Physical Sciences Paper 1: prelim problem solving

Built from the Grade 12 Physical Sciences Paper 1 prelim revision broadcast of 10 September 2026, 14:30 to 16:00, presented for the Gauteng Department of Education in collaboration with the Sci-Bono Discovery Centre. The lesson is about problem solving: not re-teaching the physics, but how to answer the questions. Every worked answer below has been recomputed independently, because the auto-caption track garbles numbers. Where the broadcast slips, the slip is flagged rather than silently fixed.

1. At a glance

Paper 1 is the physics paper, out of 150 marks. It covers four knowledge areas: mechanics, waves, sound and light (the Doppler effect), electricity and magnetism, and matter and materials (the photoelectric effect).

Covered in this broadcast	Worked example used
Newton's laws	Two crates joined by a string, pulled at an angle on a rough surface
Vertical projectile motion	One ball dropped, one projected upwards, both land together
Momentum and impulse	Crash test: a car with a crumple zone against one without
Work, energy and power	A crate pulled at 30° against friction
Doppler effect	An ambulance moving away; a red-shifted star
Electrostatics	Two point charges 0,03 m apart
Electric circuits	A battery with internal resistance and a parallel network with a switch

Not covered in this broadcast: electrodynamics (motors and generators, mentioned only in passing through P_ave = V_rmsI_rms) and the photoelectric effect. Revise those separately.

The three habits the presenter repeats all lesson.
1. Data collection first. Write down every given value before you choose a formula. The givens choose the formula for you.
2. Formula, substitution, answer, SI unit, and direction for vectors. Every calculation, every time.
3. Know the exam-guideline definitions word for word. They are free marks, and the data sheet will remind you of most of them.

2. Exam technique: reading the marks and the words

What the mark allocation is telling you

Marks	What the marker wants
Multiple choice	2 marks each, questions 1.1 to 1.10. Never leave one blank
A definition or law	Usually 2 marks. Exam-guideline wording, complete
3 marks, "calculate"	Formula, substitution, answer with its SI unit, plus direction if it is a vector
4 marks or more	A multi-step problem. Expect to calculate something first and use it
A free-body diagram	One mark per correctly labelled force. The marks tell you how many forces to look for

A formula without substitution is not marked. Copy the formula, then substitute into it. The formula mark is only given once you have shown the numbers going in. And even if you are unsure, write something logical: marks are often awarded for method.

The data sheet is your friend, for definitions too

The data sheet is not only for choosing formulas. Read a formula aloud and it reminds you of the definition:

• F_net = ma reminds you of Newton's second law: a net force on a mass causes acceleration.
• p = mv reminds you that momentum is the product of mass and velocity.
• F_netΔt = Δp reminds you that impulse is the product of the net force and the time it acts.
• E = F/q reminds you that the electric field at a point is the force per unit positive charge.

Copy formulas exactly as they appear on the data sheet, including the Doppler equation. Rearrange after you have written it down, not before.

Words in the question that give you a value

The word	What it tells you
rough surface	Friction acts
smooth / frictionless surface	No friction, or friction so small it is negligible
dropped	vi = 0 m·s⁻¹
starts from rest / initially at rest	vi = 0 m·s⁻¹
constant velocity	a = 0, so Fnet = 0. Look for this phrase in both Newton's laws and work-energy questions
forces in equilibrium / balanced	Fnet = 0: at rest OR moving at constant velocity
magnitude	Give the size only. No direction needed
isolated system	No net external force, so momentum (and, with no friction, mechanical energy) is conserved
only the horizontal forces	Leave out weight and the normal force on the free-body diagram

You may not use a highlighter in your answer book, but you may highlight on the question paper. Do it: mark every given value as you read.

Change is always final minus initial

Δv = v_f − v_i, and Δp = p_f − p_i = mv_f − mv_i. Never the other way round.

3. Graphs: gradient and area

Two rules cover every motion graph. The gradient is Δy / Δx. The area under the graph is Δy × Δx. So work out what the two axes multiply or divide to.

