AQA Physics Paper 2

Equipment: a trolley on a flat runway, a pulley fixed at the end of the runway, a length of string connecting the trolley to hanging masses over the pulley, two light gates connected to a data logger, a metre rule, and a set of masses. Method: Set up one light gate near the start of the trolley's path and a second light gate further along; measure the distance between them with the metre rule. Transfer masses from the trolley to the hanger to provide a known force — this keeps the total system mass constant, which is the control variable. Allow the trolley to accelerate from rest through both light gates. The data logger records the velocity at each gate; calculate acceleration as change in velocity divided by the time between gates. Repeat the measurement for each force setting three times and calculate the mean acceleration to improve reliability. Plot a graph of acceleration (y-axis) against force (x-axis) — a straight line through the origin confirms F = ma.

• Level 3 (5-6 marks): Method produces a valid outcome. Equipment named (trolley, runway/ramp, pulley, string, hanging masses, light gates or data logger, metre rule). Force changed by transferring masses from trolley to hanger (keeping total system mass constant). Acceleration measured from light gate timings and measured distance between gates. At least one valid reliability improvement stated (repeat and mean). Steps logically sequenced. (6m)
• Level 2 (3-4 marks): Most key steps identified but one significant element missing or not fully explained — e.g. method for changing force described but compensating masses not mentioned, OR acceleration measurement described but equipment not fully named. (4m)
• Level 1 (1-2 marks): Some relevant steps identified but method would not produce valid results — e.g. only names equipment without describing the procedure, or describes measuring force only without describing how acceleration is found. (2m)

This is Required Practical RPA7 (F = ma trolley investigation). The key insight is that masses must be transferred from the trolley to the hanger — not simply added to the hanger — so the total system mass stays constant. This makes force the only independent variable. Acceleration is calculated from light gate data: a = (v2 - v1) / t. Repeating and averaging reduces random errors. The resulting acceleration-force graph should be a straight line through the origin, confirming Newton's Second Law.

Equipment: hang a spring from a clamp stand, attach a pointer to the spring end, and use a metre rule fixed vertically alongside. Known masses are used as weights. Measure the natural length of the spring before adding any mass. The independent variable is the force applied (mass added), and the dependent variable is the extension of the spring. Add masses one at a time, recording the new length after each addition. Extension equals new length minus original length. Control variables include using the same spring throughout and keeping the temperature constant. To improve reliability, repeat each measurement at least twice and calculate a mean. Plot a graph of force (y-axis) against extension (x-axis). On the straight-line section the gradient equals the spring constant (k = F divided by extension).

• Level 3 (5-6 marks): Describes a valid method with: equipment listed (spring, ruler/metre rule, masses/weights, clamp stand, pointer), measurements described (force in N, extension in m/cm), correct identification of IV and DV, control variable mentioned (same spring/temperature), describes repeating measurements and calculating a mean, and explains how to plot a force-extension graph. (6m)
• Level 2 (3-4 marks): Describes most steps but may not fully identify variables or mention repeating, or may omit some key equipment. (4m)
• Level 1 (1-2 marks): Some relevant steps but method is incomplete or lacks logical sequence; minimal reference to measurements or variables. (2m)

A complete method: hang the spring from a clamp stand, use a metre rule to measure its natural length. Add known masses (providing known forces) one at a time and measure the new length after each addition. Extension = new length - original length. Plot a force-extension graph; the gradient gives the spring constant. Repeat readings and calculate means to improve reliability. Keep the same spring and temperature as control variables.

A scalar quantity has only magnitude (size) but no direction. A vector quantity has both magnitude and direction. An example of a scalar is speed (just a number and unit, e.g. 10 m/s). An example of a vector is velocity (magnitude and direction, e.g. 10 m/s north). Forces must be represented as vectors because forces have both magnitude and direction. When calculating a resultant force, the direction of each force matters - forces in the same direction add together while forces in opposite directions partially or fully cancel each other out. Ignoring direction would give the wrong resultant.

• Scalar: has magnitude (size) only, no direction - with correct example (speed, distance, mass, temperature, energy) (1m)
• Vector: has both magnitude and direction - with correct example (force, velocity, acceleration, displacement, momentum) (1m)
• Forces are vectors because they have direction; direction must be considered when adding forces (1m)
• Same direction forces add; opposite direction forces subtract / cancel - giving correct resultant only if direction considered (1m)

Scalars have magnitude only (e.g. speed, mass, temperature). Vectors have both magnitude and direction (e.g. force, velocity, acceleration). Forces must be treated as vectors because 10 N right + 10 N left = 0 N (they cancel), but 10 N right + 10 N right = 20 N right.

