A Level Mathematics 9709 Paper 43, Oct/Nov 2024

For either correct. Do not allow $\frac{1}{2}\times m\times (6-5{)}^2$. Note: Difference = $\frac{1}{2}\times m\times 11$.

WD against resistance $=50\times 24[=1200]$ Do not allow if errors such as e.g. $(m-24)\times 50$.

Attempt at work energy equation with 4 relevant terms; dimensionally correct. Allow sign errors. M0 for $\frac{1}{2}\times m\times (6-5{)}^2$.

$$m=\frac{341}{5.5}=62$$

Resolving either direction with correct number of relevant terms. Allow sin/cos mix. Allow sign errors. Do not allow ‘forces to the left = forces to the right’ e.g. $12\cos 30-8\sin 30=16\sin 30$ unless subsequently ‘corrected’.

$$\pm (12\sin 30+24+8\cos 30-16\cos 30)[=F_x\text{ or }F\cos \theta \text{ or }F\sin \alpha ]$$ $$F_x=\pm (30-4\sqrt{3})[=23.07\dots ]$$

$$\pm (12\cos 30-8\sin 30-16\sin 30)[=F_y\text{ or }F\sin \theta \text{ or }F\cos \alpha ]$$ $$F_y=\pm (6\sqrt{3}-12)[=-1.607\dots ]$$

Attempt to find $F$. Must have correct number of relevant terms. (Forces must have or not have components as required). All forces resolved/not resolved as appropriate, but allow consistent sin/cos muddle. Allow use of their $\theta$ provided correctly derived from equations with the correct number of relevant terms.

Attempt to find $\theta$. Must have correct number of relevant terms. (Forces must have or not have components as required). All forces resolved/not resolved as appropriate, but allow consistent sin/cos muddle. Allow upside down so ${\tan }^{-1}\left(\frac{30-4\sqrt{3}}{6\sqrt{3}-12}\right)$. Allow use of their $F$ provided correctly derived from equations with the correct number of relevant terms. Note: watch for use of ${\sin }^{-1}\left(\frac{6\sqrt{3}-12}{30-4\sqrt{3}}\right)$ or ${\sin }^{-1}\left(\frac{1.607}{23.07}\right)$ which leads to correct answer of angle ${4.0}^{\circ }$ scores M0A0.

$$F=23.1\text{ N}$$ $$\theta ={3.99}^{\circ }\text{ above the negative }x\text{-axis oe}$$ [23.1277…] Both correct Allow ${4.0}^{\circ }$ but not simply $4^{\circ }$. [3.986...] Allow answers about the direction such as ‘Above the west’, ‘north of west’ etc, or clockwise 183.99 from $x$ axis, or resultant sketch with angle indicated. If not specified in working please check original diagram to see if direction specified there instead. Allow a bearing of ${274.0}^{\circ }$. Allow explanation of direction that could be drawn uniquely. Or e.g. ${86.0}^{\circ }$ to left of the $y$-axis or ${176.0}^{\circ }$ from the positive $x$-axis.

Resolving up slope. If correct should see $\text{DF}=240+1600g\times 0.08[=240+1280=1520]$. Must have correct number of relevant terms (weight component and $240\text{ N}$ resistance). Allow sign errors. Allow $\cos 4.58$ or $\cos 4.6$. Do not allow $g$ missing. Using ${\sin }^{-1}0.08$ or $\sin 0.08$ scores M0B0A0. Must have either $0.08$ or $\sin 4.58\dots$ or $\sin 4.6$, not just $\sin \theta$.

$$\text{Power}=\text{their}(1520)\times 32$$ OE. E.g. $\frac{\text{Power}}{32}=\text{their }1520$. Allow any driving force provided it has a resistance and a weight component.

$$\text{Power}=48640\text{ W}$$ Allow $48600\text{ W}$ or $48.64\text{ kW}$ or $48.6\text{ kW}$. Must state units if given in kW.

$$\text{DF}=\frac{0.95\times \text{their }48640}{24}\text{ or }\frac{46208}{24}=\frac{5776}{3}\text{ or }1925.3\dots$$ $$\text{DF}=\frac{\text{Power}}{v}\text{ oe e.g. }0.95\times \text{their }48640=\text{DF}\times 24$$ FT their power from part (a) Do not allow if not using power from part (a) Note: candidates who use $\sin 0.08$ in part (a) should get a DF of $332.3\text{ N}$, which can score B1FT and use of ${\sin }^{-1}0.08$ should get a DF of $93299\text{ N}$, can score B1FT. If candidate uses $48600$, $\text{DF}=\frac{46170}{24}=1923.75$ Candidates who omit the weight component in part (a) should get a DF of $304\text{ N}$, and can score B1FT

$$\text{their DF}-240-1600g\times 0.08=1600a$$ N2L Must have correct number of relevant terms (weight component and $240\text{ N}$ resistance). Allow sign errors. Must be dimensionally correct. Allow without using $95\mathrm{\%}$ or with using $5\mathrm{\%}$. Must have either $0.08$ or $\sin 4.58\dots$ or $\sin 4.6$, not just $\sin \theta$ or ${\sin }^{-1}0.08$ or $\sin 0.08$.

