A Level Further Mathematics 9231 Paper 32, May/June 2025

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9231/32 · May/June 2025

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A particle $P$ of mass $8 kg$ is moving in a straight horizontal line. At time $t s$ , $P$ has displacement $x m$ from a fixed point $O$ on the line and velocity $v m s^{- 1}$ . The only horizontal force acting on $P$ has magnitude $(x^{3} + 4 x) N$ and acts in the direction $O P$ . When $t = 0$ , $x = 0$ and $v = 1$ .

Find an expression for $v$ in terms of $x$ , giving your answer in the form $v = a x^{2} + b$ , where $a$ and $b$ are constants to be determined. [3]

[3]

Find an expression for $x$ in terms of $t$ . [2]

[2]

A particle $P$ is projected with speed $24 m s^{- 1}$ at an angle $θ^{\circ}$ above the horizontal from a point $O$ on a horizontal plane, and moves freely under gravity. At a horizontal distance $35 m$ from $O$ , there is a vertical wall of height $10 m$ which is perpendicular to the vertical plane of motion of $P$ .

Determine the two values of $θ$ for which $P$ just clears the wall. [4]

[4]

Given that $P$ clears the wall, find the minimum distance from point $O$ where $P$ can land. [2]

[2]

A rough horizontal disc, centre $O$ , rotates with constant angular speed $ω rad s^{- 1}$ . A particle $P$ of mass $1.6 kg$ lies on the disc at a distance $1.5 m$ from $O$ , and is attached to a point $A$ vertically above $O$ by a light elastic string. The string has natural length $2 m$ , modulus of elasticity $32 N$ and makes an angle $α$ with the vertical $O A$ (see diagram). Particle $P$ moves in a horizontal circle also at a constant angular speed $ω rad s^{- 1}$ . Particle $P$ is on the point of slipping in the direction $O P$ . The coefficient of friction between the particle and the disc is $0.5$ .

Given that the tension in the string is $8 N$ , show that $\sin α = 0.6$ . [2]

[2]

Find the number of revolutions per minute made by the disc and the particle $P$ . [5]

[5]

An object is formed by removing a solid hemisphere, radius $2 r$ , from a uniform solid cone, radius $3 r$ and semi-vertical angle $θ$ , where $\tan θ = \frac{1}{2}$ . The axes of symmetry of the cone and the hemisphere coincide. The base of the cone and the base of the hemisphere are in the same plane as each other (see diagram).

Find, in terms of $r$ , the distance of the centre of mass of the object from its base. [4]

[4]

The object is placed such that its circular base makes contact with a rough plane which is inclined to the horizontal at an angle $α$ . The object is on the point of toppling. The plane is sufficiently rough to prevent sliding. Find the value of $α$ . [3]

[3]

One end of a light elastic string of natural length $0.5 m$ and modulus of elasticity $14 N$ is attached to a fixed point $A$ on a smooth plane. The plane makes an angle $α$ to the horizontal, where $\tan α = \frac{7}{24}$ . A particle $P$ of mass $2 kg$ is attached to the other end of the string. The string lies along a line of greatest slope of the plane. The particle $P$ is initially held on the plane above the level of $A$ , where $A P = 0.8 m$ . The particle $P$ is then released from rest. Find the maximum velocity of $P$ during the subsequent motion. [6]

[6]

Two uniform smooth spheres $A$ and $B$ of equal radii have masses $2 m$ and $m$ respectively. Sphere $A$ is moving in a straight horizontal line with speed $u$ , and sphere $B$ is stationary. Sphere $A$ collides directly with $B$ , and they both then move in the same direction with speeds $v_{A}$ and $v_{B}$ respectively. After the collision, the kinetic energy of $B$ is $\frac{9}{2}$ times the kinetic energy of $A$ .

Show that $v_{B} = \frac{6}{5} u$ . [3]

[3]

Sphere $B$ then collides with a fixed vertical barrier. Immediately before the collision, the direction of motion of $B$ makes an angle $α$ with the barrier. Immediately after the collision, the direction of motion of $B$ makes an angle $β$ with the barrier. The coefficient of restitution between $B$ and the barrier is $\frac{4}{5}$ . As a result of the collision, the velocity of $B$ is reduced to $\frac{12}{25} \sqrt{5} u$ . Find the value of $\sin (α + β)$ . [6]

[6]

A fixed hollow sphere has radius $a$ and centre $O$ . The points $A$ , $B$ and $C$ lie on the inner surface of the sphere with $O A$ and $B C$ horizontal. A portion of the sphere has been removed by a horizontal cut through points $B$ and $C$ at a vertical distance $k a$ above the centre of the sphere, where $k$ is a positive constant and $k < 1$ . The points $O$ , $A$ , $B$ and $C$ all lie in the same vertical plane. $O B$ makes an angle $θ$ with the upward vertical through $O$ (see diagram). A particle $P$ of mass $m$ is free to move on the smooth inner surface of the sphere. The particle $P$ is projected vertically downwards from $A$ with speed $u$ and begins to move in a vertical circle.

In the case where $u = \sqrt{\frac{6}{5} g a}$ , the reaction on $P$ at $B$ is half the reaction on $P$ at $A$ . Find the value of $k$ . [5]

[5]

Find an expression for $u$ , in terms of $a$ and $g$ , in the case that the particle just reaches $B$ . [1]

[1]

Find an expression for $u$ , in terms of $a$ and $g$ , in the case that the particle passes through $B$ and in its subsequent motion reaches $C$ . [4]

[4]

Mark scheme

9231/32 · May/June 2025

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1a · 3 marks

B1
Correct use of N2L.
M1
Attempt to integrate by separating variables. Must have $v \frac{d v}{d x}$ . Must include $+ c$ .
A1
Use initial conditions to find $c$ and rearrange to obtain an expression for $v$ in terms of $x$ . Condone $v = \frac{x^{2} + 4}{4}$ . Condone omitted explanation for choosing positive $v$ .