Graph	Gradient gives	Area under the graph gives
Position vs time	velocity	(not used)
Velocity vs time	acceleration	displacement
Acceleration vs time	(not used)	change in velocity, Δv = aΔt
Net force vs time	(not used)	impulse, FnetΔt = Δp

The triangle trap. If the area under the graph is a triangle, the area is ½ × base × height. Multiplying the peak force by the time without the half doubles your answer.

Worked MCQ from the broadcast: the area under an acceleration-time graph

A stone is thrown vertically downwards from the top of a building and strikes the ground t seconds later. Air friction is ignored. What does the shaded area between 0 and t seconds under its acceleration-time graph represent?

A  the final velocity of the stone
B  the change in position of the stone
C  the constant velocity of the stone
D  the change in velocity of the stone

Area = acceleration × time. Since a = Δv / Δt, cross-multiplying gives Δv = aΔt. So the area is the change in velocity. Not the final velocity: the stone was thrown, so it had an initial velocity. Not the change in position: that is the area under a velocity-time graph. Not a constant velocity: the stone is accelerating.

Answer: D

4. Newton's laws

The four laws to know word for word

Law	Statement
Newton's first law	A body will remain in its state of rest or motion at constant velocity unless a non-zero net (resultant) force acts on it.
Newton's second law	When a net force acts on an object, the object accelerates in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass of the object.
Newton's third law	When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A.
Newton's law of universal gravitation	Every particle with mass in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

First law or second law? Read for the net force

Fnet = 0: Newton's first law	Fnet ≠ 0: Newton's second law
The object is at rest, OR it is moving at constant velocity. Forces are balanced, in equilibrium.	Fnet = ma. The object accelerates in the direction of the net force.

Worked MCQ from the broadcast, question 1.1: forces in equilibrium on a moving object

Several forces act on a moving object. Which statement is correct when the forces are in equilibrium?

"In equilibrium" means F_net = 0. There are two ways F_net can be zero: the object is at rest, or it moves at constant velocity. The question says it is moving, so it cannot be at rest.

Therefore the object moves at constant velocity. Its velocity is not increasing: at constant velocity v_f = v_i, so Δv = 0, a = 0, and F_net = ma = 0.

Answer: the object is moving at constant velocity.

The problem-solving steps

1. Draw a free-body diagram for each body. A two-body system needs two diagrams. Read carefully which body the question asks for: do not draw Q when the question asks for P.
2. Represent the body as a dot and draw every force as an arrow from it, labelled.
3. Identify every force. A surface means a normal force, perpendicular to the surface. A mass means weight. A string means tension. A rough surface means friction, opposing motion. An applied force at an angle has a horizontal and a vertical component.
4. Choose a positive direction. The direction of motion is advised, but any choice works if you are consistent.
5. Write F_net = ma for one body, using only the forces along the direction of motion. On a horizontal surface, vertical forces are left out of F_net.
6. Pick the body that makes the maths easiest. If one body has an unknown angle, start with the other.

Worked example: the two crates (the full question is on the Exam tab)

Crate P (1,25 kg) is joined by a light string to crate Q (2 kg) on a rough horizontal surface. A force F of 7,5 N acts on Q at an angle θ to the horizontal. The crates accelerate at 0,1 m·s⁻² to the right. Friction on P is 1,8 N and on Q is 2,2 N.

Free-body diagram of P: four forces. Normal force N (up), weight F_g (down), tension T (to the right, along the string), friction f (to the left, opposing motion).

Tension: start with P, not Q. Q involves the unknown angle; P does not. Right is positive.
F_net = ma
T − f = ma
T − 1,8 = (1,25)(0,1)
T = 1,93 N (1,925 N)

Angle θ: now use Q. Only the horizontal component F cos θ pulls Q along the surface. Tension on Q acts to the left, back along the string, and so does friction.
F cos θ − T − f = ma
7,5 cos θ − 1,93 − 2,2 = (2)(0,1)
7,5 cos θ = 4,33
θ = 54,74°

Check: with the unrounded T = 1,925 N, θ = 54,78°. Either is accepted; the difference is only rounding.

