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    (Original post by Random1357)
    A levels are very petty sometimes
    How many marks do you think that is? 3?
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    ive seen a few people who have 1+root5 in their period answer but I think you guys made the mistake of thinking that alpha and theta were the same angles if you see my sketch below

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    (Original post by gagafacea1)
    How many marks do you think that is? 3?
    2 you just had to differentiate so M1A1
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    Easiest way for wire question
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    Correct solution to MI via integration
    Attachment 554247554249
    Attached Images
     
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    How many marks lost if I did the cylinder with top and bottom? FML
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    (Original post by CharpCharlie)
    Easiest way for wire question
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    Correct solution to MI via integration
    Attachment 554247554249
    Please view my solution. The distance from CoM to pivot is actually root10/2 since CoM lies at 1/2a,1/2a
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    (Original post by 8752)
    ive seen a few people who have 1+root5 in their period answer but I think you guys made the mistake of thinking that alpha and theta were the same angles if you see my sketch below

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    Exactly what I did. Any idea how many marks I'd lose?
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    (Original post by ombtom)
    Exactly what I did. Any idea how many marks I'd lose?
    Did you think that theta=alpha? If so then maybe only 2 or 3 marks since the method was basically there. It took me a while to realise they weren't the same angles but hey-ho, some at my school made the same mistake - seems like this question caused trouble for many people on TSR. Either they thought that alpha=theta or they thought that the distance of CoM to pivot was root2 instead of root10/2
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    (Original post by 8752)
    Did you think that theta=alpha? If so then maybe only 2 or 3 marks since the method was basically there. It took me a while to realise they weren't the same angles but hey-ho, some at my school made the same mistake - seems like this question caused trouble for many people on TSR. Either they thought that alpha=theta or they thought that the distance of CoM to pivot was root instead of root10/2
    How many for finding COM?


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    (Original post by 8752)
    Did you think that theta=alpha? If so then maybe only 2 or 3 marks since the method was basically there. It took me a while to realise they weren't the same angles but hey-ho, some at my school made the same mistake - seems like this question caused trouble for many people on TSR. Either they thought that alpha=theta or they thought that the distance of CoM to pivot was root instead of root10/2
    I didn't even see an alpha :lol:
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    (Original post by ombtom)
    I didn't even see an alpha :lol:
    there wasn't an alpha in the question but you had to realise that the masses of the two wires didn't lie on the same line so they made different angles to the downward vertical (like they usually do for M5 pendulum questions - they're normally straight objects) because the wire was kinked. Normally in M5 questions - it's just a normal straight pendulum so the angle both masses make with the downward vertical are the same because they both lie on a straight rod or whatever, however here they aren't.
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    (Original post by physicsmaths)
    How many for finding COM?


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    I think probably 2. The rest of the method is all right except M0 on CoM method mark and A0 on accuracy mark of CoM
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    (Original post by physicsmaths)
    How many for finding COM?


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    None normally: mark for subbed in eqn
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    (Original post by 8752)
    The centre of mass of the object lies on 1/2,1/2 not on the line joining the 2 masses. use sum of moments=moments of sum using B as the origin

    m(0,a)+m(a,0)=2m(x,y)
    (x,y)=1/2a,1/2a----> this does not lie on
    the line joining the two masses on your diagram.
    Hmmm yes it does actually the only thing incorrect is the use of csq=asq+bsq-abcosC, instead of the correct csq=asq+bsq-2abcosC (=5/2) hmm embarrassing
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    (Original post by CharpCharlie)
    Hmmm yes it does actually the only thing incorrect is the use of csq=asq+bsq-abcosC, instead of the correct csq=asq+bsq-2abcosC (=5/2) hmm embarrassing
    Yes you're right it does lie on the line joining the two masses. The distance should be root10/2 a however. Im curious as to how people got root 2 tho?
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    my answers:
    i diddnt do m5
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    (Original post by gagafacea1)
    My answers:

    1) < 6 , 12.5 >

    2) < sin t , cos 2t >

    3)
    a) < 2 , 3 , 0 >
    b) < 0 , -2 , -3 >

    4) M/6 ( 3r^2 + 2h^2 )

    5) Same as Random1357

    6) u ln(5) - 19.6

    7) don't remember the J thing, but got mg/8 in the lastest part.
    They seem to pretty much match mine but I'm pretty sure that the answer to 1 was -6,12.5 as it was (-1,5)+(5/2)*(-2,3)=(-6,12.5) (going from memory)
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    (Original post by Ewanclementson)
    They seem to pretty much match mine but I'm pretty sure that the answer to 1 was -6,12.5 as it was (-1,5)+(5/2)*(-2,3)=(-6,12.5) (going from memory)
    yes it was, I forgot to add that minus, I'll edit it right now. Though I'm hoping your answer to 2 was different, cuz they asked for the velocity and I only put r. ie my answer to 2 was incomplete.
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    (Original post by gagafacea1)
    yes it was, I forgot to add that minus, I'll edit it right now. Though I'm hoping your answer to 2 was different, cuz they asked for the velocity and I only put r.
    Yes I gave them the velocity equation. I cannot remember what it was but it fitted the boundary conditions and when differentiated and subbed back in, gave the required value so I'm confident with it
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    (Original post by gagafacea1)
    yes it was, I forgot to add that minus, I'll edit it right now. Though I'm hoping your answer to 2 was different, cuz they asked for the velocity and I only put r. ie my answer to 2 was incomplete.
    Having checked the answer you agreed with for Q5, I do disagree. The centre of mass was a distance of root(10)/2 from the pivot.
    m(2a)+m(a)=2mx leads to x=1.5a then the centre of mass was 1.5a below L and by symmetry, 0.5a from l which gives root(10)/2.
 
 
 
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