• State that gradient of acceleration-force graph = a/F = 1/m (from F = ma rearranged as a = F/m) (1m)
• Correct relationship: mass = 1 / gradient (1m)
• Correct substitution: mass = 1 / 0.4 (1m)
• Correct answer: mass = 2.5 kg (1m)

Rearranging F = ma gives a = (1/m) x F. A graph of a against F is a straight line through the origin with gradient = 1/m. Therefore m = 1/gradient = 1/0.4 = 2.5 kg. This graph analysis skill tests whether students can connect the mathematical form of F = ma to the physical meaning of the gradient of an experimental graph.

Elastic deformation occurs when an object is deformed by a force but returns to its original shape and size when the force is removed. Inelastic deformation occurs when an object is deformed and does not return to its original shape when the force is removed - the change in shape is permanent.

• Elastic deformation: object returns to its original shape when the force is removed (1m)
• Inelastic deformation: object does not return to its original shape / the change is permanent (1m)
• Clear comparison or reference to what happens when force is removed in both cases (1m)

The key difference is reversibility. Elastic deformation is reversible - the object returns to its original shape. Inelastic (plastic) deformation is permanent - the object stays deformed.

• Correct rearrangement: extension = force / spring constant (1m)
• Correct substitution: extension = 12 / 40 (1m)
• Correct answer: 0.3 m (1m)

Using F = ke, rearrange to e = F/k. Substituting: e = 12 / 40 = 0.3 m. The spring extends by 0.3 m (or 30 cm).

If masses are simply added to the hanger to increase the force, the total mass of the system also increases. By F = ma, acceleration depends on both force and mass, so two variables would change at once. To make it a fair test, total mass must be kept constant. This is done by transferring masses from the trolley to the hanger — the total system mass stays the same, so any change in acceleration is caused only by the change in force.

• Acceleration depends on both force and mass (F = ma), so if mass also changes, you cannot tell which variable is responsible for the change in acceleration (1m)
• Keeping total mass constant makes it a fair test / ensures only one variable (force) changes at a time (1m)
• Method: transfer masses from the trolley to the hanger so total system mass is unchanged (1m)

F = ma shows that acceleration is affected by both force and mass. If you simply add masses to the hanger, you increase both the force and the total system mass simultaneously — making the test unfair. Transferring masses from the trolley to the hanger keeps total mass constant, so force is the only independent variable.

Friction is a contact force because it acts between surfaces that are touching. Gravity is a non-contact force because it acts between objects without them needing to touch, acting across a distance. Magnetic force is a non-contact force because it acts on magnetic materials at a distance without physical contact. Contact forces require objects to be physically touching; non-contact forces act at a distance through a field.

• Any one contact force correctly named (e.g. friction, tension, normal contact force, air resistance) (1m)
• Any one non-contact force correctly named (e.g. gravity / gravitational, magnetic, electrostatic) (1m)
• Correct explanation: contact forces require objects to be touching; non-contact forces act at a distance (through a field) (1m)

Contact forces (friction, tension, normal contact force, air resistance) require physical contact between objects. Non-contact forces (gravity, magnetism, electrostatic) act through fields and do not require objects to touch.

• Correct equation recalled: Ee = 0.5 x k x e^2 (1m)
• Correct substitution: Ee = 0.5 x 250 x 0.08^2 (1m)
• Correct answer: 0.8 J (1m)

Ee = 0.5 x k x e^2 = 0.5 x 250 x (0.08)^2 = 0.5 x 250 x 0.0064 = 0.8 J.

The straight line region shows that the spring obeys Hooke's Law - the extension is directly proportional to the force applied. The gradient of this straight line equals the spring constant. The curved region shows that the spring has exceeded its limit of proportionality and Hooke's Law no longer applies. Beyond this point the spring may also undergo inelastic deformation and not return to its original shape.

• Straight line through origin: extension is directly proportional to force / Hooke's Law is obeyed / spring constant is constant (1m)
• Curved region: limit of proportionality has been exceeded / Hooke's Law no longer applies (1m)
• Beyond the limit the relationship is no longer linear / extension increases more rapidly / spring may be permanently deformed (1m)

Straight line through origin = directly proportional relationship = Hooke's Law obeyed. Where it curves = limit of proportionality exceeded = Hooke's Law no longer applies. Beyond this the extension is greater for each unit of force increase.

