Unit 1 High-Yield Topics

Unit 1 High-Yield Topics | Flowxiom

Unit 1 High-Yield Topics

Free resource by Flowxiom — Edexcel A-level Physics

Not everything. Just what’s on the paper. High-frequency topics only — covering ~80% of exam marks.

Edexcel A-level Physics | Mechanics & Materials | WPH11 & WPH12


Topic 1: Vector Resolution and Resultant Forces

Question types: Inclined plane force analysis, rope tension equilibrium, magnitude and direction of resultant.

Key Formulae — Resolving a Force

For force \(F\) at angle \(\theta\) from horizontal:

\[F_x = F\cos\theta \qquad F_y = F\sin\theta\]

Adjacent side uses cos, opposite side uses sin (θ measured from horizontal)

Four-Step Method for Resultant

Resolve all forces into horizontal and vertical components — right/up positive

Sum horizontal: \(R_x = \sum F_x\); sum vertical: \(R_y = \sum F_y\)

Magnitude: \(F_R = \sqrt{R_x^2 + R_y^2}\)

Direction: \(\tan\alpha = \dfrac{|R_y|}{|R_x|}\) — state as angle from horizontal

Equilibrium Condition

For three forces in equilibrium: they must be concurrent and form a closed triangle.

Common Mistakes

  • ❌ sin/cos confused — always draw the triangle and mark θ clearly
  • ❌ Forgetting to include weight — vertical equilibrium must also hold
  • ❌ Direction stated as a number only — must specify “angle from horizontal”

Topic 2: Motion Graphs

Question types: Read graph to identify motion; calculate displacement/acceleration from graph; sketch graph conversions.

Golden Rules

GraphGradient =Area =
\(s\)-\(t\)velocity \(v\)
\(v\)-\(t\)acceleration \(a\)displacement \(s\)
\(a\)-\(t\)change in velocity \(\Delta v\)

\[s \xrightarrow{\text{gradient}} v \xrightarrow{\text{gradient}} a \qquad a \xrightarrow{\text{area}} \Delta v \xrightarrow{\text{area}} \Delta s\]

Common v-t Graph Shapes

ShapeMeaning
Horizontal straight lineUniform velocity (\(a = 0\))
Straight line sloping upUniform acceleration
Straight line sloping downUniform deceleration
Steepening curveIncreasing acceleration
Crosses time axisVelocity reverses direction

Common Mistakes

  • ❌ Area under v-t graph = displacement (can be negative); not the same as distance
  • ❌ Curved s-t graph means changing speed, not necessarily projectile motion
  • ❌ Gradient of v-t = acceleration; area of v-t = displacement (not the other way round)

Topic 3: SUVAT Equations

Question types: Given a motion description, find velocity, displacement, time or acceleration.

The Four Equations

EquationMissing quantity
\(v = u + at\)no displacement
\(s = \dfrac{1}{2}(u+v)t\)no acceleration
\(s = ut + \dfrac{1}{2}at^2\)no final velocity
\(v^2 = u^2 + 2as\)no time

Method

List all five quantities \(s, u, v, a, t\); fill in knowns, mark unknown with ?

Identify 3 knowns + 1 target → select correct equation

Substitute — take care with signs for direction

Key Phrase Translations

Phrase in questionValue
from rest / starts from rest\(u = 0\)
comes to rest / stops\(v = 0\)
deceleration / retardation\(a\) is negative
free fall / dropped\(a = 9.81\ \text{m s}^{-2}\), \(u = 0\)
thrown upward\(a = -9.81\ \text{m s}^{-2}\) (taking up as positive)

Common Mistakes

  • ❌ Deceleration: \(a\) must be negative
  • ❌ If “thrown” not “dropped”, initial velocity is not zero
  • ❌ Mixed units — convert first (cm → m, km/h → m/s)

Topic 4: Projectile Motion

Question types: Horizontal or angled launch — find time of flight, range, velocity at landing.

Core Principle

Horizontal and vertical motions are completely independent; time is the only link.

