PAT practice · 20 questions · Free
PAT Exercise Paper 8: Physics and Mathematics
Every question carries a worked solution behind a disclosure, so you commit to an answer before you see the key.
[01] Momentum and collisions
Question 1
Two trolleys run along the same straight horizontal track. A $4.0 \ \text{kg}$ trolley moving at $6.0 \ \text{m s}^{-1}$ catches up with a $2.0 \ \text{kg}$ trolley moving in the same direction at $1.5 \ \text{m s}^{-1}$, and the two collide. Immediately afterwards the $4.0 \ \text{kg}$ trolley is still moving in the same direction, now at $3.0 \ \text{m s}^{-1}$. Take the track as horizontal and friction-free, treat the collision as happening along the line of motion so that every velocity lies on one straight line, and take all velocities in the ground frame. What is the speed of the $2.0 \ \text{kg}$ trolley immediately after the collision?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $7.5 \ \text{m s}^{-1}$ (option E)
Key idea. Total momentum is the same before and after any collision, whatever happens to the kinetic energy. Both trolleys are moving beforehand, so the total before has two terms, and the only unknown afterwards is the light trolley's velocity.
Fastest approach
1. Total momentum before.
$p = (4.0)(6.0) + (2.0)(1.5) = 24 + 3 = 27 \ \text{kg m s}^{-1}$
2. What the heavy trolley still carries.
$(4.0)(3.0) = 12 \ \text{kg m s}^{-1}$
3. The rest belongs to the light trolley.
$27 - 12 = 15 \ \text{kg m s}^{-1}$, so $v = \dfrac{15}{2.0} = 7.5 \ \text{m s}^{-1}$
Worth an audit afterwards. Kinetic energy before is $\tfrac{1}{2}(4)(6^{2}) + \tfrac{1}{2}(2)(1.5^{2}) = 74.25 \ \text{J}$, and after it is $\tfrac{1}{2}(4)(3^{2}) + \tfrac{1}{2}(2)(7.5^{2}) = 74.25 \ \text{J}$. None was lost, so these numbers describe an elastic collision. That is a consistency check on the answer, not an extra assumption: momentum alone fixed the result.
A second check costs nothing. The light trolley must end up faster than the heavy one, or the heavy trolley would be passing through it rather than falling behind it, and $7.5 > 3.0$.
Common mistake. Leaving the light trolley's own initial momentum out of the total because it was the one being hit. It was already moving at $1.5 \ \text{m s}^{-1}$ and so already carried $3 \ \text{kg m s}^{-1}$, and the sum before the collision has two terms, not one.
Why the other options are wrong
- A: Quoting the $3.0 \ \text{m s}^{-1}$ the stem gives for the heavy trolley after the collision, which is a value read off the question rather than one calculated from it. Misread Question
- B: Assuming the trolleys move off together, so that $v = 27/6.0 = 4.5 \ \text{m s}^{-1}$. That is a perfectly good speed and it contradicts the stem, which says the heavy trolley leaves at $3.0 \ \text{m s}^{-1}$: the two are not travelling at one common velocity. Conceptual Misunderstanding
- C: Leaving out the light trolley's initial momentum, giving $(24 - 12)/2.0 = 6.0 \ \text{m s}^{-1}$. Incomplete Calculation
- D: Dividing the whole momentum by the wrong mass, $27/4.0$, using the heavy trolley's mass for the light trolley's velocity. Formula Misapplication
[02] Electric circuits
Question 2
A transformer has $800$ turns on its primary coil and $40$ turns on its secondary. The primary is connected to a $240 \ \text{V}$ alternating supply and the secondary is connected to a heater that dissipates $60 \ \text{W}$. Assume the transformer is ideal, so it wastes no energy and the power drawn from the supply equals the power delivered to the heater, and take the supply voltage as fixed. What is the current in the primary coil?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $0.25 \ \text{A}$ (option A)
Key idea. An ideal transformer trades voltage for current at constant power. So the primary current follows from the primary side alone: $I_{p} = P/V_{p}$. The turns tell you what the secondary does, and here they are not needed at all.
Fastest approach
1. The power is the same on both sides.
The heater takes $60 \ \text{W}$, so the supply delivers $60 \ \text{W}$.
2. Divide by the primary voltage.
$I_{p} = \dfrac{P}{V_{p}} = \dfrac{60}{240} = 0.25 \ \text{A}$
3. Check it against the secondary.
$V_{s} = V_{p}\dfrac{N_{s}}{N_{p}} = 240 \times \dfrac{40}{800} = 12 \ \text{V}$, so $I_{s} = \dfrac{60}{12} = 5.0 \ \text{A}$.
The voltage was stepped down by twenty and the current stepped up by twenty, which is the trade the transformer makes. Three of the values on offer are a voltage, a bare ratio or a current twenty times too large, and one line of checking removes them: only $P/V_{p}$ has the units of an ampere and respects the energy the heater actually uses.
Common mistake. Applying the turns ratio to the current in the same direction as to the voltage. Turns down means volts down and amps up, because their product is fixed, so a step-down transformer draws a smaller current on the primary side and delivers a larger one on the secondary.