$$a=0.253{\text{ ms}}^{-2}$$ Allow $\frac{19}{75}$ Note: $0.25$ scores A0. If candidate uses $48600$ they must get $0.252(34\dots )$ rather than $0.253$.

For attempt at use of conservation of momentum. Must have three non-zero terms. Allow sign errors. Must have correct masses with relevant velocities. Their $v$ may be in the opposite direction. If $g$ included with the masses: Allow M1A0A1 for first three marks. Can then score M1M0A0 for final three marks.

$$3\times 8=3\times 2+6v$$ $$\text{Or }3\times 8=-3\times 2+6v$$

$$v=3[v=5]$$ Allow finding only $v=3$, but if both speeds found they must both be correct

$${\text{KE}}_{\text{after}}=0.5\times 3\times 2^2+0.5\times 6\times \text{their }{v}_B^2[=33]\text{ if correct}$$ Allow use of any $v_B$ even if $<0$ Using speed of $5$, ${\text{KE}}_{\text{after}}=0.5\times 3\times 2^2+0.5\times 6\times \text{their }{5}^2[=81]$ Candidates may work out the loss for each particle separately which only scores DM1 when losses subsequently added together.

$${\text{KE}}_{\text{loss}}=\pm (0.5\times 3\times 8^2-(0.5\times 3\times 2^2+0.5\times 6\times \text{their }{3}^2))[=96-33]\text{ if correct}$$ If two speeds found then FT their lower speed, even if later choose the wrong loss. If only one speed found then FT their speed. If the candidate thinks that the particles coalesce then score M0 here

$$\text{Loss}=63\text{ [J]}$$ If both losses found, must state which is the greater. Only award this mark if no errors – e.g. the KE loss for $v=5$ should be $15\text{ J}$. If wrong then A0. Allow $-63\text{ [J]}$. Can score full marks even if only $v=3$ is found.

Attempt at resolving in at least one direction. Correct number of relevant terms with $T$ resolved; allow sign errors; allow sin/cos mix. Can score M1 for any $F=T\cos 30$. Do not allow $g$ missing in the equation for $R$. Must have $12$, not just $m$. Could see $R$ as part of an equation for $F$. E.g. $F=0.5(12g-T\sin 30)$.

$$R+T\sin 30=12g$$ $$F=T\cos 30$$ Both correct.

Use of $F=0.5R$ to form an equation in $T$ or $R$ only. Allow sign errors in $R$; allow consistent sin/cos mix in $R$ but no other errors. Must be two term $R$ as a linear combination of weight and a component of $T$, and $F$ must be a single term which is a component of $T$. Do not allow $g$ missing. If correct $T\cos 30=0.5(120-T\sin 30)$ or $\frac{\sqrt{3}}{2}T=60-0.25T$. If no working shown to eliminate $T$ or $R$, then DM2 for getting $T$ value correct for their equations and A1 if fully correct. Could use $0.5R=T\cos 30$ and solve simultaneously.

Attempt to solve for $T$. Allow consistent sin/cos mix and allow sign errors. Must get to '$T = $'. Dependent on both previous M1s.

$$T=53.8\text{ N}$$ $[53.7622\dots ]$ Note: For sign errors: $R-T\sin 30=12g$ answer should be $97.3985\dots$ $R-T\sin 30=-12g$ answer should be $-97.3985\dots$ $R+T\sin 30=-12g$ answer should be $-53.7622\dots$ Each of the above would usually get M1A0M1M1A0.

$$T\cos 30-F=12\times 0.2$$ Attempt at N2L; correct number of relevant terms with $T$ resolved; allow sign errors; allow sin/cos mix, but can be $F$ or any reasonable attempt at friction.

