Mechanics and materials: A-level Physics AQA revision notes

3.4.1

Force, energy and momentum

Vectors and perpendicular components 3.4.1.1

Definitions

Scalar: fully described by magnitude alone — e.g. mass, energy, temperature, speed.
Vector: needs magnitude and direction — e.g. displacement, velocity, acceleration, force, momentum.

Key results

Resolving into components: $F_{x} = F cos θ$ and $F_{y} = F sin θ$ .
Recombining components: $F = F_{x}^{2} + F_{y}^{2}$ , with direction $θ = tan^{- 1} (\frac{F _{y}}{F _{x}})$ .
Equilibrium at a point: $Σ F_{x} = 0$ and $Σ F_{y} = 0$ simultaneously.

Notes

Confusing a scalar with a vector is one of the most common sources of sign errors in mechanics — a car that changes direction at constant speed has zero change in speed but a large change in velocity, because velocity is a vector.
Component addition is almost always faster than adding vectors graphically once more than two vectors are involved — sum all the $x$ -components and all the $y$ -components separately before recombining.
Weight resolved on a slope gives $m g cos θ$ perpendicular to the slope and $m g sin θ$ along it — the two dashed components together are equivalent to the single weight $m g$ , not extra forces added on top of it.
A common exam trap: forgetting the sign convention when resolving. Decide once (e.g. right and up are positive) and keep every component consistent with it throughout a single calculation.

Moments, couples and equilibrium 3.4.1.2

Definitions

Moment: $moment = F \times d$ , where $d$ is the perpendicular distance from the pivot to the force's line of action.
Couple: a pair of equal, opposite, parallel forces separated by a distance $d$ , producing a turning effect with zero resultant force.

Key results

Principle of moments (rotational equilibrium): sum of clockwise moments about any point $=$ sum of anticlockwise moments about that same point.
Moment of a couple: $F d$ .
Full rigid-body equilibrium requires both $Σ F = 0$ AND total moment $= 0$ simultaneously.

Notes

For an object in equilibrium under three forces, those forces form a closed triangle when added tip-to-tail — if the triangle does not close, the object is not in equilibrium.
A couple has zero resultant force (so it can never cause linear acceleration by itself) but a nonzero turning effect — the clearest illustration that force balance and moment balance are two independent conditions, not one implying the other.
An object can have zero resultant force yet still rotate if the moments about some point do not also balance — both conditions must hold for full static equilibrium.

Motion along a straight line 3.4.1.3

Key results

SUVAT equations (constant acceleration): $v = u + a t$ ; $s = u t + \frac{1}{2} a t^{2}$ ; $v^{2} = u^{2} + 2 a s$ ; $s = \frac{( u + v )}{2} t$ .
Displacement-time graph: tangent gradient at a point = instantaneous velocity there (average velocity over an interval is $Δ s /Δ t$ ). Velocity-time graph: gradient = acceleration, signed area under the graph = displacement. Acceleration-time graph: signed area = change in velocity.

Method

Required practical 3: measure the time for an object to fall a known height $h$ from rest; plot $h$ against $t^{2}$ (since $h = \frac{1}{2} g t^{2}$ from rest) — the gradient of the resulting straight line is $\frac{g}{2}$ .

Notes

Each SUVAT equation is derivable from the other three — which one you reach for depends on which quantity is absent from the question.
The graphical gradient/area relationships hold even when acceleration is not constant, which the SUVAT equations alone cannot handle.
Plotting $h$ against $t^{2}$ rather than $h$ against $t$ turns a curved relationship into a straight line, making $g$ readable directly from the gradient — the same 'linearise before plotting' technique used throughout this course's data-analysis topic.

Projectiles and motion through a fluid 3.4.1.4

Definitions

Terminal velocity: the constant speed reached when resistive (drag) force exactly balances the driving force (e.g. weight), so resultant force and acceleration are both zero.

Key results

Projectile launched at angle $θ$ : horizontal $v_{x} = u cos θ$ (constant, $a_{x} = 0$ ); vertical $v_{y} = u sin θ - g t$ ( $a_{y} = - g$ ).

