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

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-papers-and-mark-schemes/2013/january/AQA-MM2B-QP-JAN13.PDF; mark scheme: http://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2013/january/AQA-MM2B-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, $l = \dfrac{4r \cos 60}{\cos 30}=\dfrac{4r}{\sqrt{3}}$.

Is 4:5 an acceptable ratio? I had like √(16ag):√(20ag)
16ag:20ag
4:5

Or is that wrong

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.

Oh damn, silly me. Oh well it's only 2 marks and I think that they may be the only marks I lost

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?

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?

Is 4:5 an acceptable ratio? I had like √(16ag):√(20ag)
16ag:20ag
4:5

Or is that wrong

Any ideas what the 100ums and A* grade boundary will be?