A Level Further Mathematics 9231 Paper 32, May/June 2026

Moments equation with three terms, dimensionally correct so must be of the form $\alpha r^2{x}^2+(\beta r+\gamma x)r^3=\bar{x}\times (\lambda r^{2}x+\mu r^3)$ where $\alpha ,\beta ,\gamma ,\lambda ,\mu$ are all non-zero constants.

At least two terms correct.

All correct.

$\bar{x}=\frac{2x^2+81r^2+72xr}{4x+72r}$ OE, for example, $\frac{\frac{1}{2}\pi x^2{r}^2+\frac{81}{4}\pi r^4+18\pi r^{3}x}{\pi r^{2}x+18\pi r^3}$. All individual terms must be simplified. ISW after correct answer seen.

For $\tan \theta =\frac{\bar{x}-x}{3r}$ or $\tan \theta =\frac{x-\bar{x}}{3r}$ or $\tan \theta =\frac{3r}{\bar{x}-x}$ or $\tan \theta =\frac{3r}{x-\bar{x}}$ only (allow equivalents with $x$ replaced with $r$ or vice versa).

For $\tan \theta =\frac{\bar{x}-x}{3x}$ or $\tan \theta =\frac{x-\bar{x}}{3x}$ only with their $\bar{x}$ correctly substituted – their $\bar{x}$ must be dimensionally correct and have come from an expression of the form $\frac{a{x}^2+b{r}^2+cxr}{dx+er}$ with $a,b,c,d$ and $e$ being non-zero. Must be in terms of either $x$ or $r$ only.

$\theta ={19.1}^{\circ }$ AWRT 19.1 (19.110752…) or 0.334 radians (0.3335455…).

Correct unsimplified expression for the radial force.

Attempt to apply N2L either horizontally or vertically to form an equation. Correct number of terms, dimensionally correct, allow sign errors and sin/cos mix only. Values of sin/cos must be substituted. Allow 37 or better (36.8698…) for the angle between $AP$ and the horizontal and 23 or better (22.6198…) for the angle between $BP$ and the horizontal.

$\frac{3}{5}{T}_A=\frac{5}{13}{T}_B+7g(\Rightarrow 39T_A=25T_B+4550)$

$\frac{4}{5}{T}_A+\frac{12}{13}{T}_B=7\times 12\times 4^2(\Rightarrow 13T_A+15T_B=21840)$

$T_A=675$ and $T_B=871$ AWRT 675 and 871 only.

Correct use of Newton’s second law – acceleration must be $v\frac{\mathrm{d}v}{\mathrm{d}x}$.

Separate variables correctly (from a differential equation of the form $\alpha +\beta v^2=\gamma v\frac{\mathrm{d}v}{\mathrm{d}x}$) and obtain an equation of the form $x=k_1\ln (k_2+k_3{v}^2)$ where $k_1,k_2$ and $k_3$ are non-zero constants OR $x=A\ln (k_1+k_{2}v)+B\ln (k_3+k_{4}v)$ where $A$ and $B$ must have the same sign. Condone lack of $+c$.

$x=-6.4\ln \mathrm{\mid }8-0.5v^2\mathrm{\mid }+c$ (*) or $x=-6.4\ln \mathrm{\mid }0.5v^2-8\mathrm{\mid }+c$ (**)

Use correct initial conditions to find $c$.

$1.37{\text{ m s}}^{-1}$ WWW AWRT 1.37 (1.3711489…) – or exact, for example, $v=4\sqrt{1-{\mathrm{e}}^{-\frac{1}{8}}}$.

Correct use of Newton’s second law – acceleration must be $\frac{\mathrm{d}v}{\mathrm{d}t}$.

Separate variables correctly (from a differential equation of the form $\alpha +\beta v^2=\gamma \frac{\mathrm{d}v}{\mathrm{d}t}$) and obtain an equation of the form $t=A\ln (k_{1}v+k_2)+B\ln (k_{3}v+k_4)$ where $A,B,k_1,k_2,k_3$ and $k_4$ are non-zero constants but $A$ and $B$ must have different signs. Allow $t=\lambda \mathrm{artanh}(\mu v)$ for any $\lambda ,\mu \mathrm{\ne }0$ where $\mu \mathrm{\ne }\pm 1$.