The distance travelled is equal to the area under the velocity-time graph. The graph can be split into a triangle (under the sloped section) and a rectangle (under the horizontal section). Calculate the area of each shape separately using the correct formulas, then add the two areas together to find the total distance.

• Distance is equal to the area under the velocity-time graph (1m)
• Split the area into recognisable shapes (triangle and rectangle) (1m)
• Calculate the area of each shape and add them together to find total distance (1m)

On a velocity-time graph, distance = area under the graph. Break it into sections (triangles and rectangles), compute each area, then sum all the areas.

The total distance equals the area under the velocity-time graph. The area under the first section is a triangle: area = 0.5 x base x height = 0.5 x 5 x 10 = 25 m. The area under the second section is a rectangle: area = length x width = 10 x 10 = 100 m. Total distance = 25 + 100 = 125 m.

• States that distance = area under the velocity-time graph (1m)
• Correctly calculates area of triangle = 0.5 x 5 x 10 = 25 m AND area of rectangle = 10 x 10 = 100 m (1m)
• Total distance = 125 m (correct addition of both areas) (1m)

Distance = area under v-t graph. Triangle: 0.5 x 5 s x 10 m/s = 25 m. Rectangle: 10 s x 10 m/s = 100 m. Total = 125 m.

The speed in section A is greater than the speed in section B. The gradient of a distance-time graph represents speed, so a steeper gradient means a higher speed. In section A the gradient (and therefore speed) is greater than in section B. Both speeds are constant because both lines are straight.

• The speed in section A is greater than in section B (section A is faster) (1m)
• This is because section A has a steeper gradient, and gradient = speed on a distance-time graph (1m)
• Both sections show constant speed because both lines are straight (1m)

Gradient = speed on a d-t graph. Steeper gradient = higher speed. Both sections are straight lines, so both represent constant (but different) speeds.

The resultant force is the single force that has the same effect as all the individual forces acting on an object combined. It is found by adding all the forces together, taking into account their directions. When forces act in the same direction they add together; when they act in opposite directions they subtract.

• The resultant force is the single/overall/net force that represents the combined effect of all forces acting on the object (1m)
• Found by adding forces taking direction into account - same direction forces add, opposite direction forces subtract (1m)

The resultant force is the net single force equivalent to all forces acting on an object. For forces along a line: add forces in the same direction and subtract forces in opposite directions.

• Correct substitution: resultant = 3000 - 800 (1m)
• Correct answer: 2200 N forwards (1m)

Resultant force = driving force - friction = 3000 - 800 = 2200 N. The resultant is forwards because the driving force is larger.

During the first section the velocity increases. The straight line shows the velocity increases at a constant rate, meaning the object has a constant acceleration.

• The velocity increases / the object accelerates during the first section (1m)
• The rate of increase is constant / the acceleration is constant (shown by the straight line) (1m)

A straight line with a positive gradient on a velocity-time graph means velocity is increasing at a constant rate — this is uniform (constant) acceleration.

To find the acceleration, calculate the gradient of the straight section of the velocity-time graph. The gradient is found by dividing the change in velocity by the time taken: acceleration = change in velocity divided by time.

• State that the gradient of the velocity-time graph gives the acceleration (1m)
• Gradient = change in velocity divided by time (accept: correct reading from graph with correct division) (1m)

Acceleration = gradient of a velocity-time graph = change in velocity / time. Read off the velocity at 0 s and at 5 s, subtract, then divide by 5.

During the first section the distance increases with time. The straight line shows the object is moving at constant speed (the distance increases by the same amount each second).

• The distance is increasing / the object is moving forward (1m)
• The object is moving at constant speed (accept: the rate of increase is constant / the line is straight so speed is constant) (1m)

A straight line with a positive gradient on a distance-time graph means the object moves with constant speed. A steeper straight line means a faster constant speed.

Speed is calculated from the gradient of the distance-time graph. The gradient equals the change in distance divided by the change in time: speed = change in distance / change in time.

• The gradient of the distance-time graph gives the speed (1m)
• Gradient = change in distance divided by change in time (accept correct formula or reference to rise over run) (1m)

Speed = gradient of a distance-time graph = change in distance / change in time. Choose two clearly separated points on the straight line and read off their coordinates.