Why the other options are wrong
- B: The current in the secondary coil, $60/12 = 5.0 \ \text{A}$. That is the current through the heater, and the question asks about the primary. Misread Question
- C: The secondary voltage, $240 \times 40/800 = 12 \ \text{V}$, offered as a current. It is the right number for a quantity the question did not ask for, and it is measured in volts. Dimensional Error
- D: The turns ratio $800/40 = 20$, which is a pure number and cannot be a current at all. It is a step on the way and not an answer. Incomplete Calculation
- E: Stepping the current the same way as the voltage, $5.0 \times 20 = 100 \ \text{A}$. That is a genuine current, so nothing in the units objects, but at $240 \ \text{V}$ it would mean the supply delivering $24 \ \text{kW}$ to run a $60 \ \text{W}$ heater. Conceptual Misunderstanding
[03] Waves
Question 3
A narrow pipe of length $0.85 \ \text{m}$ is closed at one end and open at the other. Sound travels at $340 \ \text{m s}^{-1}$ in the air inside it. A standing wave forms with a displacement node at the closed end, where the air cannot move, and a displacement antinode at the open end, where it moves most freely. Ignore any end correction and take the pipe as narrow enough that only these along-the-pipe modes occur. What is the frequency of the lowest mode above the fundamental?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $300 \ \text{Hz}$ (option B)
Key idea. Sketch the pipe rather than reaching for a formula. A node at one end and an antinode at the other means the pipe holds a quarter of a wavelength, or three quarters, or five quarters, and never a half or a whole one. So $L = \frac{n\lambda}{4}$ with $n$ odd, and the mode above the fundamental is the third harmonic, not the second.
Fastest approach
1. The fundamental.
The shortest shape that fits is a quarter wavelength, so $\lambda_{1} = 4L = 3.4 \ \text{m}$ and
$f_{1} = \dfrac{v}{\lambda_{1}} = \dfrac{340}{3.4} = 100 \ \text{Hz}$
2. The next shape that fits.
Adding half a wavelength keeps the node at the closed end and the antinode at the open end, so the next mode holds three quarter wavelengths:
$\lambda = \dfrac{4L}{3} = \dfrac{3.4}{3} \ \text{m}, \qquad f = \dfrac{340 \times 3}{3.4} = 300 \ \text{Hz}$
The modes of a closed pipe therefore run $100, 300, 500, \ldots \ \text{Hz}$, three times and five times the fundamental. The even harmonics are missing because they would need an antinode where the closed end insists on a node, which is what gives a stopped organ pipe its hollow tone.
Common mistake. Treating the pipe as though it were open at both ends and doubling the fundamental. A pipe open at both ends has an antinode at each end and fits whole half-wavelengths, so its modes are $f_{1}, 2f_{1}, 3f_{1}$. One end being closed removes every even one.
Why the other options are wrong
- A: The fundamental itself, $v/4L$, rather than the mode above it. Misread Question
- C: Treating the pipe as open at both ends, where the fundamental is $v/2L = 200 \ \text{Hz}$ and the mode above it is $v/L = 400 \ \text{Hz}$. That is a perfectly good frequency for a different pipe: the closed end forces a node and removes the even harmonics entirely. Conceptual Misunderstanding
- D: The fifth harmonic, $5v/4L$. It does exist in this pipe, but it is two modes above the fundamental rather than one. Misread Question
- E: Taking the third mode to hold three whole wavelengths, $\lambda = L/3$, which gives $3v/L$. Each loop of a standing wave is half a wavelength, and the end sections here are quarters. Formula Misapplication
[04] Optics
Question 4
A narrow beam of light travelling in air meets the flat surface of a transparent liquid at $45^{\circ}$ to the normal, and inside the liquid it travels at $30^{\circ}$ to the normal. Take the refractive index of air as $1.00$, take the liquid to be uniform throughout, and take the light to be of a single colour so that dispersion can be ignored. What is the refractive index of the liquid?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\sqrt{2} \approx 1.41$ (option C)
Key idea. Snell's law compares the sines of the angles, never the angles themselves, and both angles are measured from the normal rather than from the surface. With air on the incident side, $n_{\text{liquid}} = \dfrac{\sin 45^{\circ}}{\sin 30^{\circ}}$.
Fastest approach
1. Write Snell's law with the given angles.
$n_{\text{air}}\sin 45^{\circ} = n_{\text{liquid}}\sin 30^{\circ}$
2. Rearrange and evaluate.
$n_{\text{liquid}} = \dfrac{\sin 45^{\circ}}{\sin 30^{\circ}} = \dfrac{\sqrt{2}/2}{1/2} = \sqrt{2} \approx 1.41$
The bound closes half the list before any arithmetic. The ray bends towards the normal on entering, so light travels more slowly in the liquid than in air, so the index must exceed $1$. Any value below $1$ would have light moving faster inside the liquid than outside it, which is not what a refracting medium does, and every value below $1$ on offer here comes from taking the ratio the wrong way up.
Common mistake. Taking the ratio of the angles rather than of their sines. That gives $45/30 = 1.5$, close enough to the true $1.41$ to pass unchallenged, and wrong for almost every other pair of angles: at $60^{\circ}$ and $30^{\circ}$ the two rules differ by more than fifteen per cent.
Why the other options are wrong
- A: Snell's law taken upside down, $\sin 30^{\circ}/\sin 45^{\circ}$. A refractive index below $1$ would mean light travelling faster in the liquid than in air, which contradicts the ray bending towards the normal. Conceptual Misunderstanding
- B: Measuring both angles from the surface rather than from the normal, so the ratio becomes $\sin 45^{\circ}/\sin 60^{\circ}$. Every angle in refraction is measured from the normal. Misread Question
- D: Taking the ratio of the angles themselves, $45/30$. Snell's law is a statement about sines, and the two agree here only by accident. Formula Misapplication
- E: Using tangents in place of sines, $\tan 45^{\circ}/\tan 30^{\circ} = \sqrt{3}$. Formula Misapplication
[05] Electric fields and charge
Question 5
Two large parallel metal plates are held $5.0 \ \text{cm}$ apart and connected to a $600 \ \text{V}$ supply, so that a uniform electric field fills the space between them. A small sphere carrying a charge of $40 \ \text{nC}$, where $1 \ \text{nC} = 10^{-9} \ \text{C}$, is held at rest between the plates. Take the field as uniform right up to the sphere, neglect edge effects and the field of the sphere itself, and take the sphere as small enough not to disturb the field. What is the magnitude of the electric force on the sphere?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $4.8 \times 10^{-4} \ \text{N}$ (option D)
Key idea. Between parallel plates the field is $E = V/d$, and the force on a charge sitting in a field is $F = qE$. Together, $F = qV/d$. Every prefix has to be converted once, before any of that: the separation into metres and the charge into coulombs.
