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    (Original post by Windowswind123)
    Did quite a lot better in m3 than this haha which is ironic.
    The M3 paper was honestly easier than M2. (I still managed to do worse in M3... )
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    (Original post by TheMoon)
    I did the same thing, that question was so annoyingly worded!

    I reckon that I lost 3 marks to be honest, I would have got a couple for integrating I guess. Maybe I lost 4 marks since I didn't have much working.
    We would definitely get 2/5 because on last years paper for a similar question 1 mark was for attempting to integrate 1 mark was for correct integral and the rest were for using the correct limits and answer
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    (Original post by econam)
    Probably two. One for wrong limits and another for the accuracy mark.



    If you look in a lot of the mark schemes it often only credits answer to 2 or 3 sgf.
    I see what you mean, but usually these answers are ones that have been rounded from a very long decimal or a recurring decimal.
    Rounding 10.44 to 3sf seems almost pointless as the real value is only one extra decimal point long.
    Rounding something like 3.3723184021 to 3sf would be more important and I can see why you'd lose marks for it but not in this case.
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    (Original post by onur_m)
    How many marks will i get for 1b) for integrating between 0 and 3 and getting 15m as my final answer?
    They were probably aiming to catch people out like that so probably 0 :/
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    (Original post by dylandcd)
    They were probably aiming to catch people out like that so probably 0 :/
    Really? :/ even though I integrated and subbed in 3 for t?
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    Hate to ask one of those questions but...

    In question 1 I did the right method but somehow for to +11 when subbing back in so got the wrong constants and as such got the integral wrong for part b.

    How many do you think I'll lose?
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    How did you do Q5... mans perplexed


    (Original post by RThornton)
    My answers:
    1).
    a). q=-8, p=2, r=11
    b). 11/3m


    2).
    P=4900w
    R=784N
    b). d=53.1m


    3).
    c=10/3


    4).
    x=0.83m below
    angle= 114 degrees


    5).
    d= 2.56m


    6). lambda= 30/7 (4.29)
    speed= 10.4m
    angle= 73.6 degrees to the vector i (horizontal)


    7). A= 3u/20 (left)
    B= 21u/10 (right)
    e=1/14
    no collision as B after wall moves at the same speed as A, hence never meet
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    In the work-energy question I forgot to square 12.5 in my equation so I ended up with d=4.25, but ik that the rest of the method is correct. How many marks do you think I will lose for this? Also for c o mass i got 0.99 and 114.something.
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    (Original post by onur_m)
    Really? :/ even though I integrated and subbed in 3 for t?
    Depends on how many the question is out of I guess, you'll get the first couple of marks
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    My answers are :
    1. a) acceleration = 4
    b) distance = 23/3 (but I was silly and integrated between 3 and 4 so this is wrong)
    2. a) R= 80g
    P= 500g
    b) Distance = 53.1 m
    3. c= 10/3
    4. a) Distance of centre of mass = 0.830 m
    b) Angle made = 114 degrees
    5. a) Reaction= (5g/4)(1-d(cos^2(theta)))
    Frictional force= (5g/4)d(sin(theta))(cos(theta))
    b) d= 169/66
    6. a) lambda = 30/7
    b) velocity = sqrt(109) = 10.44
    c) direction of travel: 73.3 degrees below the horizontal
    7. a) velocity of A after collision = 21/10
    velocity of B after collision = 3/20
    b) e = 1/14
    c) No collision (I put that B was moving slower than A but apparently they were moving at the same speed)
    don't necessarily trust these though!!! I'm just some guy who did the exam, not qualified in any way!! feel free to correct/mock me
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    (Original post by dylandcd)
    Depends on how many the question is out of I guess, you'll get the first couple of marks
    Hopefully!
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    For 6b) for the angle I got 73.6 degrees and then added 90 degrees, and wrote in a sentence 163.6 degrees from the j axis. Would I lose a mark?
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    I got both values of d different to you ...
    (Original post by Cong)
    How did you got e in q7
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    (Original post by onur_m)
    Hopefully!
    I reckon we'll get 2/5 - i'm optimistic they might give 3/5 due to the wording of the question that could catch people out - but i'd just assume 2.
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    (Original post by omar8888)
    How did you do Q5... mans perplexed
    Set up equations for the vertical reaction force and the frictional force, R and uR respectively.

    By taking moments around A we get;

    5gcos theta d = 4P
    P = (5gcos theta d)/4

    Now resolving upwards to get R:

    R = 5g - Pcos theta
    subbing in P we get
    R = 5g - (5gcos^2 theta d)/4
    or R = 5g(1- (cos^2theta d)/4)

    Now for horizontal:

    uR = P sin theta
    = (5gcos theta sin theta d)/4

    And then we get given that the coefficient of friction is 1/2 and tan theta = 5/12

    Work out that cos theta = 12/13 and sin theta = 5/13

    Sub in your values for these into the equations above to get two different values for R and then set them equal

    You should end up with something like 330d/169 = 0.5g

    Which gives d = 2.56

    Sorry for poor explanation im in a rush
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    Ok. What the F**K was that exam?!


    Posted from TSR Mobile
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    (Original post by callum9977)
    I reckon we'll get 2/5 - i'm optimistic they might give 3/5 due to the wording of the question that could catch people out - but i'd just assume 2.
    yeah 2 is more likely, thanks!
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    (Original post by zandneger)
    Set up equations for the vertical reaction force and the frictional force, R and uR respectively.

    By taking moments around A we get;

    5gcos theta d = 4P
    P = (5gcos theta d)/4

    Now resolving upwards to get R:

    R = 5g - Pcos theta
    subbing in P we get
    R = 5g - (5gcos^2 theta d)/4
    or R = 5g(1- (cos^2theta d)/4)

    Now for horizontal:

    uR = P sin theta
    = (5gcos theta sin theta d)/4

    And then we get given that the coefficient of friction is 1/2 and tan theta = 5/12

    Work out that cos theta = 12/13 and sin theta = 5/13

    Sub in your values for these into the equations above to get two different values for R and then set them equal

    You should end up with something like 330d/169 = 0.5g

    Which gives d = 2.56

    Sorry for poor explanation im in a rush
    Yeah thats more or less right, but note that g is on both sides, and its 5 rather than 0.5
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    (Original post by fernk3)
    Ok. What the F**K was that exam?!


    Posted from TSR Mobile
    Mechanics 2
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    hahah, thank you! I got some very complicated expressions and got 1.94... just hope I get a few working marks. Cheers tho

    (Original post by zandneger)
    Set up equations for the vertical reaction force and the frictional force, R and uR respectively.

    By taking moments around A we get;

    5gcos theta d = 4P
    P = (5gcos theta d)/4

    Now resolving upwards to get R:

    R = 5g - Pcos theta
    subbing in P we get
    R = 5g - (5gcos^2 theta d)/4
    or R = 5g(1- (cos^2theta d)/4)

    Now for horizontal:

    uR = P sin theta
    = (5gcos theta sin theta d)/4

    And then we get given that the coefficient of friction is 1/2 and tan theta = 5/12

    Work out that cos theta = 12/13 and sin theta = 5/13

    Sub in your values for these into the equations above to get two different values for R and then set them equal

    You should end up with something like 330d/169 = 0.5g

    Which gives d = 2.56

    Sorry for poor explanation im in a rush
 
 
 
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