A force is a push or pull that acts on an object. Forces can change the speed, direction or shape of an object. They are measured in newtons (N) and are vector quantities.

When forces are equal in size but opposite in direction, they are balanced. The resultant force is found by adding the forces, taking direction into account: 5 N down + 5 N up = 0 N. A resultant force of zero means the book does not accelerate.

A horizontal line on a velocity-time graph means velocity is not changing with time, so the object is travelling at constant (steady) velocity.

The flat horizontal section shows the object is travelling at constant velocity (constant speed in a constant direction).

• The object is moving at constant velocity / constant speed (the velocity is not changing) (1m)

A horizontal line on a velocity-time graph means velocity is not changing with time. The object is at constant velocity — no acceleration, no deceleration.

The gradient (slope) of a velocity-time graph equals acceleration. A steeper gradient means greater acceleration. A horizontal line (zero gradient) means zero acceleration — constant velocity.

A horizontal line on a distance-time graph means distance is not changing with time — the object is stationary. The distance stays the same, so the object is not moving.

Gravity is a non-contact force because it acts on objects without them needing to touch. Friction, tension, and normal contact force all require physical contact between objects.

Elastic deformation occurs when an object returns to its original shape and size after the deforming force is removed. Inelastic (plastic) deformation is permanent - the object does not return to its original shape.

A steeper gradient on a distance-time graph means the distance changes more quickly with time — the object is moving faster. Gradient = speed, so a steeper line = greater speed.

Stopping distance equals thinking distance plus braking distance. Thinking distance equals speed multiplied by reaction time. If speed doubles, thinking distance also doubles. Braking distance depends on how much kinetic energy must be removed by the brakes. Kinetic energy equals one-half times mass times velocity squared. If speed doubles, velocity squared quadruples (2 squared equals 4). This means kinetic energy quadruples. The brakes must do four times as much work to remove this kinetic energy. Using work done equals force times distance, if the braking force stays the same, the braking distance quadruples. So thinking distance doubles but braking distance quadruples. The total stopping distance more than doubles, confirming the statement.

• Thinking distance = speed x reaction time. Doubling speed doubles thinking distance. (1m)
• Braking distance depends on kinetic energy: Ek = 0.5mv^2 (1m)
• Doubling speed quadruples kinetic energy (2^2 = 4) (1m)
• Brakes must do 4 times more work to remove the kinetic energy (1m)
• With the same braking force, braking distance quadruples (W = Fd) (1m)
• Overall stopping distance more than doubles because the braking distance contribution quadruples, even though thinking distance only doubles (1m)

Thinking distance doubles when speed doubles. Braking distance quadruples because kinetic energy is proportional to v^2. Combined, the total stopping distance more than doubles.

• Correct kinetic energy calculation: Ek = 0.5 x 1200 x 25^2 = 375,000 J (1m)
• Equation: work done = braking force x braking distance (1m)
• Rearrangement: braking distance = kinetic energy / braking force (1m)
• Correct answer: 375,000 / 6000 = 62.5 m (1m)

Ek = 0.5 x 1200 x 625 = 375,000 J. Work done by brakes = 375,000 J. Distance = 375,000 / 6000 = 62.5 m.

The braking force is produced by friction between the brake pads and the wheel disc. The braking force acts in the opposite direction to motion, producing a deceleration (Newton's second law: F = ma, so a = F/m). A very large braking force produces a very large deceleration. This is dangerous because the large deceleration can cause the wheels to lock (stop rotating) leading to loss of control and skidding. The driver and passengers may also experience large forces due to their inertia, which could cause injury.

• Braking force opposes motion and causes deceleration (F = ma referenced) (1m)
• Larger braking force produces larger deceleration (F = ma) (1m)
• Very large deceleration can cause wheels to lock / skidding / loss of control (1m)
• Large deceleration produces large forces on passengers due to inertia / risk of injury (1m)

Braking force decelerates the car via F=ma. Very large braking forces cause very large decelerations, which can lock wheels causing skidding. Passengers are also at risk from the large forces due to inertia.

