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AQA A Level Maths MM2B Mechanics 2 Unofficial Mark Scheme 25 June 2018 watch

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    I'll start it below
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    Q1

    (a) Find the x distance. [3 marks]

    \overline{x} = \dfrac{\sum m_ix_i}{\sum m_i}, 52 cm

    (b) Find the y distance. [2 marks]

    \overline{y} = \dfrac{\sum m_iy_i}{\sum m_i}, 40.5 cm

    (c) Find the mass of the zebra. [4 marks]

    \overline{x} = 60 \Rightarrow \sum m_i x_i = 60 \sum m_i, 2.67 kg

    Q2

    (a) Calculate the KE at the top. [2 marks]

    \dfrac{1}{2} \times 21 \times 2^2 = 42\;\text{J}

    (b) (i) Calculate the KE at the bottom. [3 marks]

    42 + 21 \times 9.8 \times 16 \cos 60 = 1688.4 J = 1690 J (3 sf)

    (ii) Calculate the speed at the bottom. [2 marks]

    1688.4 = \dfrac{1}{2} \times 21 \times v^2 \Rightarrow v= \sqrt{\dfrac{1688.4}{\frac{1}{2} \times 21}} = 12.7\;\text{m s}^{-1}

    (c) Find \mu. [3 marks]

    By Newton's second law, R = 21g = 205.8\;\text{N}

    1688.4 = \mu \times 205.8 \times 18 \Rightarrow \mu = 0.456 (3 sf)

    Q3

    \textbf{v} = (12t - t^3)\textbf{i} - 6\mathrm{e}^{-2t}\textbf{j}

    (a) Find \textbf{a}. [2 marks]

    \textbf{a} = (12 - 3t^2)\textbf{i} + 12\mathrm{e}^{-2t}\textbf{j}

    (b) (i) Find F newtons. [2 marks]

    \textbf{F} = ((24 - 6t^2)\textbf{i} + 24\mathrm{e}^{-2t}\textbf{j})\;\text{N}

    (ii) Find the magnitude of F newtons at t = 0. [2 marks]
    F = (24i + 24j) N
    |F| = 33.9 N

    (c) Find t when F newtons is acting due north. [2 marks]

    24 - 6t^2 = 0 \Rightarrow t = 2

    (d) Find r, if r = ??? when t = 0. [5 marks]

    \textbf{r} = (6t^2 - \frac{1}{4}t^4)\textbf{i} + 3\mathrm{e}^{-2t}\textbf{j} + \textbf{c}

    Q4

    (a) Show that the frictional force is 1620 N. [2 marks]

    By Newton's second law, since a = \dfrac{v^2}{r}, F = \dfrac{mv^2}{r} = 1620\;\text{N}

    (b) Find the least value of \mu. [2 marks]

    By Newton's second law, R = mg

    \mu \leqslant \dfrac{1620}{900 \times 9.8} = 0.184

    Q5

    The van experience a resistive force of kv newtons, and has a maximum wattage of 32 kilowatts.

    (a) The maximum speed is 40 \text{m s}^{-1}. Show that k=20. [2 marks]

    By Newton's first law, F = kv

    So, using the wattage equation, P = Fv = kv^2 \Rightarrow 32 \times 10^3 = k \times 40^2 \Rightarrow k = 20

    (b) The wattage is reduced to zero.

    (i) Show that \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{3g - v}{30}. [3 marks]

    The wattage is zero, so the engine provides no force

    Using Newton's second law, the component of weight parallel to the slope is mg \sin \theta = 600 \times g \times 0.1 = 60g, and the resistive force is kv

    By Newton's second law, 600 \dfrac{\mathrm{d}v}{\mathrm{d}t} = 60 g - 20 v \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{60g-20v}{600} = \dfrac{3g - v}{30}

    (ii) Find t in terms of v. [5 marks]

    When the wattage is reduced to zero, the speed is still 18ms^-1, so v(0) = 18

    \displaystyle \int \frac{\mathrm{d}v}{3g - v} = \int \frac{1}{30}\;\mathrm{d}t \Rightarrow -\ln(3g-v) = \frac{1}{30}t + C