Fastest approach
1. Convert first.
$d = 5.0 \ \text{cm} = 5.0 \times 10^{-2} \ \text{m}, \qquad q = 40 \ \text{nC} = 4.0 \times 10^{-8} \ \text{C}$
2. Field between the plates.
$E = \dfrac{V}{d} = \dfrac{600}{5.0 \times 10^{-2}} = 1.2 \times 10^{4} \ \text{V m}^{-1}$
3. Force on the charge.
$F = qE = (4.0 \times 10^{-8})(1.2 \times 10^{4}) = 4.8 \times 10^{-4} \ \text{N}$
The field does not depend on where between the plates the sphere sits, which is what uniform means, so no distance from either plate is needed or given. Note also that $qV$ on its own is $2.4 \times 10^{-5} \ \text{J}$, the work the field would do on this charge over the whole gap: an energy, not a force, and the difference is the metre.
Common mistake. Substituting the separation in centimetres, which makes the field a hundred times too small. Volts per metre is what $E = V/d$ delivers, so the length has to be in metres before the division, not after it.
Why the other options are wrong
- A: Multiplying by the separation instead of dividing by it, $qVd$, which carries units of joule metres rather than newtons. Dimensional Error
- B: Leaving the separation in centimetres, so the field comes out as $600/5.0 = 120$ rather than $1.2 \times 10^{4} \ \text{V m}^{-1}$. Unit Error
- C: Quoting $qV = 2.4 \times 10^{-5}$, which is the work done moving this charge from one plate to the other. It is an energy in joules, and it becomes a force only after division by the gap. Dimensional Error
- E: Reading the separation as $5.0 \ \text{mm}$ rather than $5.0 \ \text{cm}$, which makes the field ten times too strong. Order of Magnitude Error
[06] Estimation
Question 6
Estimate the number of molecules in one breath of air. Take the volume of a breath as $5.0 \times 10^{-4} \ \text{m}^{3}$, the density of the air in it as $1.2 \ \text{kg m}^{-3}$, and the average mass of one air molecule as $3.0 \times 10^{-26} \ \text{kg}$. Assume the air is uniform and that all of it is made of molecules of that average mass.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $2.0 \times 10^{22}$ (option E)
Key idea. A count is a division: the mass of the whole divided by the mass of one. The density is what turns the volume into a mass, and it is the step most easily dropped. Handle the powers of ten separately from the digits and the arithmetic is two lines.
Fastest approach
1. Mass of the breath.
$m = \rho V = 1.2 \times (5.0 \times 10^{-4}) = 6.0 \times 10^{-4} \ \text{kg}$
2. Divide by the mass of one molecule.
$N = \dfrac{6.0 \times 10^{-4}}{3.0 \times 10^{-26}} = 2.0 \times 10^{22}$
Digits first, then exponents: $6.0 \div 3.0 = 2.0$, and $10^{-4 - (-26)} = 10^{22}$.
Worth comparing with something known. A mole is $6 \times 10^{23}$ molecules, so this breath holds about a thirtieth of a mole, which sits sensibly beside the $0.6 \ \text{g}$ of air the first line gave it. One significant figure is all this estimate can honestly carry, since every input is a rounded typical value.
Common mistake. Dividing the volume straight by the mass of a molecule and never using the density. Cubic metres divided by kilograms is not a count of anything, and the check takes a second: only a mass divided by a mass leaves a pure number.
Why the other options are wrong
- A: Multiplying the mass of the breath by the mass of one molecule instead of dividing, which gives a mass squared, and a count of far less than one molecule in a whole breath. Dimensional Error
- B: Taking the molecular mass as $3.0 \times 10^{-23} \ \text{kg}$, a thousand times too heavy. Every power of ten in an estimate has to be carried deliberately, because nothing later in the calculation will object. Order of Magnitude Error
- C: Slipping a decimal in $1.2 \times 5.0 \times 10^{-4}$ and taking the mass of the breath as $6.0 \times 10^{-5} \ \text{kg}$. Arithmetic Slip
- D: Dividing the volume by the mass of one molecule, $5.0 \times 10^{-4} / 3.0 \times 10^{-26}$, with the density left out. That is a volume per unit mass rather than a number of molecules, and it lands close enough to the true answer to pass unnoticed. Dimensional Error
[07] Problem solving from supplied information
Question 7
A laser emits a narrow beam of power $5.0 \ \text{mW}$, that is, it delivers energy at a rate of $5.0 \times 10^{-3} \ \text{J s}^{-1}$. A beam delivering energy at a rate $P$ also delivers momentum at a rate $P/c$, where $c = 3.0 \times 10^{8} \ \text{m s}^{-1}$ is the speed of light. The beam falls squarely on a black surface that absorbs all of it, so none is reflected. Take the surface as fixed and at rest, and ignore any heating of it. What force does the beam exert on the surface?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $1.7 \times 10^{-11} \ \text{N}$ (option A)
Key idea. A force is the rate at which momentum is delivered. The stem hands over that rate directly, $P/c$, so the force is that expression and nothing else has to be recalled. Absorbed means the light stops, so the momentum delivered is what the beam carried, with no doubling.
Fastest approach
1. Say what a force is.
$F = \dfrac{\Delta p}{\Delta t}$, and the stem gives the momentum arriving per second as $P/c$.