Notes

Projectile motion treats horizontal and vertical motion as independent: horizontal velocity is constant (no horizontal force, ignoring air resistance) while vertical motion obeys the SUVAT equations under gravity's constant acceleration — the two motions share only time as a common variable, usually the key to solving a projectile problem.
At the very top of a projectile's path, vertical velocity is momentarily zero, but vertical acceleration is still $- g$ throughout — a common confusion is to assume acceleration is also zero at the peak.
With significant air resistance, a real trajectory has reduced maximum height and range compared with the no-drag case, and loses its ideal parabolic symmetry (a steeper descent than ascent).
As speed increases from rest, drag grows until it exactly equals the driving force (e.g. weight for a falling object) — resultant force and acceleration then fall to zero, and the object continues at its constant terminal speed.

Newton's laws and force diagrams 3.4.1.5

Definitions

Newton's first law: an object's velocity stays constant unless a resultant force acts on it.
Newton's third law: forces come in equal and opposite pairs acting on different objects.

Key results

Newton's second law: $F = ma$ (constant mass); first law: $Σ F = 0$ gives constant velocity (including zero).

Notes

Common error: pairing a weight with a normal contact force acting on the SAME object as a Newton's-third-law pair — they are not, even though they may happen to balance in magnitude.
A genuine third-law pair always acts on two different objects, is the same type of force, and is equal in magnitude but opposite in direction.
For a uniform, regularly-shaped object (like a rectangular block), the centre of mass sits at its geometric centre — the single point through which weight can be treated as acting for force diagrams and moments.

Momentum, collisions and explosions 3.4.1.6

Definitions

Elastic collision: kinetic energy is conserved as well as momentum. Inelastic collision: kinetic energy is not conserved (some is transferred to internal energy).

Key results

Momentum: $p = m v$ , conserved in any isolated system: $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ .

Notes

In an elastic collision, both momentum and kinetic energy are the same before and after — e.g. a 1 kg mass at $4 m s^{- 1}$ striking an identical stationary 1 kg mass can exchange velocities entirely, conserving both momentum ( $4 kg m s^{- 1}$ throughout) and kinetic energy ( $8 J$ throughout).
In a perfectly inelastic collision, the two objects move off together afterwards; momentum is still conserved, but kinetic energy is not — some is transferred to internal energy (e.g. $8 J \to 4 J$ , with $4 J$ lost to internal energy, for two 1 kg masses that stick together).
In an explosion, momentum before is zero (if the system starts at rest), so the resulting momenta must be equal and opposite — kinetic energy is not conserved either, but increases, since stored energy (chemical, elastic, etc.) converts into kinetic energy of the fragments.
Momentum conservation solves collisions and explosions that would otherwise need detailed knowledge of the forces involved during the interaction — only the total momentum before and after needs to be equal.

Impulse and contact time 3.4.1.6

Definitions

Impulse: $J = F Δ t$ for a constant force, or the area under a force-time graph in general — equal to the change in momentum, $J = Δ p$ .

Key results

Newton's second law (general form): $F_{net} = \frac{d p}{d t}$ , reducing to $F = ma$ only when mass is constant.
Mean resultant force $= \frac{Δ p}{Δ t}$ .

Notes

For the same change in momentum (the same impulse), a longer contact time produces a smaller mean force, and a shorter contact time produces a larger mean force — e.g. a $10 N s$ impulse delivered over $0.10 s$ needs a mean force of $100 N$ , but delivered over $0.20 s$ needs only $50 N$ .
Extending the time of a collision (an airbag, a crumple zone, cushioning) is exactly this principle applied deliberately — it reduces the peak force experienced for the same unavoidable change in momentum.

Work, energy and power 3.4.1.7–3.4.1.8

Key results

Work done by a constant force: $W = F s cos θ$ — equivalently, the area under a graph of the force component parallel to displacement, against displacement, even when that force is not constant.
Work-energy theorem: $W = Δ E_{k} = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$ .
Power: mean power $= \frac{Δ W}{Δ t}$ ; instantaneous power of a force $F$ : $P = F v cos θ$ .
Efficiency: $η = \frac{useful energy output}{total energy input}$ (multiply by 100 for a percentage).
Principle of conservation of energy: the total energy of an isolated system is constant — energy transfers between forms (kinetic, gravitational potential, elastic potential, internal/thermal, etc.), but the total before any process equals the total after.
Gravitational potential energy: $E_{p} = m g h$ . Kinetic energy: $E_{k} = \frac{1}{2} m v^{2}$ .