Tiredness increases the driver's reaction time, which increases thinking distance. Thinking distance = speed × reaction time, so a longer reaction time at any given speed directly increases thinking distance. Tiredness does not affect braking distance (assuming road and vehicle conditions are unchanged). Icy road conditions reduce the friction force between the tyres and the road surface. A smaller friction force means a smaller deceleration force is available when braking, so the vehicle takes longer to decelerate to rest — braking distance increases. Ice does not affect the driver's reaction time, so thinking distance is unchanged. Together, a tired driver on an icy road would have both a longer thinking distance and a longer braking distance, making overall stopping distance much larger.

• Tiredness increases reaction time, which increases thinking distance (thinking distance = speed × reaction time) (1m)
• Tiredness does not affect braking distance (braking distance depends on friction/deceleration not reaction time) (1m)
• Icy roads reduce friction between tyres and road surface, reducing the braking force available (1m)
• Smaller braking force means smaller deceleration, so the car takes longer to stop / braking distance increases; ice does not affect reaction time / thinking distance (1m)

Stopping distance = thinking distance + braking distance. Thinking distance depends only on reaction time (thinking distance = speed × reaction time) — tiredness slows reaction time and so increases thinking distance. Braking distance depends on the deceleration available, which in turn depends on the friction force between tyres and road. Ice reduces this friction force, reducing deceleration and increasing braking distance. Icy conditions do not affect the driver's reaction time, so thinking distance is unaffected. Tiredness does not affect road friction, so it does not increase braking distance. The two factors therefore affect different components of stopping distance.

Kinetic energy = ½mv². This means kinetic energy is proportional to the square of speed. If speed doubles, kinetic energy quadruples. When the brakes are applied, the braking force does work to remove all of the vehicle's kinetic energy. Work done = force × distance. If the braking force stays the same, a vehicle with four times the kinetic energy needs four times the braking distance to bring it to rest. Therefore, doubling the speed quadruples the braking distance, not doubles it — braking distance increases disproportionately with speed.

• Kinetic energy = ½mv² — kinetic energy is proportional to speed squared / squaring speed multiplies KE by the square (1m)
• Doubling speed quadruples kinetic energy (or appropriate numerical example showing non-proportional increase) (1m)
• Work done by brakes = braking force × braking distance — must equal kinetic energy to bring vehicle to rest (1m)
• With constant braking force, braking distance is proportional to kinetic energy / quadrupling KE requires quadrupling braking distance / braking distance increases with speed squared (1m)

The non-proportional relationship between speed and braking distance comes directly from two equations working together: KE = ½mv² (kinetic energy proportional to speed squared) and work done = force × distance. When a vehicle brakes, the braking force does work equal to the vehicle's kinetic energy. If braking force is fixed, braking distance must equal KE ÷ force. Since KE scales as speed², braking distance also scales as speed². Doubling speed → ×4 braking distance; tripling speed → ×9 braking distance. This is why high-speed crashes are so much more severe than low-speed ones.

• Correct substitution: thinking distance = 20 x 0.7 (1m)
• Correct calculation shown (1m)
• Correct answer: 14 m (1m)

Thinking distance = speed x reaction time = 20 x 0.7 = 14 m.

A car travelling at higher speed has more kinetic energy (Ek = 0.5mv^2, so Ek increases with speed squared). The brakes must transfer all this kinetic energy to thermal energy in the brake pads and tyres. Since the braking force is the same, more work must be done by the brakes, which requires a greater distance (work done = force x distance). Therefore the braking distance is greater.

• Higher speed means greater kinetic energy (Ek is proportional to v^2) (1m)
• Brakes must do more work to remove the greater kinetic energy (1m)
• Since braking force is constant, greater distance is needed (work = force x distance) (1m)

Braking distance is greater at higher speeds because kinetic energy is proportional to v^2. Greater kinetic energy requires more work done by the brakes (W = Fd), and with the same braking force, a longer distance is needed.

On a wet road, there is less friction between the tyres and the road surface. The braking force is produced by friction between the tyres and the road. With less friction, the braking force is reduced. Since work done = force x distance, a smaller force acting over a greater distance is needed to remove the same amount of kinetic energy, so the braking distance increases.

• Wet road reduces friction between tyres and road surface (1m)
• Reduced friction means smaller braking force (1m)
• Smaller force requires greater distance to do the same work / remove the same kinetic energy (1m)

Wet road reduces friction, which reduces the braking force. Less braking force means a greater distance is needed to do the same amount of work to remove the kinetic energy.