2. Substitute once.
$F = \dfrac{P}{c} = \dfrac{5.0 \times 10^{-3}}{3.0 \times 10^{8}} = 1.7 \times 10^{-11} \ \text{N}$
The units settle the route on their own: a watt divided by a metre per second is a joule per metre, and a joule per metre is a newton. Multiplying by $c$ instead would give something a hundred million million million times larger and measured in watt metres per second, which is nothing at all.
The number is worth pausing on. It is roughly a billionth of the weight of a paperclip, which is why nobody notices being pushed by a torch beam, and also why a solar sail has to be enormous and patient.
Common mistake. Doubling the answer by treating the surface as a mirror. A reflected beam leaves with its momentum reversed, so it delivers twice as much, but this surface is black and absorbs everything: the light arrives and stops.
Why the other options are wrong
- B: Using $2P/c$, the value for a perfect mirror that sends the beam straight back. The stem says the surface absorbs all of the light, so its momentum is delivered once rather than reversed. Conceptual Misunderstanding
- C: Dividing by the $3.0$ and losing the factor of $10^{8}$ altogether. Order of Magnitude Error
- D: Quoting the beam's power, $5.0 \times 10^{-3} \ \text{W}$, as the force. A watt is a joule per second and a newton is a joule per metre, so the two differ by exactly the division the stem asks for. Dimensional Error
- E: Multiplying by the speed of light rather than dividing by it. Nothing about a milliwatt laser can produce a force of over a million newtons, which is roughly the thrust of a rocket engine. Dimensional Error
[08] Forces and equilibrium
Question 8
A block of mass $5.0 \ \text{kg}$ rests on a plane inclined at $30^{\circ}$ to the horizontal, held in place by a light rope that runs up the slope, parallel to the surface. Take $g = 10 \ \text{m s}^{-2}$, treat the plane as smooth so that no friction acts, take the rope as light and inextensible, and take the block as in equilibrium. What is the tension in the rope?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $25 \ \text{N}$ (option B)
Key idea. Resolve the weight along two axes chosen to suit the surface: one down the slope, one perpendicular to it. The rope is parallel to the slope, so it balances only the along-slope component $mg\sin\theta$; the perpendicular component $mg\cos\theta$ is balanced by the surface pushing back.
Fastest approach
1. The weight.
$W = mg = 5.0 \times 10 = 50 \ \text{N}$, straight down.
2. Its component along the slope.
$T = mg\sin\theta = 50 \times \sin 30^{\circ} = 50 \times 0.5 = 25 \ \text{N}$
Which of sine and cosine belongs here is settled by two cases whose answers are already known, and it is quicker than re-deriving the geometry. On flat ground, $\theta = 0$, the rope should be slack: $\sin 0 = 0$ and $\cos 0 = 1$, so it is the sine. On a vertical wall, $\theta = 90^{\circ}$, the rope should carry the entire weight: $\sin 90^{\circ} = 1$, and it does.
The two components are $25 \ \text{N}$ along the slope and $25\sqrt{3} \approx 43 \ \text{N}$ into it, and $\sqrt{25^{2} + 43.3^{2}} = 50 \ \text{N}$, the whole weight recovered.
Common mistake. Reaching for cosine because the angle is measured from the horizontal. The angle in the force triangle is not the angle in the picture: what settles it is that a gentle slope needs almost no tension, and only the sine does that.
Why the other options are wrong
- A: Resolving the mass rather than the weight, $5.0 \times \sin 30^{\circ}$, so $g$ never enters. Kilograms are not newtons: the answer is a mass wearing the wrong unit. Dimensional Error
- C: Using $mg\cos\theta$, which is the component pressing into the slope. It is a genuine force on the block and it is the one the surface balances, not the one the rope holds. Conceptual Misunderstanding
- D: Quoting the whole weight, as though the slope did nothing. That is the tension only when the slope is vertical. Incomplete Calculation
- E: Dividing by the sine instead of multiplying, $mg/\sin 30^{\circ}$, which makes the rope carry twice the block's weight on a gentle slope. Formula Misapplication
[09] Circular motion
Question 9
A puck of mass $1.0 \ \text{kg}$ slides on a smooth horizontal table, moving in a horizontal circle of radius $2.0 \ \text{m}$ on the end of a light string whose other end is held at the centre of the circle. The string snaps if the tension in it rises above $18 \ \text{N}$. Take $g = 10 \ \text{m s}^{-2}$, take the table as smooth so that friction is negligible, and treat the string as light and inextensible and the puck as a point mass. What is the greatest speed at which the puck can travel round the circle?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $6.0 \ \text{m s}^{-1}$ (option C)
Key idea. The string lies in the horizontal plane, so its tension is the entire centripetal force: $T = \dfrac{mv^{2}}{r}$. The weight acts vertically and the table pushes back on the puck with an equal vertical force, which is why $g$ is given here and never used.
Fastest approach
1. What supplies the centripetal force.
The only horizontal force on the puck is the tension, so $T = \dfrac{mv^{2}}{r}$.
2. Put in the largest tension the string survives.
$v_{\max}^{2} = \dfrac{T_{\max}r}{m} = \dfrac{18 \times 2.0}{1.0} = 36$, so $v_{\max} = 6.0 \ \text{m s}^{-1}$
3. Ask what happened to $g$.
It was given and never used. The puck's weight of $10 \ \text{N}$ acts straight down, the table pushes straight up with $10 \ \text{N}$, and neither has any component along the string. A quantity that drops out of an answer deserves a second's thought rather than a hunt for somewhere to put it: here it says the same puck on the same string would reach the same speed on the Moon.
Common mistake. Making the string support the weight as well as turn the puck, and so subtracting $mg$ from the breaking tension. The string is horizontal, so it has no vertical component at all, and a horizontal force cannot balance a vertical one.
