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    Will it really matter if it's push or pull for Q5, I had the same predicament so I tried to work out both cases as friction will always act in the opposite direction and the 1. 2N will still be resolved in the same way, so does it actually change things? Just depends which direction you assigned as positive, but I'm not sure : )
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    (Original post by 98Lauren98)
    Will it really matter if it's push or pull for Q5, I had the same predicament so I tried to work out both cases as friction will always act in the opposite direction and the 1. 2N will still be resolved in the same way, so does it actually change things? Just depends which direction you assigned as positive, but I'm not sure : )
    Think it does matter, if you're pushing then you'd resolve the 1.2N force down alongside the weight and if you're pulling, you'd resolve it up with the reaction. So for pushing it would be: R = 0.4g+1.2sinx and for pulling, it would be: 1.2sinx+R=0.4g (think the mass was 0.4kg anyway). Then friction would be altered as F=muR. I did pulling, hopefully they'll allow for either as I didn't see anything in the question which specified a certain way
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    (Original post by chloe-jessica)
    Think it does matter, if you're pushing then you'd resolve the 1.2N force down alongside the weight and if you're pulling, you'd resolve it up with the reaction. So for pushing it would be: R = 0.4g+1.2sinx and for pulling, it would be: 1.2sinx+R=0.4g (think the mass was 0.4kg anyway). Then friction would be altered as F=muR. I did pulling, hopefully they'll allow for either as I didn't see anything in the question which specified a certain way
    Ok, Thanks. I did push but was so ambiguous 😮
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    Name:  IMG_0462.JPG
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Size:  253.5 KB Somone correct me if I'm wrong but the friction acts in the same way as the motion is to the right (the way I've drawn it), therefore it's crucial whether its push/pull as you resolve the 1.2N differently in each case.

    Need to find the actual paper to make sure the wording was indeed ambiguous.


    (Original post by chloe-jessica)
    Think it does matter, if you're pushing then you'd resolve the 1.2N force down alongside the weight and if you're pulling, you'd resolve it up with the reaction. So for pushing it would be: R = 0.4g+1.2sinx and for pulling, it would be: 1.2sinx+R=0.4g (think the mass was 0.4kg anyway). Then friction would be altered as F=muR. I did pulling, hopefully they'll allow for either as I didn't see anything in the question which specified a certain way
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    (Original post by LukeBarnett)
    Name:  IMG_0462.JPG
Views: 285
Size:  253.5 KB Somone correct me if I'm wrong but the friction acts in the same way as the motion is to the right (the way I've drawn it), therefore it's crucial whether its push/pull as you resolve the 1.2N differently in each case.

    Need to find the actual paper to make sure the wording was indeed ambiguous.
    The question stated the angle was above the horizontal so the push force should act parallel to the pull force therefore giving you the same answer
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    In 5iv) I did something very strange.

    I wrote the acceleration was 3ms-2 UPWARDS. I specifically stated upwards, but didn't state anything for acceleration horizontally. I don't know if I will get anything for that or if I'll even get a method mark for that. What do you guys think?

    PS: 60 for an A? You're having a laugh, it was 55 last year so it will probably be 56 or 57 this year.
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    (Original post by Flather)
    The question stated the angle was above the horizontal so the push force should act parallel to the pull force therefore giving you the same answer
    Did it? I thought it just said "angle to the horizontal", in that case that makes much more sense and would indeed yield the same answers for both cases.
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    (Original post by LukeBarnett)
    Name:  IMG_0462.JPG
Views: 285
Size:  253.5 KB Somone correct me if I'm wrong but the friction acts in the same way as the motion is to the right (the way I've drawn it), therefore it's crucial whether its push/pull as you resolve the 1.2N differently in each case.

    Need to find the actual paper to make sure the wording was indeed ambiguous.
    Yeah that's what I'm trying to explain when I'm going on about push & pull. Going to hunt down my mechanics teacher tomorrow and demand an answer :lol: I can't remember the actual wording of the question so it doesn't help.
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    For the vt graph question

    For some reason I put 20km 4 hours (idk why) in part i)

    For parts ii) and iii) I used the correct method to find T and the distance but I used 4 hours instead of 5.

    Will I get ecf marks for parts ii) and iii) ?




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    (Original post by Flather)
    The question stated the angle was above the horizontal so the push force should act parallel to the pull force therefore giving you the same answer
    (Original post by LukeBarnett)
    Did it? I thought it just said "angle to the horizontal", in that case that makes much more sense and would indeed yield the same answers for both cases.
    No... R is different in both cases, and thus  \mu R is different.
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    (Original post by LukeBarnett)
    Push & pull do yield different final values however, so in this case it appears to matter.
    Only because they affect R. That's to do with the direction which was what came from confusion of the phrasing "above the horizontal" I think personally.
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    (Original post by LukeBarnett)
    Did it? I thought it just said "angle to the horizontal", in that case that makes much more sense and would indeed yield the same answers for both cases.
    That's what I was thinking - otherwise they would have to have been much clearer in their wording. Would have loved a diagram!!
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    (Original post by chloe-jessica)
    Yeah that's what I'm trying to explain when I'm going on about push & pull. Going to hunt down my mechanics teacher tomorrow and demand an answer :lol: I can't remember the actual wording of the question so it doesn't help.
    Apparently, the wording was "angle above the horizontal", in which case both the push/pull would yield the same answer, the pushing diagram I just drew is wrong in that case as its X degrees below the horizontal. fingers crossed
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    (Original post by LukeBarnett)
    Apparently, the wording was "angle above the horizontal", in which case both the push/pull would yield the same answer, the pushing diagram I just drew is wrong in that case as its X degrees below the horizontal. fingers crossed
    Your diagrams both have X above the horizontal no?
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    (Original post by AlexShelley)
    A spare M1 OCR June 2015 paper a teacher had. Sorry for what has been written on it by him. Images of the paper have been attached to this comment.
    Thank you!
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    (Original post by chloe-jessica)
    Your diagrams both have X above the horizontal no?
    If you look at the left one, the angle of 20 degrees is actually below the horizontal which means its pushing it down rather than pushing it up which above the horizontal would do.
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    (Original post by LukeBarnett)
    If you look at the left one, the angle of 20 degrees is actually below the horizontal which means its pushing it down rather than pushing it up which above the horizontal would do.
    Oh wait I get it, was having a stupid moment. Direction of the force is 20 below the horizontal. Luckily I did pull, made the diagram less cramped and generally look prettier because the number of forces acting was balanced on each side
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    Has anyone released an unofficial mark scheme yet?
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    Th friction is larger than the pulling force which leads people to think the frictional force will pull it the opposite direction (negative acceleration), but think about it, friction cannot actually move a particle, it can only slow it down, when it reaches 0ms^-2 the friction will stop acting in any direction as it is no longer fighting against any force in the opposite direction.
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    How did people do 4ii
    >.<
 
 
 

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