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Mechanics M1 8th June - Unofficial Markscheme W/Explained Answers watch

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    (Original post by jess_12)
    And on the last question it didn't say give answer as a bearing?? Did you have to anyway, or do you think you'd still get it if you give it as a normal angle?
    It just said direction. In a past paper they asked the same and accepted a drawing of it too.
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    (Original post by fishs003)
    Agree with most.q1 - yes
    q2 - yes
    q3 - I got 3Ns. You had to work out the acceleration of the particle first using F=ma (I think a=-1.225) . The with that acceleration work out v of the particle using a SUVAT. Once you had worked out v (which was 3.5 I think) you could sub it into the equation I=m(v-u), hence get Impulse of 3Ns.
    q4 - yes
    q5 - yes
    q6 - yes - I got 1.2m and think a lot of others got 1.2m too.
    q7 - (a) I got part very wrong, and drew a triangle with hypotenuse of R, and managed to work out the angle opposite it. I then subbed all this (using magnitudes and Lamda and Beta as unknowns for R magnitude (so lamda root 10) and F2 magnitude (so beta root 2) into the cosine equation and got values of beta to be 6.something or 3.something (7 marks)
    q7- (b) yes (4 marks)
    q8 - yes. I thought theta was 45 degrees so Resultant Force was 16.6N (3sf). You also had to say what direction - I put bearing of 225, and also drew a little diagram and marked it on my large diagram.

    Obviously I could have and have done (like q7a) got some questions wrong.
    Thanks for the reassurance! Seems like you have got a very high mark, definitely a solid A, well done
    Also, thanks for explaining the impulse one, I know where I have gone wrong in question, I took the velocity as 4ms, for some reason, and so got an acceleration of -1.6ms^2 not -1.225ms^2 oh well.
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    (Original post by TrueDAN)
    Hey all - going to try my best to create a nice markscheme for this one, off memory, so if I struggle, all help is appreciated! I thought it was a standard paper, but tricky in some areas and chucked in some abstract questions - what do you think?!

    Thanks to KloppOClock for clarity on each question and for many values provided in the question - insanely good memory!

    Question 1:
    A nice vectors one - bit surprising to be the very first question but anyhow.
    Part a) The bearing was 104 degrees (nearest whole number)
    You had to do Arctan(5/20) + 90
    Part b) r = ro + vt for both parts.
    I believe Q = 20i-5j and P = ?
    Part c) West implies same J component and equating them you get T = 32 seconds
    and substitute this in!
    640i + 640j
    Question 2:
    - Triangular lift question. Mass brick = 1.5kg and lift = 0.5kg with a vertical acceleration of 1/2.
    Part a) You need to look at the entire system. T-2g = 2a
    Therefore T = 19.6 + 1
    Therefore T = 20.6N
    Part B) Just look at the brick.
    R-1.5g = 1.5a
    Therefore R = 1.5g + 0.75
    Therefore R = 15.45N

    Question 3 - Impulse Question ( I messed up!!)
    I got 3/8Ns somehow haha and used I = FT and found the force - must have done it incorrectly.
    The answer is 3Ns.
    Hardest question for me - I usually like impulse haha.

    Question 4: Speed time Graph
    Part a) 4 MARKS for a diagram was very generous. Make sure it is clearly labelled and that you clearly read the question.
    Part b) 8.75 seconds
    You could determine the total time via considering N:
    20x35 = 750 975-750 = 225
    0.5 x h x b = 225
    15b = 225
    b =15
    Therefore total time is 15 + 25 (provided in the question) = 40seconds.
    So you had to equate the area and you arrived that the total time = 40seconds.
    You then had to use this 40 seconds to determine T.
    40 x 40-t / 2 + 40xt = 975
    800 - 20t + 40t = 975
    20t = 175
    t = 8.75 seconds.
    I believe you found that it was constant for 8.75s at 40ms^-1 and then it decelerated for 31.25 seconds.
    And I double checked this and it does in fact = 975m

    Question 5: Incline plane
    - u= 0.727 or 0.73
    40cos30 = 2gsin20 + uR
    Determine the value of R and I believe it was 30 something.
    Proceed to solve and you get the value of u.
    The trickiest part was taking into account the fact that the force was pushing inwards against the plane.