    Using v(0) = 18, C = -\ln(3g - 18) \Rightarrow t = 30(\ln(3g - 18) - \ln(3g - v)) = 30\ln\dfrac{3g-18}{3g-v}

    (iii) Find the time, to 3 sf, for the speed to increase to 22 ms^-1. [3 marks]

    t = 30 \ln\dfrac{3g - 18}{3g - 22} = 12.96... = 13.0 s

    Q6: I'll post workings later

    (a) Find v in terms of a and g. [6 marks]

    v=4\sqrt{ag}

    (b) Find the ratio u:v. [2 marks]

    \dfrac{2}{\sqrt{5}} (0.894 to 3 sf)

    Q7

    The string AP has modulus of elasticity 160 newtons, and natural length of 2 m; BP has modulus of elasticity 240 newtons and natural length of 3 m.

    (a) Show that the total elastic potential energy is 160 J. [2 marks]

    The EPE can be derived using Hooke's law T = \dfrac{\lambda x}{L}.

    Extension in AP is 0 m, so EPE is 0 J.

    Extension in BP is 5 - 3 = 2 m, so EPE is \dfrac{240 \times 2^2}{2 \times 3} = 160\;\text{J}.

    Total EPE = 0 + 160 = 160 J

    (b) Find v. [5 marks]

    Conservation of energy:
    160 = \dfrac{160 \times x^2}{2 \times 2} + \dfrac{240 \times (2-x)^2}{2 \times 3} + \dfrac{1}{2} \times 8 \times v^2 + 8 \times \mu \times g \times x

    160 = 40(2x^2 - 4x + 4) + 4v^2 + 8\mu g x

    v^2 = 40 - 10(2x^2 - 4x + 4) - 2\mu g x

    v = \sqrt{(40-2\mu g)x - 20x^2}

    (c) Find x when the speed is greatest. [2 marks]

    \dfrac{\mathrm{d}(v^2)}{\mathrm{  d}x} = 0 \Rightarrow 40 - 2\mu g = 40x \Rightarrow x = 1 - \dfrac{\mu g}{20}

    Q8 Find l in terms of r. [9 marks]

    This question was the same as January 2013 Q8 (paper: http://filestore.aqa.org.uk/sample-p...-QP-JAN13.PDF; mark scheme: http://filestore.aqa.org.uk/sample-p...W-MS-JAN13.PDF), but here the radius was r (not a), and the angle was specified as \theta = 30.

    Hence, using the derived formula, l = \dfrac{4r \cos 60}{\cos 30}=\dfrac{4r}{\sqrt{3}}.
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    The last question was find r in terms of l, wasn’t t
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    Agh I missed one of the forces on 7B so it was all wrong. Honestly dropped like 18 marks on those last 3 questions so I need the rest of the paper to be completely perfect for me to stand a chance at an A :facepalm:
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    if my method was 100% right for the last question but i made a mistake when subbing one equation into another, how many marks would i drop out of 9?
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    Thanks so much! Also I think HARvey1233 is right I think it was find r in terms of l so just rearrange for r.
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    Whoops on 7b forgot to divide the 8ug when simplifying so I had the v expression and the max speed distance with a 8ug instead of 2ug. But 'twas alright apart from that. I vividly remembered doing the last question on the 2013 paper, got a bit excited at that.
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    Is 4:5 an acceptable ratio? I had like √(16ag):√(20ag)
    16ag:20ag
    4:5

    Or is that wrong
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    r = 4i - 2j, for 3d) if you still need it
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    (Original post by Integer123)
    Q2

    (a) Calculate the KE at the top. [2 marks]

    \dfrac{1}{2} \times 21 \times 2^2 = 42\;\text{J}

    (b) (i) Calculate the KE at the bottom. [3 marks]

    42 + 21 \times 9.8 \times 16 \cos 60 = 1688.4 J = 1690 J (3 sf)

    (ii) Calculate the speed at the bottom. [2 marks]

    1688.4 = \dfrac{1}{2} \times 21 \times v^2 \Rightarrow v= \sqrt{\dfrac{1688.4}{\frac{1}{2} \times 21}} = 12.7\;\text{m s}^{-1}

    (c) Find \mu. [3 marks]

