question 4, xi = yj (i.e 2i = 2j) if the resultant is 045 degrees right? So (16-t^2) = (31-8t) which gives 0=t^2-8t+15. Well if you solve this quadratic, you get to positive values, however the answer is looking for one? Have I done something wrong, or is the question worded incorrectly?

    Couldn't do any of 7 or 8 after part iii.

    LOL

    *cries*
    you are correct, but if you sub the values 3 and 5 back in, you get one set of positive is and js with 3 and one set of negatives is and js with 5

    hence you take 3 as the answer
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    (Original post by Axion)
    you are correct, but if you sub the values 3 and 5 back in, you get one set of positive is and js with 3 and one set of negatives is and js with 5

    hence you take 3 as the answer
    Well s**t
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    (Original post by JG1027)
    wasnt one of the forces for part 5 -10newtons?
    I got 10 and 50, 10 if it's going right, 50 if it's going left. The acceleration didn't specify a direction, so either could be positive I thought.
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    (Original post by jelmes96)
    Well, here it is guys...

    http://db.tt/STbgP2wY
    Looks good to me, though I think you need the angle in the triangle.

    44/72 for me


    Posted from TSR Mobile
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    For question 1, is it not 9g down above the block (weight of top two blocks), and 16g up below the block (reaction force of all three blocks)?
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    (Original post by ryanm95)
    For question 4, xi = yj (i.e 2i = 2j) if the resultant is 045 degrees right? So (16-t^2) = (31-8t) which gives 0=t^2-8t+15. Well if you solve this quadratic, you get to positive values, however the answer is looking for one? Have I done something wrong, or is the question worded incorrectly?

    Couldn't do any of 7 or 8 after part iii.

    LOL

    *cries*
    t^2 -8t +15 factorises to give (t-3)(t-5). So yeah you have two positive solutions but only when t=3 will the bearing be 045 degrees as when t=5 it is in a negative direction so doesn't have this bearing.
    Hope this helps
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    (Original post by ryanm95)
    I got 10 and 50, 10 if it's going right, 50 if it's going left. The acceleration didn't specify a direction, so either could be positive I thought.
    oh so am i wrong ?
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    for the angle question for question 7 werent we supposed to do 90- the tan thing to find that angle?
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    For q2 you used 20 as u when it's 15, in the paper it says uy is 15. I remember doing that exactly the same and reading it over and thinking oh god!
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    (Original post by jelmes96)
    Name:  IMG_0007.jpg
Views: 432
Size:  368.6 KB
    How did you get 40gsin(15) for the compnent of the weight down the slope whe I drew out the triangle I kept getting 40g/sin(15) then couldn't gett the question to work.
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    This was the WORST exam I have EVER done.
    E P I C L Y failed it. I didnt even bother doing question 7, and I screwed up most of question 8 and the rest of the questions aswell. I literally am so angry right now. I really wanted an A at AS, but now I can only get a freaking C at the most.
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    Anyone know what the grade boundary for an A is normally in ters of raw marks?
 
 
 
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