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    (Original post by OturuDansay)
    Yes. I found the difference between 1 for A and 1 for B for t= 5 and t=25. Learn the "proper" method to do it tho lol because my method is confusing.

    Don't worry tho because there were a lot more people who didn't get it than whom got it
    oh thanks but did you get the right answer 👀
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    (Original post by dgkjhl)
    Did anyone else get 12.8ms^1 for the 7 marker on question 7?
    Ye i did
    I thinks its coz i made the vertical displacement zero but it was meant to hit the TOP of box so ment to be a metre
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    Does anyone remember what information we were given on Q7? Thank you.
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    my method was right for the last question but i made a stupid mistake early on and said t=10 rather than 12. the rest of the question I carried out with the right method but with the wrong value of t what. how many marks to you reckon i will drop out of 10 with this one mistake?

    thanks
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    (Original post by OturuDansay)
    Yes. I found the difference between 1 for A and 1 for B for t= 5 and t=25. Learn the "proper" method to do it tho lol because my method is confusing.

    Don't worry tho because there were a lot more people who didn't get it than whom got it
    t=12 seconds.
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    (Original post by Eddy253)
    I certainly did, 008 degrees
    I'm a bit late but I'm pretty sure it was 081 degrees as it seemed too big so I flipped the opp/adj fraction to ensure it was the most sensible answer and about 7.something degrees came out, confirming it was actually about 081

    note: there will still be a fair few method marks for getting that far.
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    (Original post by omarharoun3)
    For question 6, where you had to say 49 + Tsin50... how ****ed am i that i thought it was Tcos60...? if i did that for both parts, out of 8 how many would i lose???
    Either I made the same mistake or there was more than 1 tension Q... Anyway, that depends on how lenient they are in the MS.
    For instance, you might get up to 4 marks if it states in the MS for the examiner to use the student's values rather than the correct versions.This would be for method and going to solution only.

    Problem is, that is the most probable lenient scenario I can think of and AQA hasn't been that lenient on the Q's they've been asking (e.g. C1,C3,Chem1,Phy1,IT2 etc).
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    (Original post by gcsekid)
    Attachment 554175
    Ignore the pencil line
    How are people getting 43 newtons for T

    I've done 2 or 3 questions of this format and did it just like you did. Don't worry, I'm VERY sure your answer is correct.
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    (Original post by gcsekid)
    Attachment 554175
    Ignore the pencil line
    How are people getting 43 newtons for T
    Ahh, it might be the direction of the Tsin(60), which has caused the confusion. It's going downwards, rather than upwards like normal. Hence, R= 49+Tsin60. Damn that sucks.
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    (Original post by Conte'sBlueArmy)
    Doea anyome remember exactly what the 7 marker was on Q7? Like what were the numbers given?
    Initial velocity of the first part was 12 at an angle of 50 degrees. Length of box = 3m, height of box = 1m. Part 1 was find distance to the front of the box. Part 2 was find inital speed needed to reach back of box (needed to use answer to part 1, thats why i mentioned it), when the angle is the same.
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    (Original post by Colsop50)
    my method was right for the last question but i made a stupid mistake early on and said t=10 rather than 12. the rest of the question I carried out with the right method but with the wrong value of t what. how many marks to you reckon i will drop out of 10 with this one mistake?

    thanks
    If you got all the rest exactly right then i reckon 6 maybe... But you might be lucky and get 7 its very hard to say as there hasnt been a 10 marker before
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    (Original post by Thequickspark)
    I've done 2 or 3 questions of this format and did it just like you did. Don't worry, I'm VERY sure your answer is correct.