Why the other options are wrong
- A: Dividing by the radius rather than multiplying by it, $\sqrt{T/(mr)}$. That combination carries units of one per second, a rate rather than a speed. Dimensional Error
- B: Subtracting the weight from the breaking tension, $\sqrt{(18 - 10)r/m}$, as though part of the string's strength were spent holding the puck up. It is a genuine speed, so nothing in the units objects; what is wrong is that the table carries the weight and the string pulls only horizontally. Conceptual Misunderstanding
- D: Quoting $T/m = 18$, the greatest centripetal acceleration the string allows, as though it were a speed. It is measured in metres per second squared. Dimensional Error
- E: Forgetting the square root and reporting $Tr/m = 36$, which is a speed squared. Dimensional Error
[10] Dimensional analysis
Question 10
A block of mass $m$ hangs from a light spring of stiffness $k$ and oscillates vertically. Assume the period $T$ can depend only on $m$ and on $k$, through $T = Cm^{\alpha}k^{\beta}$, where $C$ is a dimensionless constant that dimensions cannot supply and experiment gives as $2\pi$. The stiffness is defined by $F = kx$, the force needed to stretch the spring by $x$. Take $m = 0.10 \ \text{kg}$ and $k = 40 \ \text{N m}^{-1}$, assume the spring obeys Hooke's law throughout, and neglect the mass of the spring and any air resistance. What is the period of the oscillation?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $0.31 \ \text{s}$ (option D)
Key idea. A stiffness has no units of its own until its defining equation supplies them: $F = kx$ gives $[k] = \text{M T}^{-2}$. Matching mass and time in $T = Cm^{\alpha}k^{\beta}$ then forces $\alpha = \tfrac{1}{2}$ and $\beta = -\tfrac{1}{2}$, so $T = C\sqrt{m/k}$, with the size of $C$ beyond anything dimensions can say.
Fastest approach
1. Dimensions of the stiffness.
$k = \dfrac{F}{x} \implies [k] = \dfrac{\text{M L T}^{-2}}{\text{L}} = \text{M T}^{-2}$
2. Match the exponents.
$\text{T} = \text{M}^{\alpha}\left(\text{M T}^{-2}\right)^{\beta}$
Mass: $\alpha + \beta = 0$. Time: $-2\beta = 1$.
So $\beta = -\tfrac{1}{2}$ and $\alpha = \tfrac{1}{2}$, giving $T = C\sqrt{m/k}$.
3. Substitute once.
$T = 2\pi\sqrt{\dfrac{0.10}{40}} = 2\pi\sqrt{2.5 \times 10^{-3}} = 2\pi(0.050) = 0.31 \ \text{s}$
The check to run before the arithmetic is that $\text{kg}$ divided by $\text{N m}^{-1}$ is a second squared, so the square root is what makes this a time at all. Notice too that the derivation says nothing about the amplitude, and nothing about $g$: hanging the spring vertically shifts where the block sits but not how long it takes to get back.
Common mistake. Inverting the ratio and computing $\sqrt{k/m}$. That is a perfectly real quantity, the angular frequency, but it is measured in radians per second and a period is measured in seconds, so the two cannot be swapped.
Why the other options are wrong
- A: Reporting $m/k = 2.5 \times 10^{-3}$ with neither the square root nor the constant. The quotient is a time squared, so this is not yet a time. Dimensional Error
- B: $\sqrt{m/k} = 0.050 \ \text{s}$, the right form with the constant $C = 2\pi$ left out. Dimensions never supply that factor, which is why the stem has to state it. Incomplete Calculation
- C: Using $C = \pi$ rather than $2\pi$, which is the time for half a cycle: the block going from one extreme of its motion to the other, rather than returning to where it began. Conceptual Misunderstanding
- E: Inverting the ratio, $2\pi\sqrt{k/m}$. That quantity is the angular frequency in radians per second, not a period, and a spring this stiff carrying a mass this small plainly does not take two minutes to complete one oscillation. Dimensional Error
[11] Geometry
Question 11
A solid metal sphere of radius $3 \ \text{cm}$ is melted down and recast, with no loss of material, as a solid right circular cone whose base radius is also $3 \ \text{cm}$. What is the height of the cone?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $12 \ \text{cm}$ (option E)
Key idea. Melting conserves the material, so the two volumes are equal: $\frac{4}{3}\pi r^{3} = \frac{1}{3}\pi r^{2}h$. Every $\pi$ and two powers of $r$ cancel, leaving $h = 4r$, so the height is four radii whatever the radius happens to be.
Fastest approach
### 1. Sphere: $V = \frac{4}{3}\pi r^{3}$. Cone on a base of the same radius: $V = \frac{1}{3}\pi r^{2}h$.
### 2. Nothing is lost in the recasting, so set them equal:
$\frac{4}{3}\pi r^{3} = \frac{1}{3}\pi r^{2}h$
### 3. Cancel $\frac{1}{3}\pi r^{2}$ from both sides, which is everything the two solids have in common:
$4r = h$
### 4. Only now put the number in: $h = 4 \times 3 = 12 \ \text{cm}$.
Working in symbols to the last line pays here, because $h = 4r$ answers every version of this question at once, and it makes plain that the radius of the cone matching the radius of the sphere is what allows the cancelling.
Common mistake. Cancelling the shape factors along with the $\pi$ and the $r^{2}$, as though a sphere and a cone of the same radius held the same volume per unit height. The $\frac{4}{3}$ and the $\frac{1}{3}$ are the entire content of the question.