5. Vertical projectile motion

Two definitions not to confuse

Term	Definition	The trap
Projectile	An object which has been given an initial velocity and then moves under the influence of the gravitational force only.	A projectile is an object
Free fall	A motion during which the only force acting on an object is the gravitational force.	Free fall is a motion, not an object. You will be penalised for writing it as an object

Choosing the equation

The four equations of motion are on the data sheet. Let the givens choose:

• Time is given or wanted: use an equation with Δt in it, such as Δy = v_iΔt + ½aΔt².
• Time is neither given nor wanted: use v_f² = v_i² + 2aΔy, which has no time in it.

The critical points and the symmetries

• Mark the point of launch, the maximum height (where v = 0), the point level with the launch, and the ground.
• Choose a sign convention and keep it: upward positive or downward positive.
• Time symmetry: time from launch up to the maximum height equals time from the maximum height back down to the launch level.
• Velocity symmetry: launched upward at 10 m·s⁻¹, the ball returns to the launch level at 10 m·s⁻¹ downward. Same speed, opposite direction.
• g is constant: 9,8 m·s⁻² downwards, all the way up, at the top, and all the way down.
• Consider the motion between any two points, A to B, whichever pair makes the question easiest.
• Be able to sketch and read position-time, velocity-time and acceleration-time graphs.

Worked MCQ from the broadcast: where is the diver's momentum greatest?

A person dives from a high platform into a pool. At which position is the magnitude of his momentum a maximum?

Momentum is p = mv, and the mass does not change, so the momentum is greatest where the speed is greatest. At the maximum height the velocity is zero, so the momentum is zero. By velocity symmetry, his speed when he passes the level of the platform on the way down equals his take-off speed. From there he keeps accelerating downwards at 9,8 m·s⁻².

Answer: just before he enters the water, where his speed is greatest.

Worked example: the two balls (the full question is on the Exam tab)

Ball A is dropped from the top of a 15,2 m building. After A has fallen 3,2 m, ball B is projected vertically upwards from the ground. Both balls strike the ground at the same time. Upward is positive.

Time for A to reach the ground. Data: v_i = 0 (dropped), Δy = −15,2 m, a = −9,8 m·s⁻², Δt = ?
Δy = v_iΔt + ½aΔt²
−15,2 = (0)Δt + ½(−9,8)Δt²
Δt = 1,76 s

B's launch speed: first find how long B is in the air. B only starts when A has already fallen 3,2 m.
Time for A to fall 3,2 m: −3,2 = ½(−9,8)Δt², so Δt = 0,81 s.
Time B is in the air = 1,76 − 0,81 = 0,95 s.

B goes up and comes back to the ground, so its displacement is Δy = 0.
0 = v_i(0,95) + ½(−9,8)(0,95)²
v_i = 4,66 m·s⁻¹

The question asks for the magnitude, so no direction is needed. Check: without rounding the times, v_i = 4,67 m·s⁻¹. Either is accepted.

Position-time graph, both balls on one set of axes, ground as the zero position:
• A starts at 15,2 m at t = 0 and curves down to 0 m.
• B starts at 0 m at t = 0,81 s, rises to about 1,1 m, and falls back.
• Both reach 0 m at t = 1,76 s.
Marks go to the starting time of each ball, the initial position of each ball, the time both strike the ground, the shapes of the two curves, and the labels A and B.

6. Momentum and impulse

The definitions

Term	Definition
Momentum	The product of an object's mass and its velocity. p = mv. A vector.
Newton's second law in terms of momentum	The net force acting on an object is equal to the rate of change of momentum of the object (in the direction of the net force). Fnet = Δp / Δt
Impulse	The product of the net force acting on an object and the time the net force acts on the object. FnetΔt
Isolated system	A system on which there is no net external force.
Principle of conservation of linear momentum	The total linear momentum of an isolated system remains constant (is conserved).
Elastic collision	Both total momentum and total kinetic energy are conserved.
Inelastic collision	Total momentum is conserved, but total kinetic energy is not: the total Ek before is not equal to the total Ek after.