Use of $F=0.5R$ to form an equation in $T$ and solve. Must be a two term $R$ as a linear combination of weight and a component of $T$. Allow sign errors and consistent sin/cos mix. Must get to '$T=$'. The equations if correct should be $T\cos 30-0.5(120-T\sin 30)=12\times 0.2$ or $\frac{\sqrt{3}}{2}T-60+0.25T=2.4$ or $(\frac{\sqrt{3}}{2}+0.25)T=62.4$ and these must be solved. Any use of $T$ or $R$ from part (a) scores DM0 here.

$$T=55.9\text{ N}$$ $T=55.9127\dots$ Note: For sign errors: $R-T\sin 30=12g$ answer should be $101.294\dots$ $R-T\sin 30=-12g$ answer should be $-93.5026\dots$ $R+T\sin 30=-12g$ answer should be $-51.6117\dots$ Each of the above would usually get M1M1A0.

$$v=\int 0.6t\text{ d}t=\frac{0.6}{2}{t}^2[=0.3t^2][+0]$$ $$[t=4]V=0.3\times 4^2\text{ or }0.3\times 16=4.8$$ AG Must see $\frac{0.6}{2}{t}^2$ or $0.3t^2$ and $4$ must be actually shown substituted. Merely stating $t=4$ is not enough to score this mark. Do not allow use of $s=ut+\frac{1}{2}a{t}^2$ which leads to $4.8$ if $a=0.6$ is used.

Quadratic with correct curvature starting from $(0,0)$ to $(4,4.8)$. Note: the grid is for reference – not shown in QP. Their graph does not need to be to scale.

Horizontal line at from $(4,4.8)$ to $(15,4.8)$. The points should be specified somehow, but for an accurate sketch allow a line just below 5 without specifying 4.8.

Line from $(15,4.8)$ to $(20,0)$. The points should be specified somehow, but for an accurate sketch allow a line just below 5 without specifying 4.8. Allow all 3 marks if using $V$ instead of 4.8. ISW any extra out of the range for $t$ of 0 to 20. If no marks scored then SC B1 for a correct shaped graph with no numbers. If using a value of $v\mathrm{\ne }4.8$, allow SC B1 for the first section correct and SC B1 for the second and third both correct.

Attempt to find acceleration. For calculation $\frac{0-4.8}{20-15}\text{ oe }[=-0.96]$. Allow $+0.96$.

$$[v=]19.2-0.96t$$ Oe e.g. Allow $[v=]4.8-0.96(t-15)$.

$$\left[{\int }_0^{4}0.3t^2\text{d}t\right]\frac{0.3}{3}{t}^3[=0.1t^3]$$ Attempt to integrate their $v$ from part (a) provided this came from integration, but allow a restart here. The power of $t$ must increase by 1 with a change of coefficient. Use of $s=vt$ scores M0. No need for limits. If no integration seen allow SCM1 for answer of 6.4 in place of M1M1.

$$0.1\times 4^3[-0.1\times 0^3]$$ Correct use of correct limit(s) (expect 6.4).

$$4.8\times 11+0.5\times 4.8\times 5[=52.8+12=64.8]$$ Both correct and added. May be done in one go using a trapezium $\frac{(11+16)\times 4.8}{2}$. Could do the last stage by integration. Maximum B1 for final answer 74.4 from thinking the first section is also straight.

$$\text{Distance}=71.2\text{ m}$$

For one correct equation: $T-3g=\pm 3a$, $5g-T=\pm 5a$, $5g-3g=\pm (3+5)a$.

For any two correct consistent equations. If tensions $T_A$ and $T_B$ both stated must at some point state or imply that they are equal to score the second B1.

Attempt to solve for $T$. May find $a$ first $[a=2.5]$. Must get to '$T = $'. Dep on both B marks.

$$T=37.5\text{N or }\frac{75}{2}\text{ N}$$ Allow without working.

Correct use of suvat with their $a$ and solve for $t$. E.g. $2=0.5\times 2.5\times t^2$. Must get to '$t=$'. Dep on both B marks or the B1 for the equation $5g-3g=\pm (3+5)a$ if this used to find $a$, but not on first M1 mark. Could find $v=\sqrt{10}$ then use $2=\frac{0+\text{their }\sqrt{10}}{2}t$. (Or Alternative: Use of energy to find velocity at plane and then suvat to find $t$: $\text{PE loss}=\text{KE gain}:5g\times 2-3g\times 2=\frac{1}{2}\times 5v^2+\frac{1}{2}\times 3v^2$, or $\text{PE loss}-\text{WD by tension}=\text{KE gain}:5g\times 2-\text{their }37.5\times 2=\frac{1}{2}\times 5v^2$, or $\text{WD by tension}-\text{PE gain}=\text{KE gain}:\text{their }37.5\times 2-3g\times 2=\frac{1}{2}\times 3v^2$, $v=\sqrt{10}$ or $v=3.16$ then $2=\frac{0+v}{2}t$, dependent on both B marks only if candidate uses tension, but otherwise not dependent on either B mark.)

$$t=1.26\text{ or }\frac{\sqrt{40}}{5}\text{ or }\frac{2\sqrt{10}}{5}$$ $t=1.2649\dots$ If candidates do not try to find $T$ but do attempt to find the time, they can score B1B1M0A0M1A1. Do not allow if also give negative answer and do not discard. Note: $t=\sqrt{1.6}$ only, scores A0.