    By Newton's second law, R = 21g = 205.8\;\text{N}

    1688.4 = \mu \times 205.8 \times 18 \Rightarrow \mu = 0.456 (3 sf)

    Q3

    \textbf{v} = (12t - t^3)\textbf{i} - 6\mathrm{e}^{-2t}\textbf{j}

    (a) Find \textbf{a}. [2 marks]

    \textbf{a} = (12 - 3t^2)\textbf{i} + 12\mathrm{e}^{-2t}\textbf{j}

    (b) (i) Find F newtons. [2 marks]

    \textbf{F} = ((24 - 6t^2)\textbf{i} + 24\mathrm{e}^{-2t}\textbf{j})\;\text{N}

    (ii) Find the magnitude of F newtons at t = 0. [2 marks]
    F = (24i + 24j) N
    |F| = 33.9 N

    (c) Find t when F newtons is acting due north. [2 marks]

    24 - 6t^2 = 0 \Rightarrow t = 2

    (d) Find r, if r = ??? when t = 0. [5 marks]

    \textbf{r} = (6t^2 - \frac{1}{4}t^4)\textbf{i} + 3\mathrm{e}^{-2t}\textbf{j} + \textbf{c}

    Q4

    (a) Show that the frictional force is 1620 N. [2 marks]

    By Newton's second law, since a = \dfrac{v^2}{r}, F = \dfrac{mv^2}{r} = 1620\;\text{N}

    (b) Find the least value of \mu. [2 marks]

    By Newton's second law, R = mg

    \mu \leqslant \dfrac{1620}{900 \times 9.8} = 0.184

    Q5

    The van experience a resistive force of kv newtons, and has a maximum wattage of 32 kilowatts.

    (a) The maximum speed is 40 \text{m s}^{-1}. Show that k=20. [2 marks]

    By Newton's first law, F = kv

    So, using the wattage equation, P = Fv = kv^2 \Rightarrow 32 \times 10^3 = k \times 40^2 \Rightarrow k = 20

    (b) The wattage is reduced to zero.

    (i) Show that \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{3g - v}{30}. [3 marks]

    The wattage is zero, so the engine provides no force

    Using Newton's second law, the component of weight parallel to the slope is mg \sin \theta = 600 \times g \times 0.1 = 60g, and the resistive force is kv

    By Newton's second law, 600 \dfrac{\mathrm{d}v}{\mathrm{d}t} = 60 g - 20 v \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{60g-20v}{600} = \dfrac{3g - v}{30}

    (ii) Find t in terms of v. [5 marks]

    When the wattage is reduced to zero, the speed is still 18ms^-1, so v(0) = 18

    \displaystyle \int \frac{\mathrm{d}v}{3g - v} = \int \frac{1}{30}\;\mathrm{d}t \Rightarrow -\ln(3g-v) = \frac{1}{30}t + C

    Using v(0) = 18, C = -\ln(3g - 18) \Rightarrow t = 30(\ln(3g - 18) - \ln(3g - v)) = 30\ln\dfrac{3g-18}{3g-v}

    (iii) Find the time, to 3 sf, for the speed to increase to 22 ms^-1. [3 marks]

    t = 30 \ln\dfrac{3g - 18}{3g - 22} = 12.96... = 13.0 s

    Q6: I'll post workings later

    (a) Find v in terms of a and g. [6 marks]

    v=4\sqrt{ag}

    (b) Find the ratio u:v. [2 marks]

    \dfrac{2}{\sqrt{5}} (0.894 to 3 sf)

    Q7

    The string AP has modulus of elasticity 160 newtons, and natural length of 2 m; BP has modulus of elasticity 240 newtons and natural length of 3 m.

    (a) Show that the total elastic potential energy is 160 J. [2 marks]

    The EPE can be derived using Hooke's law T = \dfrac{\lambda x}{L}.

    Extension in AP is 0 m, so EPE is 0 J.

    Extension in BP is 5 - 3 = 2 m, so EPE is \dfrac{240 \times 2^2}{2 \times 3} = 160\;\text{J}.