The impulse trap. Do not define impulse as "the change in momentum". Impulse equals the change in momentum (F_netΔt = Δp), but that is the impulse-momentum theorem, not the definition. The definition is the product of the net force and the time for which it acts.

Link it all together: Newton's third law (equal and opposite forces on the two colliding objects), the change in momentum of each object, the average force each exerts, and Newton's second law in terms of momentum. Velocity, momentum, change in momentum and force are all vectors, so every answer about them needs a direction.

Why a crumple zone saves lives: the three-mark answer

1. The law and its formula: F_net = Δp / Δt, Newton's second law in terms of momentum.
2. The proportionality: for the same change in momentum, the net force is inversely proportional to the contact time.
3. The application: the crumple zone increases the contact time, so it decreases the net force on the occupants, which reduces injuries.

Airbags and seatbelts work on exactly the same principle.

Worked example: the crash test (the full question is on the Exam tab)

Car A (no crumple zone) and car B (crumple zone) both hit a wall at 15 m·s⁻¹ and stop. The cars are not identical. The force-time graphs are triangles.

Impulse on car A = area under its graph = ½ × base × height
= ½ (1,85)(16 500)
= 15 262,5 N·s, directed away from the wall, opposite to the car's motion.

Mass of car B. Impulse = area under B's graph = ½ (2,83)(11 850) = 16 767,75 N·s.
F_netΔt = Δp = mv_f − mv_i
Taking the car's direction of motion as positive, the impulse is −16 767,75 N·s:
−16 767,75 = m(0 − 15)
m = 1 117,85 kg

Notice what the graphs show: car B's crumple zone stretched the collision to 2,83 s and the peak force fell to 11 850 N, even though car B is the heavier car and had more momentum to lose.

A caution: the graphs were not visible in the transcript. The values 16 500 N, 1,85 s, 11 850 N and 2,83 s are as read aloud in the broadcast. They are confirmed by the result: they give car masses of about 1 000 kg, which is what real cars weigh.

7. Work, energy and power

This is a three-year course. The Grade 10 energy work still counts in Grade 12.

The definitions

Term	Definition
Gravitational potential energy	The energy an object has because of its position in the gravitational field relative to a reference point. Ep = mgh
Kinetic energy	The energy an object has because of its motion. Ek = ½mv²
Mechanical energy	The sum of the gravitational potential energy and the kinetic energy.
Principle of conservation of mechanical energy	The total mechanical energy in an isolated system remains constant.
Work done	W = FΔx cos θ, where θ is the angle between the force and the displacement.
Work-energy theorem	The work done by the net force on an object is equal to the change in the kinetic energy of the object. Wnet = ΔEk
Conservative force	A force for which the work done in moving an object between two points is independent of the path taken. Example: gravity, Fg.
Non-conservative force	A force for which the work done in moving an object between two points depends on the path taken. Examples: friction, tension, an applied force.
Power	The rate at which work is done, or the rate at which energy is transferred. P = W / Δt

Positive, negative or zero work

Force and displacement are ...	Angle	Work done
in the same direction	cos 0° = 1	positive
in opposite directions	cos 180° = −1	negative. Friction always does negative work on a moving object
perpendicular	cos 90° = 0	zero. The normal force and weight do no work on horizontal motion

Which energy principle, when

Principle	Use it when
Wnet = ΔEk	Any situation. Add the work done by every force
Wnc = ΔEk + ΔEp	Non-conservative forces (friction, an applied force) act. Leave gravity out of Wnc; it is already in ΔEp
(Ep + Ek)A = (Ep + Ek)B	Only in an isolated system, with no friction and no other non-conservative force. Only gravity acts
Pave = Fvave	The object moves at constant velocity

The calculation steps

1. Draw a free-body diagram to see which forces do positive work and which do negative work.
2. For a force at an angle, only the component along the displacement does work: F cos θ.
3. Convert to SI units. Divide km·h⁻¹ by 3,6 to get m·s⁻¹. Convert cm and mm to m.
4. Read the instruction: "draw the free-body diagram showing only the horizontal forces" means no weight and no normal force.