Notes

A force perpendicular to motion does no work at all — a satellite in a circular orbit experiences a constant gravitational force, but that force does zero work on it.
Energy is always conserved overall, even when it is not conserved in a single convenient form — a falling object with air resistance converts its gravitational PE partly into kinetic energy and partly into internal (thermal) energy of the object and the air, but the total stays equal to the PE lost.
Conservation of energy applies equally to qualitative reasoning (tracking which forms energy moves between and why) and to quantitative calculation (setting total initial energy equal to total final energy to find an unknown speed, height, or heat generated) — both are routinely examined, often within the same question.
The work-energy theorem lets a change in kinetic energy be found without ever calculating acceleration or time — useful whenever a problem gives distances and forces but not a clean SUVAT setup.
A motor lifting a load at steady speed has useful power output $\frac{m g h}{t}$ — a direct, frequently examined application of both the power and gravitational PE equations together.

Worked examples

Worked example 3.4.1 · 4 marks

A crate of weight 250 N rests in equilibrium on a smooth plane inclined at 20° to the horizontal, held in place by a rope running parallel to the plane.

Calculate the tension in the rope and the normal reaction force from the plane.

Show worked solution

Resolving the weight along the plane:

250 sin 20° \approx 85.5 N

which equals the tension since the plane is smooth and the crate is in equilibrium along the plane, so $T \approx 85.5 N$ .

Resolving perpendicular to the plane:

N = 250 cos 20° \approx 234.9 N \approx 235 N

Mark scheme · 4 marks

Resolves the weight's component parallel to the plane: $250 sin 20° \approx 85.5 N$ 1 mark
Recognises tension equals this parallel component for equilibrium along a smooth plane, giving $T \approx 85.5 N$ 1 mark
Resolves the weight's component perpendicular to the plane: $250 cos 20° \approx 234.9 N$ 1 mark
States the normal reaction $N \approx 235 N$ , equal and opposite to the perpendicular weight component 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.4.1 · 5 marks

A uniform plank of length 4.0 m and weight 120 N rests horizontally on supports at each end, A and B.

A person of weight 600 N stands 1.0 m from support A.

Calculate the reaction force at each support.

Show worked solution

The plank's weight (120 N) acts at its centre, 2.0 m from A.

Taking moments about A:

R_{B} \times 4.0 = 120 \times 2.0 + 600 \times 1.0 = 840 N m

so $R_{B} = 210 N$ .

Vertical equilibrium:

R_{A} + R_{B} = 120 + 600 = 720 N

so:

R_{A} = 720 - 210 = 510 N

Mark scheme · 5 marks

Takes moments about support A 1 mark
Includes the plank's own weight (120 N) acting at its centre, 2.0 m from A 1 mark
Solves $R_{B} \times 4.0 = 120 \times 2.0 + 600 \times 1.0 = 840$ to get $R_{B} = 210 N$ 1 mark
Applies vertical equilibrium $R_{A} + R_{B} = 120 + 600 = 720 N$ 1 mark
Calculates $R_{A} = 720 - 210 = 510 N$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.4.1 · 5 marks

$h$ / m	$t$ / s	$t^{2}$ / s²
0.200	0.202	0.0408
0.400	0.286	0.0815
0.600	0.350	0.1223
0.800	0.404	0.1631
1.000	0.452	0.2039

In Required Practical 3, a student releases a ball from rest and measures the mean time $t$ taken to fall each height $h$ , repeating each measurement to reduce random error.

The results are shown in the table.

Use the data to determine $g$ , and suggest one improvement to reduce random error.

Show worked solution

Since:

h = \frac{1}{2} g t^{2}

from rest, a graph of $h$ against $t^{2}$ is a straight line through the origin with gradient $\frac{g}{2}$ .

Using the first and last data points, gradient:

= \frac{1.000 - 0.200}{0.2039 - 0.0408} \approx 4.91 m s^{- 2}

so:

g \approx 2 \times 4.91 \approx 9.81 m s^{- 2}

in close agreement with the accepted value.

Reaction time in starting/stopping a stopwatch is a significant random error here; repeating each drop several times and averaging $t$ (or using light gates instead of a stopwatch) would reduce it.