    Question 6: Moments
    - Mass = 42 Kg
    You had to use the idea that if point about tilting, then there must be no reaction at the other pivot.
    - Distance - me and other people in my class each got different answers even though we all got the value for the mass to be the same hahaha.
    I got 0.7m, others got 0.8m and 1.2m
    Clarity on this one would be great! I'm assuming 1.2m

    Question 7: Vectors
    1. I managed to get this one I believe. You were told *lies in the direction of* - this basically implies that it is a *multiple* of the given vectors.
    F1 + F2 = R and we were told that R lies in direction of (?)
    You had to introduce 2 unknowns - I used psi and lambda and solve simultaneously. Quite nasty actually, edexcel really want to distinguish people - if you did not know how to do it, then that is 7 marks gone. Equate I and j components and you find lambda = 3/2 .
    Using this value, F2 = 5/2i + 5/2j

    2. Much more easier question, suvat with vectors.
    Use V= u +at
    V = 12i-5j
    Therefore speed = 13ms^-1
    Remember, it wanted speed which has no direction therefore it is the modulus of the velocity vector.

    Question 8: pulley
    - Create 2 equations and you find that a = 5.88ms^-2
    - Therefore T = 11.76
    Last part:
    2Tcostheta = 16.6N
    Don't forget the bearing like me! 225 degrees

    I think it was a decent paper, but don't be fooled by it still. It was quite demanding and you could drop marks very easily. 56-62ish for an A? I guess we will see xD
    There was not any pure Suvat Questions (kinematics) which was nice.
    They really wanted to test knowledge on vectors.
    The moments one was a pain because you had to draw it twice for 2 different scenarios.
    The impulse one was abstract.
    The triangular lift one was a standard lift question in disguise.
    The rest was quite standard.
    Please feel free to correct me/ update / provide extra information for certain questions! It is just nice to get some clarity and get a rough idea as to how you have done so you are not worrying over the summer! Thanks
    Are error carried forward marks allowed in this? If i got the tension in part 8a wrong but got part 8b correct with my incorrect answer do i get the marks for part b? And do they want part 1 b to be simplified (Q and R or something in terms of t - the position vector)
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    (Original post by TrueDAN)
    ...
    Added a bit of detail, hope it helps.

    Q1)

    Part a:

    -Arctan(5/20) gives you and answer which rounds to 14 (nearest whole number)

    -Add this to 90 for a bearing, which gives you an answer of 104 degrees.

    Part b:

    Initially P is at 400i, Q is at 800j.

    Velocity of P=15i+20j Q=20i-5j

    Sub into r = ro + vt to get:

    (i) p = 400i + t(15i+20j)

    simplified: p=(400+15t)i + 20tj

    (ii) q = 800j + t(20i-5j)

    simplified: 20ti + (800-5t)j

    Part c:

    Due west therefore, j component of Q - j component of P = 0

    so, 800-5t-20t=0

    Which gives you t=32Sub into equation "q=" to get 640i+640j

    Q2)-Triangular lift question.

    Mass of brick = 1.5kg and lift = 0.5kg with a vertical acceleration of 1/2.


    Part a:Resolve upwards for whole system

    so,

    T-1.5g-0.5g=0.5 x 2

    Therefore, T=20.6N


    Part b:

    Due to Newton's third law, the force exerted on the brick by the scale pan (The reaction, R) is equal to the force exerted on the scale pan by the brick.