HorizontalVertical
Acceleration\(a_x = 0\)\(a_y = -9.81\ \text{m s}^{-2}\)
Velocityconstant: \(v_x = u\cos\theta\)changes — use SUVAT
Displacement\(s_x = v_x \times t\)use SUVAT

Solution Framework

Vertical (solve first)          Horizontal (solve after)
s_y = ?                         s_x = ?
u_y = u sinθ                    u_x = u cosθ
a_y = −9.81                     a_x = 0
→ find t                        s_x = u_x × t

Common Mistakes

  • ❌ Horizontal velocity is constant — do not apply SUVAT to it
  • ❌ Landing speed requires combining both components: \(v = \sqrt{v_x^2 + v_y^2}\)
  • ❌ Horizontal component uses \(\cos\theta\); vertical uses \(\sin\theta\)

Topic 5: Momentum and Impulse

Question types: Collision calculations, bouncing ball, F-t graph, explosion and recoil.

Key Formulae

\[p = mv \qquad \text{(momentum — vector)}\]

\[\text{Impulse} = F\Delta t = \Delta p = mv – mu\]

Conservation of Momentum

Net external force = 0 (isolated system) → total momentum is conserved:

\[m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\]

Elastic vs Inelastic Collisions

MomentumKinetic energy
Elasticconserved ✓conserved ✓
Inelasticconserved ✓not conserved ✗ (converted to heat/sound)
Perfectly inelasticconserved ✓maximum loss

F-t Graph

\[\text{Area under graph} = \text{impulse} = \Delta p\]

Common Mistakes

  • Sign errors for direction — most common mark loss

Example: ball hits wall at \(10\ \text{m/s}\), bounces back at \(8\ \text{m/s}\):

\(\Delta p = m(-8) – m(+10) = -18m\), not \(-2m\)

  • ❌ After a perfectly inelastic collision, both objects share the same final velocity
  • ❌ Momentum is always conserved in collisions; kinetic energy is only conserved in elastic collisions

Topic 6: Work, Energy and Power

Question types: Work calculations, energy conservation, power-velocity relationship.

Key Formulae

\[W = Fs\cos\theta \qquad E_k = \frac{1}{2}mv^2 \qquad \Delta E_p = mg\Delta h\]

\[P = \frac{W}{t} = Fv \qquad \text{efficiency} = \frac{\text{useful output}}{\text{total input}} \times 100\%\]

When Work Done = 0

Force perpendicular to displacement (\(\theta = 90°\)) → \(W = 0\)

Example: carrying an object horizontally — gravity does no work.

Energy Conservation

  • No friction/air resistance → mechanical energy conserved: \(E_k + E_p = \text{constant}\)
  • With friction → energy lost = friction force × distance (converted to thermal energy)

F-s Graph

\[\text{Area under graph} = \text{work done} = W\]

Common Mistakes

  • ❌ Omitting \(\cos\theta\) when force and displacement are not parallel
  • ❌ In \(P = Fv\), \(v\) must be the component of velocity in the direction of force
  • ❌ Efficiency fraction inverted

Topic 7: Newton’s Laws and Moments

Question types: Free-body diagrams, equilibrium conditions, moment calculations.

Conditions for Equilibrium (both must hold simultaneously)

Resultant force = 0 in all directions

Sum of moments = 0 about any point

Moment Pivot Trick

Take moments about the point where an unknown force acts — its moment = 0, eliminating one unknown from the equation.

Common Mistakes

  • ❌ Omitting the moment of one of the forces
  • ❌ Not consistently assigning + or − to clockwise vs anticlockwise

Topic 8: Solid Materials and Young Modulus

Question types: Calculate Young modulus, identify key points on stress-strain graph, classify material type.