Why the other options are wrong
- A: Cancelling the $\frac{4}{3}$ against the $\frac{1}{3}$ as though they were the same factor, which leaves $h = r$. Formula Misapplication
- B: Using the cylinder volume $\pi r^{2}h$ for the cone, which loses the factor of a third and gives $h = \frac{4r}{3}$. Formula Misapplication
- C: Melting only a hemisphere, $\frac{2}{3}\pi r^{3}$, which gives $h = 2r$. Misread Question
- D: Writing the sphere's volume as $\pi r^{3}$ with the $\frac{4}{3}$ dropped, which gives $h = 3r$. Formula Misapplication
[12] Differentiation
Question 12
The curve $y = x^{2}(4-x)$ meets the $x$-axis where $x = 0$ and where $x = 4$. What is the gradient of the curve at the point where $x = 4$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $-16$ (option A)
Key idea. Expand before differentiating: $y = 4x^{2} - x^{3}$, so $\frac{\mathrm{d}y}{\mathrm{d}x} = 8x - 3x^{2}$. The derivative is a function of $x$, and the question wants its value at one point, not the value of $y$ there.
Fastest approach
### 1. Expand the product rather than reaching for the product rule:
$y = 4x^{2} - x^{3}$
### 2. Differentiate term by term:
$\frac{\mathrm{d}y}{\mathrm{d}x} = 8x - 3x^{2}$
### 3. Substitute $x = 4$:
$32 - 48 = -16$
Check the sign before anything else. For $x$ a little beyond $4$ the cubic term dominates and $y$ has already gone negative, so the curve is falling steeply there and a negative gradient is exactly what the shape demands. A positive answer here is wrong before its size is even considered.
Common mistake. Differentiating the two factors separately and multiplying them, $2x \times (-1)$, which is not the product rule. The product rule adds two terms, and expanding the brackets first avoids the question entirely.
Why the other options are wrong
- B: Multiplying the derivatives of the two factors, $2x \times (-1)$, in place of the product rule. Formula Misapplication
- C: Evaluating the derivative at the other point where the curve meets the axis, $x = 0$. Misread Question
- D: Sign slip on the cubic term, giving $3x^{2} - 8x$ and so $48 - 32$. Sign Error
- E: Differentiating $4x^{2}$ and leaving $-x^{3}$ untouched, giving $8x$ at $x = 4$. Incomplete Calculation
[13] Integration
Question 13
Evaluate $\displaystyle\int_{-2}^{2}\left(x^{3}+3x^{2}+5x+1\right)\,\mathrm{d}x$.
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $20$ (option B)
Key idea. Over an interval symmetric about zero, every odd power integrates to zero because the two halves cancel, and every even power contributes twice its value over the half interval. Here $x^{3}$ and $5x$ are both odd, so only $3x^{2}+1$ survives.
Fastest approach
### 1. The interval $[-2, 2]$ is symmetric about zero, so split the integrand by parity before integrating anything.
### 2. Odd part: $x^{3} + 5x$. Replacing $x$ by $-x$ flips its sign, so the two halves cancel exactly and it contributes nothing.
### 3. Even part: $3x^{2} + 1$. Its integral is twice the integral over $[0, 2]$.
### 4. $2\int_{0}^{2}\left(3x^{2}+1\right)\,\mathrm{d}x = 2\left[x^{3}+x\right]_{0}^{2} = 2(8+2) = 20$
Integrating all four terms directly gives $20$ as well, with two more antiderivatives to evaluate and two more chances to slip a sign.
Common mistake. Discarding $x^{3}$ as odd but keeping $5x$, which is odd as well. A term is odd exactly when replacing $x$ by $-x$ flips its sign, and that is true of every odd power, not only of the highest one.
Why the other options are wrong
- A: Integrating the even part over $[0, 2]$ and forgetting to double it. Incomplete Calculation
- C: Integrating the whole expression over $[0, 2]$ instead of over $[-2, 2]$. Misread Question
- D: Discarding $x^{3}$ as odd but keeping $5x$, giving $2\left[x^{3}+\frac{5x^{2}}{2}+x\right]_{0}^{2}$. Conceptual Misunderstanding
- E: Treating every term as even and doubling the integral of the whole expression over $[0, 2]$. Conceptual Misunderstanding
[14] Quadratics
Question 14
The curve $y = x^{2}-6x+4$ crosses the $x$-axis at two points. What is the larger of the two $x$-coordinates?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $3+\sqrt{5}$ (option C)
Key idea. The expression does not factorise over the integers, so complete the square. $x^{2}-6x+4 = (x-3)^{2}-5$, which puts the two roots symmetrically at a distance $\sqrt{5}$ either side of $x = 3$.
Fastest approach
### 1. Complete the square:
$x^{2}-6x+4 = (x-3)^{2} - 9 + 4 = (x-3)^{2} - 5$
### 2. Set it to zero: $(x-3)^{2} = 5$, so $x - 3 = \pm\sqrt{5}$.
### 3. $x = 3 \pm \sqrt{5}$, and the larger root is $3+\sqrt{5}$.
The two roots always straddle the line of symmetry $x = -\frac{b}{2a} = 3$, so their mean has to be $3$. That is a free check on any pair of roots, and it fails immediately for a numerator left undivided.
Common mistake. Rounding $\sqrt{5}$ to $2.24$ at the first opportunity and carrying a decimal through the remaining steps. Keeping the surd exact costs nothing, makes the symmetry about $x = 3$ visible, and leaves the answer in the form the question wants.