1b · 2 marks

M1
Separate variables and integrate to $\tan^{- 1}$ form. Only FT their $v$ of the form $a x^{2} + b$ .
A1
Must have $+ C$ and statement that $C = 0$ .

2a · 4 marks

B1
Use trajectory equation to set up quadratic in $t = \tan θ$ using $(35, 10)$ .
M1
Form three-term quadratic in $t$ and equate to 0. OE, for example $1225 t^{2} - 4032 t + 2377 = 0$ .
A1
$\tan θ = 0.769$ leading to $θ = {37.6}^{\circ}$ .
A1
$\tan θ = 2.52$ leading to $θ = {68.4}^{\circ}$ .

2b · 2 marks

M1
Complete method to find the range, $x$ , using the larger of their two values found in part 2(a).
A1
$39.5$

3a · 2 marks

M1
Use $T = \frac{λ x}{l}$ to find $x$ . May be implied by correctly derived value for $\sin α = 0.6$ .
A1
Hence, $∣ P A ∣ = 2.5$ and $\sin α = \frac{1.5}{2.5} = 0.6$ .

3b · 5 marks

1 mark
For either equation (resolving horizontally or vertically). Allow sin/cos wrong way round. Correct number of terms and correct signs. Dimensionally correct. Numbers or algebraic.
A1
Both equations correct.
1 mark
Eliminate $R$ to obtain equation involving $ω^{2}$ with no other unknown quantities.
A1
$ω^{2} = 4$ leading to $ω = 2$
A1
Number of revolutions per minute = $\frac{60}{π} = 19.1$ .

4a · 4 marks

M1
Allow one error for cone or hemisphere. SOI, may be seen in moments equation.
M1
A moments equation with 3 terms, allow sign errors, must be dimensionally correct.
A1
Fully correct equation, does not need to be simplified.
A1
$\overset{ˉ}{y} = \frac{69}{38} r$

4b · 3 marks

M1
For attempt at $\tan α$ using $\frac{3 r}{their \overset{ˉ}{y}}$ or $\frac{their \overset{ˉ}{y}}{3 r}$ .
A1
FT their $\overset{ˉ}{y}$ .
A1
$α = {58.8}^{\circ}$

5 · 6 marks

M1
Allow $T = m g \sin α$ .
A1
Find extension at maximum velocity.
B1
One correct EPE term seen anywhere. FT only on their $x$ found using a correct method (dependent on first M mark).
M1
Dimensionally correct, including KE, GPE, and at least one EPE.
A1
Correct equation. FT only on their $x$ found using a correct method (dependent on first M mark).
A1
$v = 3.016620625 \dots$

6a · 3 marks

B1
KE equation linking $v_{A}$ and $v_{B}$ . Needs to be KE equation with $\frac{1}{2}$ and $m$ .
B1
Conservation of momentum. Must see $m$ present.
B1
AG, shown convincingly. If a quadratic equation is formed, must show method for solving and obtain a single answer.

6b · 6 marks

B1
Perpendicular to wall: $v_{B} \cos α = w \cos β$ . OE, for example $\sqrt{5} \cos α = 2 \cos β$ .
B1
Parallel to wall: $e v_{B} \sin α = w \sin β$ . OE, for example $2 \sin α = \sqrt{5} \sin β$ .
1 mark
Obtain an equation involving only one of $α, β$ . Eliminate $α$ or $β$ by squaring both sides and adding.
1 mark
Use $\sin^{2} θ + \cos^{2} θ = 1$ to reduce to equation with just sine or cosine of 1 variable.
A1
Award for any of the following: $\cos α = \frac{2}{3}$ or $\cos β = \frac{1}{3} \sqrt{5}$ or $\sin β = \frac{2}{3}$ .
A1
Exact answer: $\sin (α + β) = 1$ .

7a · 5 marks

B1
Both equations correct. Use of N2L towards centre at $A$ and $B$ . May see $u^{2} = \frac{6}{5} a g$ already substituted. May see $k$ or $\cos θ$ .
1 mark
Use of $R_{B} = \frac{1}{2} R_{A}$ . Condone use of $R_{A} = \frac{1}{2} R_{B}$ for this mark. Two terms required in their expression for $R_{B}$ . May see $u^{2} = \frac{6}{5} a g$ already substituted. May see $k$ or $\cos θ$ .
1 mark
Energy equation, dimensionally correct.
1 mark
Dependent on both previous M marks. Combine N2L and energy equations to obtain an equation that simplifies to involve $k$ only.
A1
$k = \frac{1}{5}$

7b · 1 mark

B1
FT their $\sqrt{2 a g k}$ .

7c · 4 marks

M1
Complete method to find $v^{2}$ . Must use $2 a \sin θ$ and not $2 a$ (or $a \sin θ$ and not $a$ for turning point method). May quote range formula. FT their $k$ from part 7(a). $\cos θ = their k, \sin θ = \sqrt{1 - their k^{2}}$
A1
$v^{2} = \frac{a g}{\cos θ}$ . May see $v^{2} = \frac{a g}{k}$ or $v^{2} = 5 a g$ .
M1
Use their energy equation from part 7(a) ( $v^{2} = u^{2} - 2 a g k$ ). Must substitute for $v^{2}$ .
A1
$u^{2} = \frac{27}{5} a g, u = \sqrt{\frac{27}{5} a g}$

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