Correct use of suvat before B hits the plane or when A has risen 2 m, to attempt to find velocity of A when string becomes slack. Using their $a$ and/or their $t$. $v^2=0+2\times 2.5\times 2$. Must use $s=2$ and $u=0$, OR $v=0+2.5\times \sqrt{1.6}$ Must use $u=0$, OR $2=\frac{1}{2}(0+v)\sqrt{1.6}$ Must use $s=2$ and $u=0$. Must be complete method to find $v$ or $v^2$ $[v=\sqrt{10}]$. Could have found $v=\sqrt{10}$ in part (a) and give M1 if used in part (b). (Or energy method: $\text{PE loss}=\text{KE gain}:5g\times 2-3g\times 2=\frac{1}{2}\times 5v^2+\frac{1}{2}\times 3v^2$, or $\text{PE loss}-\text{WD by tension}=\text{KE gain}:5g\times 2-\text{their }37.5\times 2=\frac{1}{2}\times 5v^2$, or $\text{WD by tension}-\text{PE gain}=\text{KE gain}:\text{their }37.5\times 2-3g\times 2=\frac{1}{2}\times 3v^2$. Must be complete method to find $v$ or $v^2$ $[v=\sqrt{10}\text{ or }3.162\dots ]$. Do not allow sign errors.)

Correct use of suvat for motion of A (between height of 3m and 3.25 m), to form an equation in $t$, using $a=-g$ $$3.25-3=(\text{their }\sqrt{10})t+0.5\times (-10)\times t^2$$ (Or finding max height: $0^2=(\text{their }\sqrt{10}{)}^2+2\times (-10)s$ where $s$ is distance above 3 metres (which leads to a maximum height of 3.5 m). Or finding velocity at height of 3.25 m: $w^2=\text{their }{\sqrt{10}}^2+2\times (-g)\times 0.25[w^2=5]\text{ or }[w=\sqrt{5}]$. Or using energy: $\text{PE gain}=\text{KE loss}:3g\times 0.25=\frac{1}{2}\times 3\times {\sqrt{10}}^2-\frac{1}{2}\times 3w^2$. Must be complete method to find $w$ or $w^2$ $w=\sqrt{5}\text{ or }2.236\dots$. Do not allow sign errors.)

Solving a 3 term quadratic for $t$ to get at least one (unsimplified) value using their 2.5 (or using any other correct method). If correct should get $t=0.093,0.540$ or $\frac{\sqrt{10}-\sqrt{5}}{10}$, $\frac{\sqrt{10}+\sqrt{5}}{10}$, $0.09262\dots$ or $0.53978\dots$ Could use formula and realise that the time for at least 3.25 m, $=2\frac{\sqrt{b^2-4ac}}{2a}=2\frac{\sqrt{10-4\times 5\times 0.25}}{2\times 5}$ which gets DM1. (Or from max height: For attempt to find time from .25 m below top to top: $t=\pm 0.224$ or $\frac{\sqrt{5}}{10}$ [$0.22360\dots$] probably from $[3.5-3.25=0t+0.5\times 10t^2]$. Dep on both previous M1s. Or from velocity at 3.25 m: For correct use of suvat to find time to max height: $0=\sqrt{5}+(-g)\times t[t=\frac{\sqrt{5}}{10}]$. Or from energy: For attempt to find time from .25 m below top to top: $t=\pm 0.224$ or $\frac{\sqrt{5}}{10}$ [$0.22360\dots$] probably from $0=\sqrt{5}-10t$. Dep on both previous M1s. Do not allow sign errors.)

$$\text{Time}=0.447\text{ s or }\frac{\sqrt{5}}{5}\text{ s or }\frac{1}{\sqrt{5}}$$ $\text{Time}=0.53978\cdots -0.09262\cdots =0.44721\dots$ $2(\frac{\sqrt{10}}{10}-\frac{\sqrt{10}-\sqrt{5}}{10})=0.44721\dots$ (Or for doubling in alternative methods).