    So, resolving for the brick:

    R-1.5g=0.5 x 1.6

    To get, R=15.45N

    Q3)

    -Impulse Question

    -Firstly, find the reaction in order to work out friction,

    R-1.5g=0.5 x 1.5 So, R=0.4g

    -Resolving to the left (after the ball has rebounded):

    (-0.4g x 1/8)=0.4 x aso,

    a=-1.225

    -Using

    SUVAT,

    S=5

    U=?

    V=0

    A=-1.225

    so, 0^2=u^2 + 2as
    so, u=3.5A


    -Then work out the impulse,

    I=0.4(3.5+4)

    I=3Ns


    Might do some more later if you want me to.
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    (Original post by getgotgetgot)
    Are error carried forward marks allowed in this? If i got the tension in part 8a wrong but got part 8b correct with my incorrect answer do i get the marks for part b? And do they want part 1 b to be simplified (Q and R or something in terms of t - the position vector)
    In that case, you would have picked some marks, but not full. Also if you have simplified q and r position vectors, it is absolutely fine, as long as your values are correct and you used R = Ro+vt
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    (Original post by getgotgetgot)
    Are error carried forward marks allowed in this? If i got the tension in part 8a wrong but got part 8b correct with my incorrect answer do i get the marks for part b? And do they want part 1 b to be simplified (Q and R or something in terms of t - the position vector)
    Yes yes yes, of course! You will only get penalised for making a mistake once - even if you messed up somewhere, if the rest of the method is correct then you pick up method marks. So for 8 you will definitely get an error carried forward and could still get 4/4 on the force exerted on pulley question.
    Question 1 - in terms of t I believe it was
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    I got the moments one wrong but I don't know what I did. I ended up with the mass being 66kg and the distance as 1.6m
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    (Original post by BRBBIT)
    Added a bit of detail, hope it helps.Q1)Part a:-Arctan(5/20) gives you and answer which rounds to 14 (nearest whole number)-Add this to 90 for a bearing, which gives you an answer of 104 degrees.Part b:Initial + t x VelosityInitially P is at 400i, Q is at 800j.Velocity of P=15i+20j Q=20i-5jSub into r = ro + vt to geti) p = 400i + t(15i+20j) simplified: p=(400+15t)i + 20tj(ii) q = 800j + t(20i-5j) simplified: 20ti + (800-5t)jPart cue west therefore, j component of Q - j component of P = 0so, 800-5t-20t=0Which gives you t=32Sub into equation "q=" to get 640i+640jQ2)-Triangular lift question. Mass brick = 1.5kg and lift = 0.5kg with a vertical acceleration of 1/2.
    Part a:Resolve upwards for whole system so, T-1.5g-0.5g=0.5 x 2Therefore, T=20.6N
    Part Bue to Newton's third law, the force exerted on the brick by the scale pan (The reaction, R) is equal to the force exerted on the scale pan by the brick.So, resolving for the brick: R-1.5g=0.5 x 1.6To get, R=15.45NQ3)-Impulse Question-Firstly, find the reaction in order to work out friction, R-1.5g=0.5 x 1.5 So, R=0.4g-Resolving to the left (after the ball has rebounded): (-0.4g x 1/8)=0.4 x aso, a=-1.225-Using SUVAT,S=5U=? 0^2=u^2 + 2asV=0 so, u=3.5A=-1.225-Then work out the impulse, I=0.4(3.5+4) I=3NsMight do some more later if you want me to.
    That is amazing thank you so so much for doing this!! Appreciate it
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    (Original post by TrueDAN)
    That is amazing thank you so so much for doing this!! Appreciate it
    I fixed the formatting, check again.
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    how many marks is each question please
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    (Original post by fishs003)
    The bearing of the resultant force was 225, but the actual degree you had to use In the calculation was 45. They may have said the resultant force was 45 degrees down from the negative x axis or something.
    45 is the pulley on the string instead of the string on the pulley
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    (Original post by BRBBIT)
    Added a bit of detail, hope it helps.

    Q1)

    Part a:

    -Arctan(5/20) gives you and answer which rounds to 14 (nearest whole number)

    -Add this to 90 for a bearing, which gives you an answer of 104 degrees.