Key Formulae

\[\sigma = \frac{F}{A} \quad \text{(stress, Pa)} \qquad \varepsilon = \frac{\Delta L}{L} \quad \text{(strain — dimensionless)}\]

\[E = \frac{\sigma}{\varepsilon} = \frac{FL}{A\Delta L} \quad \text{(Young modulus, Pa)}\]

\[F = k\Delta x \qquad E_{el} = \frac{1}{2}F\Delta x = \frac{1}{2}k(\Delta x)^2\]

Area under F-x graph = elastic potential energy stored

Key Points on the Stress-Strain Graph

PointNameMeaning
End of straight lineLimit of ProportionalityAbove this, Hooke’s law no longer holds
First point after curveElastic LimitAbove this, permanent deformation occurs
Sudden large extensionYield PointSudden large plastic deformation
Highest pointUTSMaximum stress the material can withstand
End of graphBreaking PointFracture

Read Young modulus from the gradient of the straight-line section: \(E = \Delta\sigma / \Delta\varepsilon\)

Three Material Types

TypeGraph featureExamples
DuctileLarge plastic region before fractureCopper, mild steel
BrittleFractures at or near elastic limitGlass, cast iron
PolymericNon-linear; loading and unloading paths differRubber

Common Mistakes

  • ❌ Strain has no unit — it is a dimensionless ratio
  • ❌ Use extension \(\Delta L\), not total length \(L\)
  • ❌ Area = \(\pi r^2\) — remember to halve the diameter to get radius
  • ❌ Elastic limit and limit of proportionality are different points (close but distinct)

Practice Questions

Q1. A wire of original length \(2.0\ \text{m}\) and diameter \(1.0\ \text{mm}\) extends by \(0.50\ \text{mm}\) under a force of \(80\ \text{N}\). Find the Young modulus.

Answer

\(A = \pi(0.5\times10^{-3})^2 = 7.85\times10^{-7}\ \text{m}^2\)

\(\sigma = 80 / 7.85\times10^{-7} = 1.02\times10^8\ \text{Pa}\)

\(\varepsilon = 0.50\times10^{-3} / 2.0 = 2.5\times10^{-4}\)

\(E = \sigma/\varepsilon = \mathbf{4.1\times10^{11}\ \text{Pa}}\)


Q2. A \(5.0\ \text{N}\) force acts at \(37°\) to the horizontal. Find the horizontal and vertical components. (\(\sin37°=0.60\), \(\cos37°=0.80\))

Answer

\(F_x = 5.0\cos37° = \mathbf{4.0\ \text{N}}\)

\(F_y = 5.0\sin37° = \mathbf{3.0\ \text{N}}\)


Q3. A v-t graph shows a straight line from \(v=12\ \text{m/s}\) at \(t=0\) to \(v=0\) at \(t=4\ \text{s}\). Find the acceleration and displacement.

Answer

Acceleration = gradient \(= (0-12)/4 = \mathbf{-3.0\ \text{m s}^{-2}}\) (deceleration)

Displacement = area of triangle \(= \frac{1}{2}\times4\times12 = \mathbf{24\ \text{m}}\)


Q4. A car brakes from \(30\ \text{m/s}\) to rest with deceleration \(5.0\ \text{m/s}^2\). Find the braking distance.

Answer

\(v=0\), \(u=30\), \(a=-5.0\). Use \(v^2 = u^2 + 2as\):

\(0 = 900 + 2(-5.0)s \Rightarrow s = \mathbf{90\ \text{m}}\)


Q5. A ball is launched horizontally at \(6\ \text{m/s}\) from a height of \(20\ \text{m}\). Find the time of flight and horizontal range. (\(g = 10\ \text{m s}^{-2}\))

Answer

Vertical: \(20 = \frac{1}{2}(10)t^2 \Rightarrow t = 2.0\ \text{s}\)

Horizontal: \(s_x = 6 \times 2.0 = \mathbf{12\ \text{m}}\)


Q6. A \(0.2\ \text{kg}\) ball hits a wall at \(5\ \text{m/s}\) and bounces back at \(3\ \text{m/s}\). Find the magnitude of the impulse.

Answer

Taking initial direction as positive:

\(\Delta p = 0.2 \times (-3-5) = -1.6\ \text{N s}\)

Magnitude \(= \mathbf{1.6\ \text{N s}}\)


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Free resource by Flowxiom — Edexcel A-level Physics
High-frequency topics only, covering ~80% of exam marks.