One person holds a ruler vertically with the 0 cm mark at the bottom. The other person holds their fingers near the bottom of the ruler without touching it. Without warning, the first person releases the ruler and the second person catches it as quickly as possible. The distance the ruler falls is measured. This distance is used to calculate the reaction time using the equation: d = 0.5 x g x t^2, rearranged to t = sqrt(2d/g).

• Ruler held vertically at zero end, subject's fingers positioned to catch without touching (1m)
• Ruler released without warning; distance the ruler falls before being caught is measured (1m)
• Reaction time calculated using d = 0.5 x g x t^2 rearranged as t = sqrt(2d/g) (1m)

In the falling ruler test, the distance fallen before catching is measured and t = sqrt(2d/g) gives the reaction time.

• Correct substitution: stopping distance = 9 + 14 (1m)
• Correct answer: 23 m (1m)

Stopping distance = thinking distance + braking distance = 9 + 14 = 23 m.

A driver's reaction time can be increased by tiredness (fatigue) and by the influence of alcohol or drugs.

• Tiredness / fatigue / being distracted (1m)
• Alcohol / drugs / medication (1m)

Factors that increase reaction time include tiredness/fatigue, alcohol, drugs, distractions (e.g. mobile phone), and some medications.

Stopping distance = thinking distance + braking distance. Thinking distance is the distance travelled during the driver's reaction time. Braking distance is the distance travelled from when the brakes are applied until the car stops.

Thinking distance = speed x reaction time. A higher speed means more distance is covered during the same reaction time, so thinking distance increases. Wet roads and worn tyres/brakes affect braking distance, not thinking distance.

Kinetic energy = 0.5mv^2. Doubling speed quadruples kinetic energy (2^2 = 4). Since braking force is constant, the brakes must do 4 times more work, requiring 4 times the distance. So braking distance quadruples.

Reaction time affects thinking distance (distance = speed x reaction time). If reaction time increases, thinking distance increases. Braking distance depends on speed, braking force, and road/tyre conditions, not reaction time.

When the skydiver first jumps, weight is greater than air resistance. There is a resultant force downwards so the skydiver accelerates downwards (Newton's Second Law: F = ma). As speed increases, air resistance increases because drag depends on speed. The resultant force decreases because the difference between weight and air resistance gets smaller. Since resultant force decreases, the acceleration also decreases (Newton's Second Law). The skydiver continues to speed up but with decreasing acceleration. Eventually air resistance becomes equal to weight. The resultant force is now zero, so acceleration is zero. The skydiver moves at a constant velocity — this is terminal velocity (Newton's First Law).

• Level 3 (5-6 marks): Logically sequenced account covering: (1) Initially weight > air resistance so resultant force is downward and skydiver accelerates. (2) As speed increases, air resistance increases. (3) The resultant force decreases as air resistance approaches weight. (4) Acceleration decreases as resultant force decreases. (5) At terminal velocity, weight = air resistance, forces balanced, resultant = 0, acceleration = 0, constant velocity. Uses Newton's Laws throughout. (6m)
• Level 2 (3-4 marks): Identifies most stages but may not fully explain the changing resultant force or may not clearly link force changes to acceleration changes. (4m)
• Level 1 (1-2 marks): Some relevant statements about forces or motion but not logically sequenced or linked. (2m)

Stage 1: Weight > air resistance. Resultant force downwards. Skydiver accelerates downwards (Newton's 2nd Law). Stage 2: As speed increases, air resistance increases. Resultant force decreases. Acceleration decreases. Stage 3: Terminal velocity reached when air resistance = weight. Resultant force = 0. Acceleration = 0. Constant velocity (Newton's 1st Law).

• Correct calculation of weight: W = mg = 2000 x 9.8 = 19600 N (1m)
• Correct calculation of resultant force: F = ma = 2000 x 15 = 30000 N (1m)
• Correct rearrangement: thrust = resultant + weight = 30000 + 19600 (1m)
• Correct answer: 49600 N (1m)

Weight = mg = 2000 x 9.8 = 19600 N. Resultant force = ma = 2000 x 15 = 30000 N. Since resultant = thrust - weight: thrust = 30000 + 19600 = 49600 N.

Stopping distance is the sum of thinking distance and braking distance. Thinking distance is the distance the car travels while the driver reacts and applies the brakes. It increases if the driver's reaction time is longer - caused by tiredness, alcohol, drugs, or distractions. Braking distance is the distance the car travels after the brakes are applied until it stops. It increases with greater speed, reduced friction (wet/icy roads, worn tyres), greater mass of the car, and reduced braking force.