    Part b:

    Initially P is at 400i, Q is at 800j.

    Velocity of P=15i+20j Q=20i-5j

    Sub into r = ro + vt to get:

    (i) p = 400i + t(15i+20j)

    simplified: p=(400+15t)i + 20tj

    (ii) q = 800j + t(20i-5j)

    simplified: 20ti + (800-5t)j

    Part c:

    Due west therefore, j component of Q - j component of P = 0

    so, 800-5t-20t=0

    Which gives you t=32Sub into equation "q=" to get 640i+640j

    Q2)-Triangular lift question.

    Mass of brick = 1.5kg and lift = 0.5kg with a vertical acceleration of 1/2.


    Part a:Resolve upwards for whole system

    so,

    T-1.5g-0.5g=0.5 x 2

    Therefore, T=20.6N


    Part b:

    Due to Newton's third law, the force exerted on the brick by the scale pan (The reaction, R) is equal to the force exerted on the scale pan by the brick.

    So, resolving for the brick:

    R-1.5g=0.5 x 1.6

    To get, R=15.45N

    Q3)

    -Impulse Question

    -Firstly, find the reaction in order to work out friction,

    R-1.5g=0.5 x 1.5 So, R=0.4g

    -Resolving to the left (after the ball has rebounded):

    (-0.4g x 1/8)=0.4 x aso,

    a=-1.225

    -Using

    SUVAT,

    S=5

    U=?

    V=0

    A=-1.225

    so, 0^2=u^2 + 2as
    so, u=3.5A


    -Then work out the impulse,

    I=0.4(3.5+4)

    I=3Ns


    Might do some more later if you want me to.
    Great format - very clear. Thanks so much for putting so much effort into this! I am currently busy but will update it later on tonight and give you full credit - hope the exam went well for you
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    Decent paper but I have a feeling grade boundaries will be low since there was a lot of room for mistakes
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    (Original post by TrueDAN)
    That is amazing thank you so so much for doing this!! Appreciate it
    Did some more.

    Q4)

    Speed time Graph

    Part a:

    Name:  Untitled.png
Views: 190
Size:  3.2 KB

    Part b:

    Work out the total time (y) using a trapezium for the 30m/s one:

    ((25+y)/2) x 30 = 975 So, y=40

    Do the same for the 40m/s one to get T:

    ((T+40)/2) x 40 = 975 So, T=8.75



    Q5)
    Incline plane question.

    Resolve perpendicular plane to get value for reaction(R):

    R - 2gCos20 - 40Sin30 = 0

    So, R=2gCos20 + 40Sin30

    Resolve down the plane to get an equation for the friction (F):

    F + 2gSin20 - 40Cos30 = 0

    So, F = 40Cos30 - 2gSin20

    Sub into "F=μR" to get μ=0.727
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    got 100% if they accepted "diagonally down to the right" instead of a bearing for the last question
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    I feel like it went okay So so so so so so so frustrated though because on the forces one i put 2.5i+ 2.5j then crossed it out and did something completely different with different workings out and everything!!!! I think I ended up working out the resultant which is just stupid. Oh well, hope you all did well and if your doing any other exams soon good luck
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    Genuinely don't know what on earth i did to get 0.35 for the coefficient question.. i must've gotten every m1 and d1 mark them butchered the easiest step (entering figures into a calculator). So pissed off. I probably only got about 62 now. I needed a high A since I'm 100% failing D1
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    What is everyone predicting the graded boudaries to be for an A???
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    On the final question I got the correct resultant force but wrote 'at a bearing of 45 degrees' with a diagram. How many marks will i get out of 4? Also on the moments question I got d=1.6 and M=66 with a lot of working out and diagrams. How many marks can I expect to get on this also? Thank you for any help
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    (Original post by hongse123)
    got 100% if they accepted "diagonally down to the right" instead of a bearing for the last question
    left dont you mean?
 
 
 
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