$t=1.6\ln \mid \frac{4+v}{4-v}\mid +c$ (*) or $t=-1.6(\ln \mathrm{\mid }v-4\mathrm{\mid }-\ln \mathrm{\mid }v+4\mathrm{\mid })+c$ (**)

Find $c$: $t=0$ when $v=0\Rightarrow c=0$ $t=1.6\ln \mid \frac{4+v}{4-v}\mid$ (*) or $t=1.6\ln \mid \frac{4+v}{v-4}\mid$ (**)

$v_1=\frac{12}{13}eu$ and $v_2=\frac{5}{13}u$ or $v\sin \beta =\frac{12}{13}eu$ and $v\cos \beta =\frac{5}{13}u$

Correct equation relating kinetic energies: $\frac{1}{2}m{v}^2=\frac{4}{13}\times \frac{1}{2}m{u}^2[\Rightarrow v^2=\frac{4}{13}{u}^2]$.

Correct use of $v^2{\sin }^2\beta +v^2{\cos }^2\beta =v^2$ to form an equation in $e$ (and possibly $u$) only.

$e=\frac{1}{4}\sqrt{3}$ ONLY AWRT 0.433 or better (0.4330127…).

FT their $e$ provided $0<e<1$ to find an expression for $\tan \beta$ or $\beta$.

Follow through their $\tan \beta$ or $\beta$ (the angle between $P$ and the first barrier after the first collision). If exact and correct, may see $\tan (60-\beta )=\frac{1}{7}\sqrt{3}$. OR for finding the angle between the second barrier and $P$ before the second collision – if correct, this angle is 13.897… May be seen on a diagram.

$w\sin \theta =ev\sin (60-\beta )$ $w\cos \theta =v\cos (60-\beta )$

$[\tan \theta =e\tan (60-\beta )=\frac{1}{7}\sqrt{3}\times \frac{1}{4}\sqrt{3}=\frac{3}{28}\Rightarrow ]\theta ={6.12}^{\circ }$ AWRT 6.12 (6.1155035…) or 6.11 WWW.

Three term energy equation (OE). Must be dimensionally correct. Allow sign errors and sin/cos mix only. Mass must be present.

N2L radially: $T-mg\cos \theta =\frac{m{v}^2}{a}$.

Combine 2 equations (N2L radially and energy). Their N2L radially can have sign errors and sin/cos mix only. Dimensionally correct.

$[T=]mg(3\cos \theta +2)$ OE

Setting their dimensionally correct expression for $T$ from 5(a) to zero (OE from a restart, so may see $\cos \alpha =\frac{2}{3}$ being derived). Their expression for $T$ must be two terms only with only one term containing either sine or cosine.

Find an expression (possibly unsimplified) for $v^2$ or $v$ when the string goes slack in terms of $a$ (and possibly $g$). Their expression for $v^2$ must be two terms only with only one term containing either sine or cosine.

For $v\sin \theta$ following through their $v$ and their value of $\cos \theta$.

For use of $v^2=u^2+2as$ (or equivalent complete method) with $v=0,a=-g$ and their $v_y$ for $u$.

$\frac{50}{27}a$

Correct expression for the EPE.

Conservation of energy with exactly 3 terms (GPE and KE terms must be correct). Dimensionally correct – allow sign errors and slips in the EPE term (but must be of the form $k_1{x}^2$).

AG – at least one line of working from application of conservation of energy to given answer. Any error seen is A0.

6.83 only

[Maximum occurs when] $x=2.5$

$[v^2=75\Rightarrow ]v=8.66{\text{ m s}}^{-1}$

For both unsimplified distances – may see these two values on a diagram: $[s_{\text{hor}}=]10+5\times \frac{3}{5}[\Rightarrow 13]$ and $[s_{\text{ver}}=]\frac{4}{5}\times 5[\Rightarrow 4]$.

Applying $s=ut+\frac{1}{2}a{t}^2$ both horizontally and vertically. Allow sin/cos mix only. No sign errors, must be using $a=-g$. Dependent on previous B1 mark.

Use perpendicular condition to form a correct equation in $u,t$ and $\theta$: $\frac{u\cos \theta }{gt-u\sin \theta }=\frac{4}{3}$ or $\frac{u\sin \theta -gt}{u\cos \theta }=-\frac{3}{4}$.

Eliminate two of $u,t,\theta$ to obtain an equation in one variable only – dependent on all previous marks.

$[u\cos \theta =\frac{26}{11}\sqrt{11},u\sin \theta =\frac{71}{22}\sqrt{11}\Rightarrow ]\theta ={53.8}^{\circ }$ AWRT 53.8 (53.78116…) or 0.939 (0.938658…).

$u=13.3$ AWRT 13.3 (13.26735…) or $\sqrt{\frac{7745}{44}}$.