Worked example: the crate pulled at 30° (the full question is on the Exam tab)

A constant force F at 30° to the horizontal pulls a 10 kg crate from rest at A to B, 2 m along a horizontal surface. Friction is 10 N. The speed at B is 2 m·s⁻¹. "Using energy principles only", find F.

Option 1: the work-energy theorem. Only F and friction do work; the normal force and weight are perpendicular to the motion.
W_net = ΔE_k
FΔx cos 30° + fΔx cos 180° = ½mv_f² − ½mv_i²
F(2) cos 30° + (10)(2)(−1) = ½(10)(2)² − 0
1,732F − 20 = 20
F = 23,09 N

Option 2: W_nc = ΔE_k + ΔE_p. The surface is horizontal, so ΔE_p = 0, and the same equation results. F = 23,09 N.

A 2 kg object is now placed in the crate. What happens to the work done by F from A to B?
It remains the same. W = FΔx cos θ, and F, Δx and θ are all unchanged. The mass of the crate does not appear in the work done by F.

8. The Doppler effect

The presenter's warning: learners treat the Doppler effect as not worth much. "Please let us take it very much important." It is quick marks, with a definition, a formula straight off the data sheet, and a red-shift explanation.

The definition

The Doppler effect is the change in frequency (or pitch) of the sound detected by a listener because the sound source and the listener have different velocities relative to the medium of sound propagation.

Read that definition for what it implies: if the source and the listener have the same velocity, there is no Doppler effect.

The equation

f_L = (v ± v_L) / (v ± v_s) × f_s. Copy it exactly as it appears on the data sheet. It covers four scenarios, and in each one exactly one of the two is moving:

• a moving source approaching a stationary listener
• a moving source moving away from a stationary listener
• a moving listener approaching a stationary source
• a moving listener moving away from a stationary source

Motion	What the listener detects
Source and listener approaching	fL greater than fs: higher pitch
Source and listener moving apart	fL less than fs: lower pitch

Waves, pitch and the two wave equations

• For sound, use the wave equation v = fλ. For light, as in the photoelectric effect, use c = fλ.
• At a constant wave speed, frequency and wavelength are inversely proportional.
• Pitch and frequency go together. A higher-frequency sound has a higher pitch and a shorter wavelength.
• The speed of sound depends only on the medium, so it does not change when the source speeds up.

Worked example: the ambulance (the full question is on the Exam tab)

An ambulance moves away from a stationary listener at a constant 25 m·s⁻¹. Its siren emits 550 Hz. The listener detects 512,64 Hz. Calculate the speed of sound in air.

First, sanity-check the data. The listener hears 512,64 Hz, which is less than the 550 Hz being emitted. A lower detected frequency means the source is moving away, exactly as the question says.

Stationary listener: v_L = 0. Source moving away: + v_s in the denominator.
f_L = v / (v + v_s) × f_s
512,64 = v / (v + 25) × 550
512,64v + 12 816 = 550v
v = 343,04 m·s⁻¹

The ambulance now moves away faster than 25 m·s⁻¹. Increases, decreases or remains the same?
• The speed of sound in air: remains the same. It depends on the medium only.
• The frequency emitted by the siren: remains the same. The source still produces the same sound.
• The frequency detected by the listener: decreases. The faster the source moves away, the lower f_L.

Red shift and the expanding universe

The exam guideline covers red shift only, not blue shift. You must be able to use the Doppler effect to explain why we conclude that the universe is expanding.

Worked example: the red-shifted star

The spectrum of a distant star, viewed from Earth, is red-shifted. Is the star moving towards or away from Earth? Use the Doppler effect to explain.

Moving away from Earth.
• When a light source moves away from an observer, the observer detects a longer wavelength, which is a lower frequency.
• The spectral lines are therefore shifted towards the red end of the spectrum.
• Light from distant galaxies is red-shifted in every direction, so the galaxies are moving away from us: the universe is expanding.

9. Electrostatics

This builds on Grade 11: conservation of charge, Coulomb's law, electric fields, and the electric field at a point.