    Total EPE = 0 + 160 = 160 J

    (b) Find v. [5 marks]

    Conservation of energy:
    160 = \dfrac{160 \times x^2}{2 \times 2} + \dfrac{240 \times (2-x)^2}{2 \times 3} + \dfrac{1}{2} \times 8 \times v^2 + 8 \times \mu \times g \times x

    160 = 40(2x^2 - 4x + 4) + 4v^2 + 8\mu g x

    v^2 = 40 - 10(2x^2 - 4x + 4) - 2\mu g x

    v = \sqrt{(40-2\mu g)x - 20x^2}

    (c) Find x when the speed is greatest. [2 marks]

    \dfrac{\mathrm{d}(v^2)}{\mathrm{  d}x} = 0 \Rightarrow 40 - 2\mu g = 40x \Rightarrow x = 1 - \dfrac{\mu g}{20}

    Q8 Find l in terms of r. [9 marks]

    This question was the same as January 2013 Q8 (paper: http://filestore.aqa.org.uk/sample-p...-QP-JAN13.PDF; mark scheme: http://filestore.aqa.org.uk/sample-p...W-MS-JAN13.PDF), but here the radius was r (not a), and the angle was specified as \theta = 30.

    Hence, using the derived formula, l = \dfrac{4r \cos 60}{\cos 30}=\dfrac{4r}{\sqrt{3}}.
    I think it's when t=0, r =4i-2j for question 1
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    (Original post by HarryGCSE)
    Is 4:5 an acceptable ratio? I had like √(16ag):√(20ag)
    16ag:20ag
    4:5

    Or is that wrong
    I mean you can't just go from root16:root20 to 4:5 because root20 is not 5
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    (Original post by HarryGCSE)
    Is 4:5 an acceptable ratio? I had like √(16ag):√(20ag)
    16ag:20ag
    4:5

    Or is that wrong
    Unfortunately you can't go from √(16ag):√(20ag) to 16ag:20ag because √(16ag) is not equal to √(20ag). If it was √(16ag) = √(20ag), squaring both sides would be valid.
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    (Original post by OllyBishop)
    Unfortunately you can't go from √(16ag):√(20ag) to 16ag:20ag because √(16ag) is not equal to √(20ag). If it was √(16ag) = √(20ag), squaring both sides would be valid.

    Oh damn, silly me. Oh well it's only 2 marks and I think that they may be the only marks I lost
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    (Original post by HarryGCSE)
    Oh damn, silly me. Oh well it's only 2 marks and I think that they may be the only marks I lost
    Yeah, and to be honest you'll probably get a method mark for getting that far. I think I only got 7c wrong so hopefully only two marks dropped too!
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    for 6b would it be ok to put the ratio as 4 : 2root5 without cancelling the 2? and i think i forgot to square root the v^2 expression in 7b lmao does that lose 1 or 2 marks?
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    (Original post by levi ackerman)
    for 6b would it be ok to put the ratio as 4 : 2root5 without cancelling the 2? and i think i forgot to square root the v^2 expression in 7b lmao does that lose 1 or 2 marks?
    Not simplifying the ratio further might lose you 1 mark, not square rooting the v^2 expression will lose you 1 mark (as it will just be an answer mark). So 1-2 marks lost.
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    (Original post by ppdtg)
    if my method was 100% right for the last question but i made a mistake when subbing one equation into another, how many marks would i drop out of 9?
    hmm not sure tbh, because the question from the 2013 paper was 5 marks and this one was 9 marks i think? but based on past exam questions on moments, the majority of the marks come from a. taking moments about a point and b. resolving in 2 directions, so either vertically + horizontally or parallel + perpendicular to the rod. i'm guessing there's also marks for knowing which way the reaction forces act. so maybe you only drop 1-2/3 marks? basically not many i shouldn't think!
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    (Original post by HarryGCSE)
    Is 4:5 an acceptable ratio? I had like √(16ag):√(20ag)
    16ag:20ag
    4:5

    Or is that wrong
    That's wrong, the ratio between the squares was 16:20 so between u:v it will be root both of them
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    Any ideas what the 100ums and A* grade boundary will be?
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    (Original post by wa17)
    Any ideas what the 100ums and A* grade boundary will be?
    tbh probably 75 for 100 again :/
 
 
 
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