Mark scheme · 5 marks

States $h = \frac{1}{2} g t^{2}$ means a graph of $h$ against $t^{2}$ is a straight line through the origin with gradient $\frac{g}{2}$ 1 mark
Calculates the gradient from two widely-separated data points, e.g. $\frac{1.000 - 0.200}{0.2039 - 0.0408} \approx 4.91 m s^{- 2}$ 1 mark
Doubles the gradient to find $g \approx 9.81 m s^{- 2}$ 1 mark
Notes this agrees closely with the accepted value of $g$ 1 mark
Identifies reaction-time error as a key random error and proposes repeating/averaging each drop or using light gates to reduce it 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.4.1 · 4 marks

A ball is kicked horizontally at 12 m s⁻¹ from the top of a cliff 20 m high.

Calculate the time taken to reach the ground, the horizontal distance travelled, and the ball's speed as it lands. (:

g = 9.81 m s^{- 2}

)

Show worked solution

Vertically:

20 = \frac{1}{2} (9.81) t^{2} \Rightarrow t \approx 2.02 s

Horizontally (constant velocity):

x = 12 \times 2.02 \approx 24.2 m

Vertical landing speed:

v_{y} = g t = 9.81 \times 2.02 \approx 19.8 m s^{- 1}

Landing speed:

v = 1 2^{2} + 19. 8^{2} \approx 23.2 m s^{- 1}

Mark scheme · 4 marks

Uses vertical SUVAT $20 = \frac{1}{2} (9.81) t^{2}$ to find $t \approx 2.02 s$ 1 mark
Calculates horizontal range using constant horizontal velocity: $x = 12 \times 2.02 \approx 24.2 m$ 1 mark
Calculates vertical landing speed $v_{y} = g t \approx 19.8 m s^{- 1}$ 1 mark
Combines components with Pythagoras to get landing speed $\approx 23.2 m s^{- 1}$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.4.1 · 5 marks

A 1200 kg car travelling at 15 m s⁻¹ collides with a stationary 800 kg car, and the two vehicles lock together on impact.

Calculate their common velocity immediately after the collision and the kinetic energy lost in the collision.

Show worked solution

Momentum conservation:

1200 \times 15 + 800 \times 0 = (1200 + 800) v \Rightarrow v = \frac{18000}{2000} = 9.0 m s^{- 1}

$E_{k}$ before:

= \frac{1}{2} (1200) (15)^{2} = 135, 000 J

$E_{k}$ after:

= \frac{1}{2} (2000) (9.0)^{2} = 81, 000 J

Energy lost:

= 135, 000 - 81, 000 = 54, 000 J

(54 kJ), transferred to internal/heat/sound energy since the collision is inelastic.

Mark scheme · 5 marks

Applies conservation of momentum: $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v$ 1 mark
Calculates common velocity $v = \frac{1200 \times 15}{2000} = 9.0 m s^{- 1}$ 1 mark
Calculates kinetic energy before collision: $\frac{1}{2} (1200) (15)^{2} = 135, 000 J$ 1 mark
Calculates kinetic energy after collision: $\frac{1}{2} (2000) (9.0)^{2} = 81, 000 J$ 1 mark
Calculates energy lost $= 54, 000 J$ , transferred to internal/heat/sound energy as the collision is inelastic 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.4.2

Materials

Density, stress and strain 3.4.2.1

Definitions

Density: $ρ = \frac{m}{V}$ , measured in $kg m^{- 3}$ .
Stress: $σ = \frac{F}{A}$ , force per unit cross-sectional area.
Strain: $ε = \frac{Δ L}{L _{0}}$ , extension per unit original length — strain has no unit.

Notes

Stress and strain both normalise force and extension so that samples of different sizes can be compared fairly.
$F$ here is the tension in the specimen itself, not the sum of the two equal-and-opposite forces pulling at each end — for a specimen in equilibrium under tension, both end forces share the same magnitude $F$ .
Always use the ORIGINAL (unloaded) cross-sectional area $A$ and original length $L_{0}$ when calculating stress and strain, even once the specimen has extended — this is what makes them well-defined, comparable quantities rather than ones that silently redefine themselves as the specimen deforms.

Elastic and plastic behaviour 3.4.2.1

Definitions

Limit of proportionality: the point beyond which force is no longer proportional to extension (Hooke's law stops holding), though the deformation may still be elastic just beyond it.
Elastic limit: the point beyond which the material no longer returns to its original length when unloaded — deformation beyond here is at least partly plastic.