Term	Definition	Data sheet
Coulomb's law	The magnitude of the electrostatic force exerted by one point charge on another point charge is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.	F = kQ1Q2 / r²
Electric field	A region of space in which an electric charge experiences a force.
Electric field at a point	The electrostatic force experienced per unit positive charge placed at that point.	E = F / q, and E = kQ / r²

Drawing field patterns: the three marks

1. Correct shape. Unlike charges: lines run from one charge to the other. Like charges: lines bend away, with a gap between the charges.
2. Correct direction. Arrows point away from positive and towards negative.
3. Lines touch the charges, never each other, and never cross.

Electric field is a vector: add it like one

At a point with two or more charges nearby, find each field separately with E = kQ / r², decide its direction (away from a positive charge, towards a negative one), then add them as vectors. Fields in the same direction add; fields in opposite directions subtract. That is why a minus sign appears between the terms when the two fields point opposite ways.

Worked example: the two point charges (the full question is on the Exam tab)

X = −7,2 × 10⁻⁹ C and Y = +7,2 × 10⁻⁹ C are 0,03 m apart.

Field pattern: unlike charges, so the lines run from Y to X. The arrows point from Y towards X: away from the positive charge, towards the negative one.

Force that Y exerts on X:
F = kQ₁Q₂ / r²
= (9 × 10⁹)(7,2 × 10⁻⁹)(7,2 × 10⁻⁹) / (0,03)²
= 5,18 × 10⁻⁴ N, attraction, directed from X towards Y.

The distance was given in metres here. If it is given as 30 mm or 3 cm, convert it first.

Two cautions. The auto-caption garbled the charge values; they are read here as equal in magnitude, with X negative, which is what the field direction the presenter gives (from Y to X) requires. The broadcast then introduces a third charge Z and a resultant field at X of 4,91 × 10⁵ N·C⁻¹, and asks for the charge on Z. Z's position was only visible on screen, so no numerical answer is given for that part here. The method is on the Exam tab at question 7.4.

10. Electric circuits

Term	Definition
Potential difference	The work done per unit positive charge. V = W / q
Emf	The maximum energy provided (work done) by a battery per coulomb of charge passing through it.
Internal resistance	The resistance inside the battery itself. It causes the "lost volts", Vinternal = Ir.
Ohm's law	The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature.
Power	The rate at which work is done. P = W / Δt = VI = I²R = V² / R

ε = I(R + r), so ε = V_external + V_internal. Copy the resistor formulas from the data sheet even for a simple series sum: the formula mark is free.

Where P = I²R and P = V²/R come from: substitute V = IR into P = VI. In electrodynamics the same equations appear with rms values, P_ave = I_rms²R = V_rms²/R. They are not new formulas.

Resolving a network: work from the inside out

1. Resolve the innermost parallel group first. For two resistors, product over sum: R_p = R₁R₂ / (R₁ + R₂).
2. Add anything now in series with it.
3. Resolve the next parallel group outwards, and repeat.
4. Add the internal resistance only when you need the total resistance of the whole circuit.

Worked example: the circuit (the full question is on the Exam tab)

Battery: ε = 12 V, r = 0,5 Ω. R₁ = R₂ = 10 Ω in parallel, in series with a 10 Ω bulb. That branch is in parallel with R₃ = 15 Ω, which is in series with switch S. S is closed.

Total external resistance.
R₁ ∥ R₂ = (10 × 10) / (10 + 10) = 5 Ω
In series with the bulb: 5 + 10 = 15 Ω
In parallel with R₃: (15 × 15) / (15 + 15) = 7,5 Ω

Reading on A₁, the total current. Use totals only.
ε = I(R + r)
12 = I(7,5 + 0,5)
I = 1,5 A

Power dissipated in R₃. The two parallel branches are both 15 Ω, so the 1,5 A splits equally: 0,75 A each.
P = I²R = (0,75)²(15) = 8,44 W

S is now opened. What happens to the brightness of the bulb? It increases. The reasoning chain:
• Opening S removes a parallel branch, so the total resistance increases.
• So the total current decreases.
• So the lost volts, V_internal = Ir, decrease.
• So V_external = ε − Ir increases. The bulb's branch now has the whole external voltage.
• So the power delivered to the bulb, P = V²/R, increases, and the bulb is brighter.