Why the other options are wrong
- A: The smaller root, where the question asks for the larger. Misread Question
- B: The line of symmetry, $x = \frac{6}{2}$, which is the mean of the two roots rather than either of them. Conceptual Misunderstanding
- D: Sign error inside the square root: using $b^{2}+4ac = 36+16$ instead of $b^{2}-4ac = 36-16$, giving $\frac{6+\sqrt{52}}{2}$. Sign Error
- E: Using the quadratic formula but dividing only part of the numerator by $2a$, so the whole numerator $6+2\sqrt{5}$ is reported. Formula Misapplication
[15] Probability
Question 15
A factory makes one component on two machines. Machine A makes $60\%$ of the components and $5\%$ of what it makes is defective; machine B makes the other $40\%$ and $10\%$ of what it makes is defective. A component is taken at random from the day's output and found to be defective. What is the probability that it came from machine A?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{3}{7}$ (option D)
Key idea. The condition changes the denominator. Once the component is known to be defective, the sample space is no longer the whole day's output but only the defective part of it, so the answer is machine A's defective output divided by the total defective output.
Fastest approach
### 1. Predict first. Machine A makes half again as many components as machine B but spoils them at half the rate, so the two defective streams are close in size and the answer should land a little below a half.
### 2. Take $1000$ components, which keeps every count a whole number. Machine A makes $600$, of which $30$ are defective. Machine B makes $400$, of which $40$ are defective.
### 3. Defective altogether: $30 + 40 = 70$.
### 4. $P(\text{machine A} \mid \text{defective}) = \frac{30}{70} = \frac{3}{7}$
That is just under a half, which is what the prediction called for, so the two agree and neither had to be trusted alone.
Common mistake. Quoting $5\%$, which is the probability that a component is defective given that machine A made it. That is the conditional the other way round, and the two agree only when the two machines contribute equally to the defective output, which is exactly what is not the case here.
Why the other options are wrong
- A: Stopping at $0.6 \times 0.05$, the probability that a component is both from machine A and defective, without dividing by the probability that it is defective at all. Incomplete Calculation
- B: Quoting the $5\%$ defect rate, which is the probability of a defect given machine A rather than of machine A given a defect. Conceptual Misunderstanding
- C: Reporting $0.03+0.04$, the probability that a randomly chosen component is defective, which is the denominator rather than the answer. Misread Question
- E: Computing the probability that the defective component came from machine B instead. Misread Question
[16] Graphs
Question 16
Two quantities are believed to satisfy $y = ax^{n}$, where $a$ and $n$ are constants. Plotting $\log_{10} y$ against $\log_{10} x$ gives a straight line of gradient $3$ whose intercept on the vertical axis is $2$. What is the value of $y$ when $x = 100$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $10^{8}$ (option E)
Key idea. Taking logarithms of $y = ax^{n}$ gives $\log_{10} y = n\log_{10} x + \log_{10} a$, which is a straight line in the variables $\log_{10} x$ and $\log_{10} y$. So the gradient is the power $n$, and the intercept is $\log_{10} a$ rather than $a$ itself.
Fastest approach
### 1. Take logarithms of the proposed relationship:
$\log_{10} y = n\log_{10} x + \log_{10} a$
Compare with $Y = mX + c$: the gradient is $n$ and the intercept is $\log_{10} a$.
### 2. So $n = 3$ and $\log_{10} a = 2$, which gives $a = 10^{2} = 100$.
### 3. The relationship is $y = 100x^{3}$, so at $x = 100$:
$y = 100 \times 10^{6} = 10^{8}$
The same calculation done entirely in logarithms reads $\log_{10} y = 3(2) + 2 = 8$, which is why a log plot is worth drawing: the arithmetic collapses into adding two numbers.
Common mistake. Reading the intercept as $a$ rather than as $\log_{10} a$. The straight line is a statement about the logarithms, so every number read off it is a logarithm and has to be undone before it says anything about $y$.
Why the other options are wrong
- A: Treating the straight line as a relationship between $y$ and $x$ themselves, giving $y = 3x+2$. Conceptual Misunderstanding
- B: Taking $n = 3$ but ignoring the intercept, as though $a$ were $1$. Incomplete Calculation
- C: Reading the intercept as $a = 2$ rather than as $\log_{10} a = 2$, giving $y = 2x^{3}$. Conceptual Misunderstanding
- D: Swapping the roles of the gradient and the intercept, giving $y = 10^{3}x^{2}$. Formula Misapplication
[17] Geometry
Question 17
A solid right circular cone has base radius $6 \ \text{cm}$ and slant height $10 \ \text{cm}$. What is its volume?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $96\pi \ \text{cm}^{3}$ (option A)
Key idea. The volume formula $\frac{1}{3}\pi r^{2}h$ wants the perpendicular height, and the question supplies the slant height instead. A sketch of the cross section through the axis turns that into a right angled triangle with legs $r$ and $h$ and the slant height as hypotenuse.
Fastest approach
### 1. Sketch the cross section through the axis: an isosceles triangle whose base is the diameter and whose two sloping sides are each the slant height.
### 2. Half of that sketch is a right angled triangle with legs $6$ and $h$ and hypotenuse $10$, so
$h = \sqrt{10^{2}-6^{2}} = \sqrt{64} = 8 \ \text{cm}$
The $6$, $8$, $10$ triangle is the tell, and the sketch is what makes it visible.
### 3. $V = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi(36)(8) = 96\pi \ \text{cm}^{3}$
Common mistake. Adding inside Pythagoras rather than subtracting, which gives $h = \sqrt{136}$. The slant height is the hypotenuse, so it is the largest of the three lengths and the perpendicular height has to come out smaller than $10 \ \text{cm}$; an irrational height here is a signal that the subtraction went the wrong way.
