First-quadrant conductor and image source
Question 1
A unit conductor occupies the first quadrant. Let 𝑧0 = 1 + 𝑖 and
ℎ(𝑧) = − 𝑚
2𝜋 log
𝑧2 − 𝑧2
0
𝑧2 − 𝑧0 2
!
, 𝜙 = ℜℎ(𝑧).
(a) Show that both coordinate axes are grounded. (4 marks)
(b) Identify the source inside the conductor. (4 marks)
(c) Find the complex current density. (4 marks)
(d) Find the outward normal-current density on each grounded axis. (4 marks)
(e) Calculate the flux through each axis and verify current conservation. (4 marks) (Total: 20 marks)
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MATH40007 Revised Historical-Coverage Practice Pack Questions
Worked solution and marking guidance
Question 1
On either coordinate axis, 𝑧2 is real and the numerator and denominator are complex conjugates, so their ratio has modulus 1
and 𝜙 = 0. Inside the first quadrant the only finite logarithmic singularity is at𝑧0 = 1 +𝑖, a source of strength𝑚. Differentiation
gives
𝐽𝑥 − 𝑖𝐽 𝑦 = 𝑚𝑧
𝜋
1
𝑧2 − 𝑧2
0
− 1
𝑧2 − 𝑧 2
0
!
.
On the positive 𝑥-axis,
𝐽𝑥 = 0, 𝐽 𝑦 = − 4𝑚𝑥
𝜋(𝑥4 + 4) ,
so the outward density is 4𝑚𝑥/[ 𝜋(𝑥4 + 4)]. On the positive 𝑦-axis,
𝐽𝑥 = − 4𝑚𝑦
𝜋(𝑦4 + 4) , 𝐽 𝑦 = 0,
and the outward density is 4𝑚𝑦 /[ 𝜋(𝑦4 + 4)]. Each flux is
∫ ∞
0
4𝑚𝑠
𝜋(𝑠4 + 4) 𝑑𝑠 = 𝑚
2 ,
so the total outward flux is 𝑚, matching the source strength. Question difficulty allocation: 𝐴 = 8, 𝐵 = 4, 𝐶 = 4, 𝐷 = 4.
Three-mass spring chain and modal response
Question 2
Three unit masses move between two fixed walls. Unit springs join the left wall to mass 1, successive masses, and
mass 3 to the right wall.
1 2 3
(a) Write the stiffness matrix and find an orthonormal eigensystem and the natural frequencies. (8 marks)
(b) A constant force 𝑓 = (1, 0, 1)𝑇 is applied. Find the equilibrium. (4 marks)
(c) The force is removed at 𝑡 = 0 with initial displacement equal to the equilibrium and initial velocity (0, 1, 0)𝑇.
Find the free motion in modal form. (8 marks) (Total: 20 marks)
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MATH40007 Revised Historical-Coverage Practice Pack Questions
Worked solution and marking guidance
Question 2
The stiffness matrix is
𝐾 =
[
2 −1 0
−1 2 −1
0 −1 2
]
.
An orthonormal eigensystem is
𝜙1 = (1/2, 1/
√
2, 1/2)𝑇 , 𝜆 1 = 2 −
√
2,
𝜙2 = (1/
√
2, 0, −1/
√
2)𝑇 , 𝜆 2 = 2,
𝜙3 = (1/2, −1/
√
2, 1/2)𝑇 , 𝜆 3 = 2 +
√
2.
The frequencies are 𝜔 𝑗 = √︁𝜆 𝑗. Since 𝐾 (1, 1, 1)𝑇 = (1, 0, 1)𝑇, the equilibrium is 𝑥𝑒 = (1, 1, 1)𝑇. Let 𝑎 𝑗 = 𝜙𝑇
𝑗 𝑥𝑒 and
𝑏 𝑗 = 𝜙𝑇
𝑗 (0, 1, 0)𝑇. Then
𝑎1 = 1 + 1√
2
, 𝑎 2 = 0, 𝑎 3 = 1 − 1√
2
; 𝑏1 = 1√
2
, 𝑏 2 = 0, 𝑏 3 = − 1√
2
.