Key results

Hooke's law: $F = k Δ L$ , up to the limit of proportionality; $k$ (the spring constant) is a property of the specific spring or wire, not a universal material constant.

Notes

The limit of proportionality and the elastic limit are two distinct points, often close together but not identical — a material can, in principle, still return to its original length (be elastic) slightly beyond the point where $F$ against $Δ L$ stops being a straight line.
Elastic deformation is fully recovered on unloading — the material returns to its original length — and does not have to be linear (Hooke's law) to still count as elastic.
Plastic deformation leaves a permanent extension once the load is removed — the unloading line on a force-extension graph runs parallel to, but offset from, the original loading line.

Elastic strain energy and energy transfers 3.4.2.1

Key results

Elastic strain energy stored (area under a straight-line force-extension graph): $E = \frac{1}{2} F Δ L = \frac{1}{2} k (Δ L)^{2}$ — valid for a linear elastic spring or wire.

Notes

$\frac{1}{2} F Δ L$ no longer gives the strain energy beyond the elastic limit, because the loading and unloading paths on the graph separate.
Elastic strain energy converting entirely to kinetic energy (a compressed spring released, negligible losses): $\frac{1}{2} k (Δ L)^{2} = \frac{1}{2} m v^{2}$ .
Elastic strain energy converting entirely to gravitational PE (a spring launching a mass upward, negligible losses): $\frac{1}{2} k (Δ L)^{2} = m g h$ .
In an energy-absorbing structure (a car's crumple zone), the work done is deliberately mostly plastic, not elastic — plastic work is not recovered on unloading, and a longer crumple distance lowers the mean resisting force needed to remove the same kinetic energy, the same 'spread the impulse over more time/distance' principle as impulse and contact time.

Stress, strain and material properties 3.4.2.1–3.4.2.2

Definitions

Young modulus: $E = \frac{σ}{ε} = \frac{F L _{0}}{A Δ L}$ , a genuine material constant, unlike the spring constant $k$ .

Key results

Spring constant of a uniform wire in terms of material and geometry: $k = \frac{E A}{L _{0}}$ — doubling $L_{0}$ halves $k$ ; doubling $A$ doubles $k$ .
Breaking stress $= \frac{F _{fracture}}{A}$ — this can be lower than the maximum tensile stress the material briefly sustained.

Notes

The Young modulus is found from the gradient of the initial, straight-line section of a stress-strain graph — steeper gradient means a stiffer material.
Stiff: large $E$ , little strain for a given stress. Strong: a high stress is needed to cause fracture. Brittle: little plastic strain occurs before fracture — three genuinely independent material properties; a material can be strong without being stiff, or stiff without being strong.
A brittle material's stress-strain curve rises close to a straight line almost all the way to fracture, with very little plastic curvature beforehand; a ductile material shows a substantial curved, plastic region before it finally breaks.

Measuring the Young modulus of a wire 3.4.2.2

Method

Required practical 4: clamp a long, thin test wire rigidly at one end, apply a small preload to straighten it, and record its unstretched length $L$ from the clamp to a reference marker.
Add known masses in steps; at each step, record the added force $Δ F = Δ m g$ and the corresponding extension $Δ L$ (using a travelling microscope or similar to measure the marker's displacement precisely).
Measure the wire's diameter $d$ with a micrometer at several positions and orientations along its length, and average, to find $A = \frac{π d ^{2}}{4}$ .
Plot $Δ F$ against $Δ L$ : the gradient of the linear region gives the wire's spring constant, and $E = \frac{L}{A} \times gradient$ .
Unload the wire at the end to check it has remained within its elastic limit throughout (i.e. returns to its original length) — if not, the measurements are unreliable.

Notes

Converting all extension measurements from millimetres to metres before calculating $E$ is a common and easily-missed source of error, given $E$ 's SI unit (pascals) requires SI inputs throughout.
Example result: $L = 2.00 m$ , $d = 0.400 mm$ giving $A \approx 1.26 \times 1 0^{- 7} m^{2}$ , and a graph gradient of $12.5 N mm^{- 1} = 1.25 \times 1 0^{4} N m^{- 1}$ , giving $E = \frac{L}{A} \times gradient \approx 1.99 \times 1 0^{11} Pa$ .

Worked examples

Worked example 3.4.2 · 4 marks

A copper wire of original length 1.50 m and diameter 0.60 mm is stretched by a force of 15 N, producing an extension of 1.2 mm.