Why the other options are wrong
- B: Using the slant height as the perpendicular height: $\frac{1}{3}\pi(36)(10)$. Conceptual Misunderstanding
- C: Omitting the factor of a third, which is the volume of the cylinder on the same base and height rather than of the cone. Formula Misapplication
- D: Both slips together: no factor of a third, and the slant height used as the perpendicular height. Formula Misapplication
- E: Using the diameter $12 \ \text{cm}$ in place of the radius: $\frac{1}{3}\pi(144)(8)$. Misread Question
[18] Quadratics
Question 18
A rectangular sheet of card measures $10 \ \text{cm}$ by $8 \ \text{cm}$. A square of side $x \ \text{cm}$ is cut from each of the four corners and the four flaps are folded up to make an open box. The base of the box has area $24 \ \text{cm}^{2}$. What is $x$?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $2$ (option B)
Key idea. A square is cut from both ends of every side, so each dimension loses $2x$ and not $x$. The base measures $(10-2x)$ by $(8-2x)$, and setting that product to $24$ gives a quadratic that factorises.
Fastest approach
### 1. Write the base dimensions: $(10-2x)$ by $(8-2x)$.
### 2. $(10-2x)(8-2x) = 24$, so $80 - 36x + 4x^{2} = 24$.
### 3. Divide by $4$: $x^{2}-9x+14 = 0$, which factorises as $(x-2)(x-7) = 0$.
### 4. So $x = 2$ or $x = 7$. Cutting $7 \ \text{cm}$ squares would remove $14 \ \text{cm}$ from a side only $8 \ \text{cm}$ long, so that root describes no box at all and $x = 2 \ \text{cm}$.
Check: the base is $6 \times 4 = 24 \ \text{cm}^{2}$, as required.
Common mistake. Offering both roots. The algebra knows nothing about the card, so a root has to be tested against the situation that produced it: here every dimension of the base must stay positive, which forces $x$ below $4 \ \text{cm}$.
Why the other options are wrong
- A: Setting the perimeter of the base to $24$ rather than its area: $2(18-4x) = 24$. Misread Question
- C: Removing $x$ rather than $2x$ from each dimension, giving $x^{2}-18x+56 = 0$, and taking its smaller root. Conceptual Misunderstanding
- D: The root the card forbids: $7 \ \text{cm}$ squares would consume $14 \ \text{cm}$ of an $8 \ \text{cm}$ side. Conceptual Misunderstanding
- E: Removing $x$ rather than $2x$ from each dimension and then taking the larger root of $x^{2}-18x+56 = 0$. Formula Misapplication
[19] Probability
Question 19
In a year group of $200$ students, $120$ study physics, $80$ study music and $30$ study both. A student is chosen at random from the year group and turns out to study music. What is the probability that this student also studies physics?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $\frac{3}{8}$ (option C)
Key idea. Knowing that the student studies music discards the other $120$ students outright. The denominator is therefore $80$, not $200$: the question has become one about the music students alone, $30$ of whom also study physics.
Fastest approach
### 1. Fill in all four cells. Both subjects: $30$. Music only: $80-30 = 50$. Physics only: $120-30 = 90$. Neither: $200-30-50-90 = 30$.
### 2. The music row holds $80$ students, and that row is now the whole sample space.
### 3. $P(\text{physics} \mid \text{music}) = \frac{30}{80} = \frac{3}{8}$
Written as a ratio of probabilities it is $\frac{P(\text{both})}{P(\text{music})} = \frac{30/200}{80/200}$, the same division with the $200$ cancelling. Naming the denominator before dividing is the whole discipline here.
Common mistake. Dividing the $30$ by the whole year group of $200$. That answers a different question, namely the chance that a student picked from the entire year studies both subjects, and it throws away the information the question has just supplied.
Why the other options are wrong
- A: Dividing $30$ by the whole year group of $200$, which ignores the conditioning information entirely. Conceptual Misunderstanding
- B: Dividing $30$ by the $120$ physics students, which is the probability of music given physics. Conceptual Misunderstanding
- D: Reporting $\frac{120}{200}$, the share of the whole year group studying physics, as though studying music told you nothing. Misread Question
- E: Using the $50$ students who study music but not physics, $\frac{50}{80}$, which is the complement of the answer. Misread Question
[20] Numerical reasoning
Question 20
Each person in a room states the month of their birthday. Assume nothing about how birthdays are distributed: the answer must hold however the months happen to fall. What is the smallest number of people in the room that guarantees at least three of them share a birth month?
Work this one out as a number before you read the options.
Reveal the answer and worked solution. Commit to an option first.
Answer: $25$ (option D)
Key idea. Guarantee means the worst arrangement, not the likely one. Bracket the answer: twelve months holding two people each is $24$ people with no month holding three, so $24$ is certainly not enough, and one more person has nowhere to go but a month that already holds two.
Fastest approach
### 1. Turn the question round: how many people can be in the room with no month holding three of them?
### 2. At most two per month, across twelve months, is $2 \times 12 = 24$ people. That arrangement is genuinely possible, so $24$ is a lower bound that fails: the answer is certainly more than $24$.
### 3. Add one person. Whichever month they name already holds two, so that month now holds three, and this happens whatever they say. So $25$ works: the answer is certainly no more than $25$.
### 4. The two bounds meet, so the smallest guaranteeing number is $25$.
The general pattern for $k$ people sharing one of $n$ categories is $(k-1)n + 1$, and it is worth rebuilding this way each time rather than recalling it, because the worst case is the near miss and the answer is always one step past it.
Common mistake. Answering $13$, which is the smallest number guaranteeing that two people share a month. That is the same argument with two in place of three, and here the worst case holds two per month rather than one, so the near miss is $24$ and not $12$.
Why the other options are wrong
- A: The number of months, as though filling every month once already forced a repeat. Conceptual Misunderstanding
- B: The answer to the two people version of the question, $12+1$. Misread Question
- C: The largest number that can avoid a triple, $2\times12$, without the extra person who forces one. Incomplete Calculation
- E: Multiplying by three, the number wanted in a single month, rather than by two, the number the worst case can hold in each. Formula Misapplication