    Sorry to disappoint, but its wrong
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    (Original post by Eisobdxhsonw)
    For all those people crying about the last question, I too got it wrong as I used T=25 instead of T=5, but one thing it said in the question was that they are both on the SAME horizontal plane throughout their motion(same vertical distance apart), so if you're working out when they're parallel you don't have to piss around with all that gradient **** that you learn in C4, you simply have to work out when the vertical components for velocity are equal to one another (when t=5) and sub T=5 into your distance equation for both plane A and B and take them away from one another to get the distance between them, I might be wrong but the big misconception is that people failed to realise it stated that the planes stayed on their horizontal planes through
    SAME horizontal plane only means that they are at the same height, i.e. don't move up or down relative to the ground. The question specifically said "instant" at which the velocities are parallel, which straight away means there's only one value for t - which is 12. The big misconception is really that most people didn't realise that when two velocities are parallel, one is a multiple of the other but they are not necessarily the same magnitude. With your method, if you put t=5 into the expressions for both components of both velocities, you would find that the velocities are neither the same, nor multiples of each other, so could not possibly be parallel.

    The key knowledge that AQA were looking for here was that multiples of vectors are parallel, which is GCSE knowledge, not Core 4 (although it is used briefly again in Core 4 with vectors in i, j and k) and that is what made this a 10 mark question when everything else it was asking had come up in 6-8 markers in previous years.

    I'm not saying it wasn't a tough question, it was, but that is what they were testing for.
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    (Original post by jtebbbs)
    SAME horizontal plane only means that they are at the same height, i.e. don't move up or down relative to the ground. The question specifically said "instant" at which the velocities are parallel, which straight away means there's only one value for t - which is 12. The big misconception is really that most people didn't realise that when two velocities are parallel, one is a multiple of the other but they are not necessarily the same magnitude. With your method, if you put t=5 into the expressions for both components of both velocities, you would find that the velocities are neither the same, nor multiples of each other, so could not possibly be parallel.

    The key knowledge that AQA were looking for here was that multiples of vectors are parallel, which is GCSE knowledge, not Core 4 (although it is used briefly again in Core 4 with vectors in i, j and k) and that is what made this a 10 mark question when everything else it was asking had come up in 6-8 markers in previous years.

    I'm not saying it wasn't a tough question, it was, but that is what they were testing for.
    Also if you're thinking of the planes going from 'left to right' and up and down in 2D, that's wrong unfortunately. They go 'forwards and backwards (j), left and right (i)' in the same horizontal plane. The question could've asked about particles on a frictionless table and been the same.
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    (Original post by Colsop50)
    my method was right for the last question but i made a stupid mistake early on and said t=10 rather than 12. the rest of the question I carried out with the right method but with the wrong value of t what. how many marks to you reckon i will drop out of 10 with this one mistake?

    thanks
    Did the same thing haha. I'm sure will get some follow through marks.


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    I did what a lot of people did and made the velocities equal and getting 5 and 25 then using the R equations to get the position and doing Rb -Ra and then pythag to find the distance for each t=T and t=25, which was like 80.5m and 160.9m, I know this is wrong but will a get any marks here and there?


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    (Original post by wil_is_he)
    If you got all the rest exactly right then i reckon 6 maybe... But you might be lucky and get 7 its very hard to say as there hasnt been a 10 marker before
    ok thanks, I'm a further maths student and managed to do 6x4 = 32 instead of 24 :-(
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    (Original post by Olsmarto)
    I did what a lot of people did and made the velocities equal and getting 5 and 25 then using the R equations to get the position and doing Rb -Ra and then pythag to find the distance for each t=T and t=25, which was like 80.5m and 160.9m, I know this is wrong but will a get any marks here and there?


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    It would be extremely harsh if they didn't award marks for this method. If the A grade boundary is very low, this will be why.
    But there will be at least 6M1 marks at a guess, but really depends on the mark scheme. It could be really linear :/
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    (Original post by OturuDansay)
    It would be extremely harsh if they didn't award marks for this method. If the A grade boundary is very low, this will be why.
    But there will be at least 6M1 marks at a guess, but really depends on the mark scheme. It could be really linear :/
    But the velocities were not equal, only parallel. You cannot find a value of t such that  \mathbf{V}_a=\mathbf{V}_b .
    However you can find value of k such that  \mathbf{V}_a=k\mathbf{V}_b where k is a scalar constant.
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    you need to think about the components of the velocities.

    I component/divided by J component leads directly to direction

    so this is the same when the velocities are parallel

    Still very hard for M1 and the AS students
 
 
 
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