Calculate the stress and strain in the wire.

Show worked solution

A = \frac{π d ^{2}}{4} = \frac{π ( 0.60 \times 1 0 ^{- 3} ) ^{2}}{4} \approx 2.83 \times 1 0^{- 7} m^{2}

Stress:

σ = \frac{F}{A} = \frac{15}{2.83 \times 1 0 ^{- 7}} \approx 5.31 \times 1 0^{7} Pa

Strain:

ε = \frac{Δ L}{L _{0}} = \frac{1.2 \times 1 0 ^{- 3}}{1.50} \approx 8.0 \times 1 0^{- 4}

(no units).

Mark scheme · 4 marks

Calculates cross-sectional area $A = \frac{π d ^{2}}{4} \approx 2.83 \times 1 0^{- 7} m^{2}$ 1 mark
Calculates stress $σ = \frac{F}{A} \approx 5.31 \times 1 0^{7} Pa$ 1 mark
Calculates strain $ε = \frac{Δ L}{L _{0}} \approx 8.0 \times 1 0^{- 4}$ 1 mark
States that strain is dimensionless (no units), being a ratio of two lengths 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.4.2 · 4 marks

A spring of spring constant 250 N m⁻¹ is compressed by 0.12 m and used to launch a 0.050 kg ball vertically upward.

Assuming all the elastic strain energy converts to gravitational potential energy, calculate the maximum height reached by the ball.

Show worked solution

Elastic strain energy:

E = \frac{1}{2} k (Δ L)^{2} = \frac{1}{2} (250) (0.12)^{2} = 1.8 J

Setting this equal to the gravitational PE gained:

\frac{1}{2} k (Δ L)^{2} = m g h \Rightarrow h = \frac{1.8}{0.050 \times 9.81} \approx 3.7 m

Mark scheme · 4 marks

Calculates elastic strain energy stored: $E = \frac{1}{2} k (Δ L)^{2} = \frac{1}{2} (250) (0.12)^{2} = 1.8 J$ 1 mark
States that all elastic strain energy converts to gravitational PE at maximum height (energy conservation, no losses) 1 mark
Sets $\frac{1}{2} k (Δ L)^{2} = m g h$ and rearranges for $h$ 1 mark
Calculates $h = \frac{1.8}{0.050 \times 9.81} \approx 3.7 m$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.4.2 · 6 marks

$F$ / N	$Δ L$ / mm
10.0	1.23
20.0	2.46
30.0	3.68
40.0	4.91
50.0	6.14

In Required Practical 4, a student measures the extension $Δ L$ of a metal wire of original length $L_{0} = 2.50 m$ and diameter $d = 0.36 mm$ for increasing load $F$ , staying within the elastic region throughout.

The results are shown in the table.

Use the data to determine the Young modulus of the wire, and state one check the student should carry out to confirm the wire remained within its elastic limit.

Show worked solution

A = \frac{π d ^{2}}{4} = \frac{π ( 0.36 \times 1 0 ^{- 3} ) ^{2}}{4} \approx 1.02 \times 1 0^{- 7} m^{2}

Gradient of $F$ against $Δ L$ (in metres): using the first and last points,

\frac{50.0 - 10.0}{( 6.14 - 1.23 ) \times 1 0 ^{- 3}} \approx 8.15 \times 1 0^{3} N m^{- 1}

Young modulus:

E = \frac{gradient \times L _{0}}{A} = \frac{8.15 \times 1 0 ^{3} \times 2.50}{1.02 \times 1 0 ^{- 7}} \approx 2.0 \times 1 0^{11} Pa

Mark scheme · 6 marks

Calculates cross-sectional area $A = \frac{π d ^{2}}{4} \approx 1.02 \times 1 0^{- 7} m^{2}$ 1 mark
Converts extension values from mm to m before using them in the gradient calculation 1 mark
Calculates the gradient of $F$ against $Δ L$ using two widely-separated data points, $\approx 8.15 \times 1 0^{3} N m^{- 1}$ 1 mark
Uses $E = \frac{gradient \times L _{0}}{A}$ 1 mark
Calculates $E \approx 2.0 \times 1 0^{11} Pa$ 1 mark
States that the wire should be unloaded at the end to check it returns to its original length, confirming the elastic limit was not exceeded 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.