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    (Original post by partycooper)
    4) Since B is moving towards the ground its P.E is decreasing, and since A is moving up the slope its P.E is increasing.
    This means that, the P.E lost by the whole system = P.E lost by B - P.E gained by A
    = 2mgh - mghsin(alpha)
    We have used "hsin(alpha)" since we are only looking at the VERTICAL displacement of each particle when we consider change in P.E

    8) We know from part (c) that P is "instantaneously at rest" at t=8. What this actually means is that the particle is changing direction.
    By subing in t=8 into the displacement equation we get s=48 i.e. after travelling 48m the particle changes direction and starts to move back whe way it came. Since we are asked how far the particle has travelled after "10 seconds" we must now sub in t=10 which gives a displacement of s=44. However whilst this is the particle's "displacement from the origin" after 10 seconds, it is NOT the "total distance" the particle has travelled. The particle has travelled 48m forwards in the first 8 seconds and then 4m backwards in the following 2 seconds in order to get to the point where it is 44m from the origin. Hence it has travelled (48 + (48 - 44)) m = 52m
    Thank you for answering this....but in Q4 I dint understand b part where it's written
    1/2*m*v^2 + 1/2*2m*v^2 = ???????
    I dont understand the question marks part..where it came from...??

    I understood Q8 now
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    (Original post by cazzy-joe)
    Thank you for answering this....but in Q4 I dint understand b part where it's written
    1/2*m*v^2 + 1/2*2m*v^2 = ???????
    I dont understand the question marks part..where it came from...??

    I understood Q8 now
    Oops i accidentally explained part a instead of part b, i've edited it now but let me know if you're still not sure
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    (Original post by partycooper)
    Oops i accidentally explained part a instead of part b, i've edited it now but let me know if you're still not sure
    I understand noe
    Thanksss a lot..
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    I thought i was doing pretty well in preparation but i just did the jan 2011 paper and found it horrible! >.<
    Question 4, i haven't seen an impulse question like that before , also in part B, when i used Tan = vertical/horizontal. I found the smaller angle to the line AB but in the answer they had the larger angle, does that matter in exams? since they are both to the line AB aren't they?


    Also, for the last question can anybody tell me how part b was done? I don't get the mark scheme at all D:

    the raw grade to get an A was 53/75 so i guess a lot of people who did it also found it hard but this has made my confidence levels drop so much
    (not that they were high to start off with! i'm not the biggest fan of mechanics XD)
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  1. File Type: pdf M2_Jan 10.pdf (288.8 KB, 162 views)
  2. File Type: pdf M2_Jan 10_answers.pdf (116.8 KB, 104 views)
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    Anyone got grade boundaries for the past papers?
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    CRAP. Only done 2 past papers for this.
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    I've done 1 xD - got too many exams this week =\ . 91% first try, but I cheated myself a little bit.
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    (Original post by atofu)
    I thought i was doing pretty well in preparation but i just did the jan 2011 paper and found it horrible! >.<
    Question 4, i haven't seen an impulse question like that before , also in part B, when i used Tan = vertical/horizontal. I found the smaller angle to the line AB but in the answer they had the larger angle, does that matter in exams? since they are both to the line AB aren't they?


    Also, for the last question can anybody tell me how part b was done? I don't get the mark scheme at all D:

    the raw grade to get an A was 53/75 so i guess a lot of people who did it also found it hard but this has made my confidence levels drop so much
    (not that they were high to start off with! i'm not the biggest fan of mechanics XD)
    For question 4, it matters which angle you give since the question asks for the angle that the direction of this impulse makes with "the original direction of motion of AB". So you have to do Theta = 180 - arctan(vertical/horizontal) But other questions might not be so specific so as long as you state which direction your angle goes in (e.g. 106 degrees in the positive x direction) then you'll be fine

    For 8(b) you need to use everything you've been told to substitute into the equation found in the previous part. Since it strikes the ground at A we know that y=0. We are given u=7 and since OA = R we can let x=R.
    This should give you: 0 = cR - ((4.9R^2)/49) = R (C - (R/10)) and Hence R=10c
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    hoping for an easy collisions question
    a centre of mass question which doesn't involve moments
    a hard projectile question
    standard statics question
    standard energy and power questions

    and has anyone seen this question where they take moments from a line in the M/S , but I've never seen it
    before. Was a centre of mass question

    question is June 05 number 2. part B
    can someone explain this to me.
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  3. File Type: pdf M2 2005-06.pdf (143.3 KB, 235 views)
  4. File Type: pdf M2 2005-06 MS.pdf (508.8 KB, 610 views)
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    (Original post by CapsLocke)
    hoping for an easy collisions question
    a centre of mass question which doesn't involve moments
    a hard projectile question
    standard statics question
    standard energy and power questions

    and has anyone seen this question where they take moments from a line in the M/S , but I've never seen it
    before. Was a centre of mass question

    question is June 05 number 2. part B
    can someone explain this to me.
    I've never used moments like they have in the M.S for these kind of questions. This is what I did but if you were taught to use moments then maybe you should stick with that.