Therefore
𝑥(𝑡) =
3∑︁
𝑗=1
(
𝑎 𝑗 cos 𝜔 𝑗𝑡 + 𝑏 𝑗
𝜔 𝑗
sin 𝜔 𝑗𝑡
)
𝜙 𝑗 .
Question difficulty allocation: 𝐴 = 8, 𝐵 = 6, 𝐶 = 2, 𝐷 = 4.
Graph hitting probability and terminal voltage
Question 3
Consider the graph below, with terminals 𝑊 and 𝐸.
𝑊
𝐴
𝐵
𝐶
𝑋
𝐸
(a) Give the dimensions and nullities of an incidence matrix. (4 marks)
(b) A walker starts at 𝑊, leaves it and stops on reaching 𝐸 or returning to 𝑊. Find the probability of reaching 𝐸
first. (8 marks)
(c) If 𝑊 is held at voltage 1 and 𝐸 grounded, find the voltage at 𝑋. (8 marks) (Total: 20 marks)
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MATH40007 Revised Historical-Coverage Practice Pack Questions
Worked solution and marking guidance
Question 3
There are 10 edges and 6 vertices. The incidence matrix is 10 × 6, rank 5, right-null dimension 1 and left-null dimension 5.
With boundary values 𝐸 = 1, 𝑊 = 0, harmonic equations give
𝑉𝐴 = 41
79 , 𝑉 𝐵 = 44
79 , 𝑉 𝐶 = 45
79 , 𝑉 𝑋 = 56
79 .
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MATH40007 Revised Historical-Coverage Practice Pack Course-method solutions
After leaving 𝑊, the walker chooses 𝐴, 𝐵, 𝐶 equally, hence
𝑃(𝐸 before return 𝑊) = 41 + 44 + 45
3 ·79 = 130
237 .
With 𝑊 = 1, 𝐸 = 0, the voltage is the complement, so
𝑉𝑋 = 1 − 56
79 = 23
79 .
Question difficulty allocation: 𝐴 = 4, 𝐵 = 6, 𝐶 = 6, 𝐷 = 4.
Grounded Laplacian and effective resistance
Question 4
Nodes 0 and 4 are grounded. Internal nodes 1,2,3 are connected by the path 0 − 1 − 2 − 3 − 4 and there is one extra
unit edge 1 − 3.
0 1 2 3 4
ground ground
(a) Write the reduced Laplacian 𝐾. (4 marks)
(b) Find the Green matrix 𝑀 = 𝐾 −1. (8 marks)
(c) A unit current is injected at node 2 and removed through the grounds. Find the internal voltages and verify that
the total ground current is one. (4 marks)
(d) A unit current is injected at node 1 and removed at node 3. Find the effective resistance and conductance between
nodes 1 and 3. (4 marks) (Total: 20 marks)
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Worked solution and marking guidance
Question 4
The reduced Laplacian is
𝐾 =
[
3 −1 −1
−1 2 −1
−1 −1 3
]
.
Direct inversion gives
𝑀 = 𝐾 −1 =
[
5/8 1 /2 3 /8
1/2 1 1 /2
3/8 1 /2 5 /8
]
.
For 𝑓 = 𝑒2,
𝑥 = 𝑀𝑒 2 = (1/2, 1, 1/2)𝑇 .
The ground currents through 0 − 1 and 3 − 4 are 1/2 and 1/2, totalling one. For 𝑓 = 𝑒1 − 𝑒3,
𝑥 = 𝑀 𝑓 = (1/4, 0, −1/4)𝑇 .
The voltage difference is 1/2, so
𝑅13 = 1
2 , 𝐶 13 = 2 .
Question difficulty allocation: 𝐴 = 12, 𝐵 = 4, 𝐶 = 0, 𝐷 = 4.
Pack-level difficulty audit
Every paper has 80 marks and the same aggregate official-style distribution as the 2026 paper:
𝐴 = 32, 𝐵 = 20, 𝐶 = 12, 𝐷 = 16.
The content and question order vary across historical ability families; the equality is at paper-level difficulty, not at fixed 2026
topic slots.
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