Check: with S open, the current through the bulb is 12 / (15 + 0,5) = 0,77 A, up from 0,75 A.

A caution about the diagram: it is reconstructed from the working in the broadcast. The position of S, in the R₃ branch, is inferred from the answer: it is the only position in which opening S makes the bulb brighter. Had S been beside R₁ or R₂, the bulb would have dimmed.

11. Exam-day rules

1. Manage your time. Do not spend too long on one question. Move on and come back.
2. Start with the questions you enjoy. "Pay yourself first": bank the marks you are sure of, and the confidence that comes with them.
3. Show all your steps. Marks are awarded for method, even when the final answer is wrong.
4. Data collection, then formula, then substitution, then answer with the SI unit and direction.
5. Never leave a multiple-choice question or a calculation blank. Write something for every one of 1.1 to 1.10. For every "calculate", at least write the formula and substitute into it.
6. Stay calm. You have studied. Trust your preparation and your reasoning.

"Physics is not about memorising. It is about understanding how the world works." And the lesson's closing line, from Einstein: "Anyone who has never made a mistake has never tried anything new."

12. Where the broadcast slips

Each of these is reproduced faithfully above and corrected here, so that nothing on the screen misleads you.

What was said	What is correct
In the ambulance example: "the listener is receiving 512 ... so the frequency of the listener is greater than the frequency of the source, which means this is approaching".	512,64 Hz is less than 550 Hz, so the source is moving away, as the question itself states. The rule the presenter gives next is the right one: moving away, fL is less than fs.
In the circuit example: "V internal decreases because I is increased" and "if V internal increases the power output increases".	The current decreases, so Vinternal decreases, so Vexternal increases, so the bulb's power increases. The conclusion, brighter, is right.
On equilibrium: "the second condition for Newton's second law is that the object will move with a constant velocity".	Both conditions, at rest and constant velocity, belong to Newton's first law, where Fnet = 0.
On mechanical energy: "if you have to apply the principle, it must be when we have non-conservative force".	The opposite: apply conservation of mechanical energy only when there are no non-conservative forces, as he says correctly a moment earlier.
On the two balls' graph: "ball B will start up there at 15,2".	Ball A starts at 15,2 m at t = 0. Ball B starts on the ground, at 0 m, at t = 0,81 s.
The ambulance's listener frequency is heard as "5 ... 12 ... 64 Hz".	512,64 Hz. Confirmed by the answer: it is the only reading that gives the stated 343,04 m·s⁻¹.

13. One-page summary

Topic	Remember
Every calculation	Data collection. Formula. Substitute. Answer. SI unit. Direction for vectors
Mark allocation	3 marks = formula, substitution, answer. 4+ = multi-step. FBD: one mark per force
Graphs	Gradient = Δy/Δx, area = Δy × Δx. a-t area = Δv. F-t area = impulse. Triangles: ½bh
Newton	Fnet = 0: at rest or constant velocity. Fnet = ma otherwise. Rough = friction
Projectiles	Free fall is a motion; a projectile is an object. g = 9,8 m·s⁻² down, always. Dropped: vi = 0
Momentum	Fnet = Δp/Δt. Impulse is FnetΔt, which equals Δp. Crumple zone: more time, less force
Collisions	Elastic: p and Ek conserved. Inelastic: only p conserved
Work and energy	W = FΔx cos θ. Wnet = ΔEk. Wnc = ΔEk + ΔEp. Mechanical energy conserved only with no friction
Power	P = W/Δt. Pave = Fvave at constant velocity. km·h⁻¹ ÷ 3,6 = m·s⁻¹
Doppler	Approaching: fL up. Moving apart: fL down. Speed of sound depends on the medium only. Red shift only
Electrostatics	F = kQ1Q2/r², E = kQ/r². Field lines away from +, towards −, never crossing
Circuits	ε = I(R + r). Innermost parallel first. Fewer parallel paths: R up, I down, Ir down, Vext up