    If the frame hangs in equilibrium when suspended from the midpoint of BC then the C.o.M of the system must run through this same midpoint so x=3.5 for the overall C.o.M
    From part (a) the C.o.M of the frame (without the added mass of kM) was 3cm from AB.
    And for the mass kM which is attached at C, its C.o.M (without the frame) is 7cm from AB
    Using the C.o.M formula with x-coordinates only:
    M(3) + kM(7) = (M + kM)(3.5)
    Cancel 'M' and rearrange to find k = 1/7
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    If anyone has the past papers on them, im struggling to understand how to do
    Q2 (b) on the June 2005 paper.

    An explanation would be awesome
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    (Original post by Bav Singh)
    If anyone has the past papers on them, im struggling to understand how to do
    Q2 (b) on the June 2005 paper.

    An explanation would be awesome
    partcooper has the answer here:

    If the frame hangs in equilibrium when suspended from the midpoint of BC then the C.o.M of the system must run through this same midpoint so x=3.5 for the overall C.o.M
    From part (a) the C.o.M of the frame (without the added mass of kM) was 3cm from AB.
    And for the mass kM which is attached at C, its C.o.M (without the frame) is 7cm from AB
    Using the C.o.M formula with x-coordinates only:
    M(3) + kM(7) = (M + kM)(3.5)
    Cancel 'M' and rearrange to find k = 1/7
    Or you could take moments about the midpoint of BC.

    The horizontal distance from AB to C is 3.5. The horizontal distance to the COM of the frame is (3.5-distance from AB) = (3.5-3) = 0.5

    Then: 3.5*kMg=0.5*Mg
    k=1/7
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    What are the grade boundaries like for m2?

    Also when finding the centre of mass for an object what do you do when it gets folded? For example you fold the corner of a uniform rectangle.
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    i have only done two mock papers because they were the only ones i was given.
    could anyone post some links to some more mock papers? or email me them? (i have done jan 10 and may 09)
    thanks
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    (Original post by partycooper)
    For question 4, it matters which angle you give since the question asks for the angle that the direction of this impulse makes with "the original direction of motion of AB". So you have to do Theta = 180 - arctan(vertical/horizontal) But other questions might not be so specific so as long as you state which direction your angle goes in (e.g. 106 degrees in the positive x direction) then you'll be fine

    For 8(b) you need to use everything you've been told to substitute into the equation found in the previous part. Since it strikes the ground at A we know that y=0. We are given u=7 and since OA = R we can let x=R.
    This should give you: 0 = cR - ((4.9R^2)/49) = R (C - (R/10)) and Hence R=10c
    Ah thanks for the explanation! hopefully the grade boundaries for this exam won't be too high xD
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    In work-energy questions, suppose a particle weight 3g is moving down a slope of length 8m which makes 30 degree with horizontal and slope is rough.
    Total lost in energy is 80 J . Now we have to find frictional force.
    Why dont we consider component of weight i.e. 3g sin 30

    See attachment

    (Original post by partycooper)
    ...
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    (Original post by cazzy-joe)
    In work-energy questions, suppose a particle weight 3g is moving down a slope of length 8m which makes 30 degree with horizontal and slope is rough.
    Total lost in energy is 80 J . Now we have to find frictional force.
    Why dont we consider component of weight i.e. 3g sin 30

    See attachment
    We don't consider gravity because it isn't a 'resistance'. Gravity here is helping the particle down the slope as it is acting in the direction of motion. When energy is lost, it is only the forces which 'oppose motion' that need to be considered (In this case just Friction)

    So using: Work Done against resistances = Total loss of energy
    where, Work Done against Friction = Frictional force x Distance = 8F
    and, Total Loss of Energy = 80

    8F = 80

    Hope that helps
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    sub
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    Does anyone know what s.f. you should give answers to? Mark schemes sometimes say 2 and at other times say 3, so I'm not too sure...
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    Pretty worried for tomorrow, been focussing on my other exams too much D:

    For an A* in maths is it 90% across C3 C4 and M2 or just the core modules
 
 
 
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