• Stopping distance = thinking distance + braking distance (or both terms correctly defined) (1m)
• Factors affecting thinking distance: reaction time increased by tiredness / alcohol / drugs / distractions (any one valid factor) (1m)
• Factors affecting braking distance: increased speed / reduced friction (wet/icy roads, worn tyres) (any one valid factor) (1m)
• Correct explanation of why the factor increases stopping distance (e.g. less friction means less deceleration so car takes longer to stop) (1m)

Stopping distance = thinking distance + braking distance. Thinking distance increases with longer reaction time (tiredness, alcohol, drugs, distractions). Braking distance increases with higher speed, reduced friction (ice, wet roads, worn tyres), and higher vehicle mass.

• Convert radius to metres: r = 7000 x 10^3 = 7 x 10^6 m; convert period to seconds: T = 100 x 60 = 6000 s (1m)
• Calculate speed: v = 2 x pi x 7 x 10^6 / 6000 = 7330 m/s (accept 7330 to 7340 m/s) (1m)
• Substitute into F = mv^2 / r: F = 500 x 7330^2 / 7 x 10^6 (1m)
• Correct answer: F ≈ 3838 N (accept 3800–3850 N depending on π approximation) (1m)

Convert: r = 7 × 10^6 m, T = 6000 s. Calculate speed: v = 2π × 7 × 10^6 / 6000 = 7330 m/s. Calculate force: F = mv²/r = 500 × 7330² / 7,000,000 = 500 × 53,728,900 / 7,000,000 ≈ 3838 N. The centripetal force is directed toward Earth's centre. Accept answers in the range 3800–3850 N depending on how π is approximated.

• Correct calculation of deceleration: a = (v - u) / t = (0 - 20) / 4 = -5 m/s^2 (or 5 m/s^2 deceleration) (1m)
• Correct application of F = ma: F = 1500 x 5 (1m)
• Correct answer: 7500 N (1m)
• Correct direction stated (braking force acts backwards/in opposite direction to motion) (1m)

a = (0 - 20) / 4 = -5 m/s^2 (deceleration of 5 m/s^2). F = ma = 1500 x 5 = 7500 N (acting backwards to oppose motion).

• Part a: Correct substitution into v^2 = u^2 + 2as: 0 = 400 + 2a x 50, giving 100a = -400 (1m)
• Part a: Correct deceleration: a = -4 m/s^2 (magnitude 4 m/s^2) (1m)
• Part b: Correct substitution into v = u + at: 0 = 20 + (-4)t, giving 4t = 20 (1m)
• Part b: Correct time: t = 5 s (accept answer to part b only if part a method is correct) (1m)

Part a: Using v² = u² + 2as: 0 = 20² + 2a × 50, so 0 = 400 + 100a, giving a = -4 m/s² (deceleration = 4 m/s²). Part b: Using v = u + at: 0 = 20 + (-4)t, so 4t = 20, giving t = 5 s. Take care with signs: deceleration means acceleration is negative when taking the initial direction as positive.

• Correct rearrangement: a = F / m (1m)
• Correct substitution: a = 600 / 1200 (1m)
• Correct answer: 0.5 m/s^2 (1m)

Rearranging F = ma: a = F / m = 600 / 1200 = 0.5 m/s^2.

• Correct substitution: F = 80 x 2.5 (1m)
• Correct calculation: F = 200 (1m)
• Correct answer with unit: 200 N (1m)

F = ma = 80 x 2.5 = 200 N.

• Correct identification: u = 0 m/s, a = 2.5 m/s^2, t = 8 s and correct substitution into s = ut + 1/2 at^2 (1m)
• Correct working: s = 0 + 1/2 x 2.5 x 64 = 1.25 x 64 (1m)
• Correct answer: s = 80 m (1m)

u = 0 (starts from rest). s = ut + ½at^2 = 0 + ½ x 2.5 x 8^2 = ½ x 2.5 x 64 = 1.25 x 64 = 80 m.

Newton's Third Law states that when object A exerts a force on object B, object B exerts an equal and opposite force on object A. For a person standing on the ground: the person exerts a downward force (their weight) on the ground; the ground exerts an equal force upward on the person (the normal contact force). These forces do not cancel out because they act on different objects - the weight acts on the ground and the normal contact force acts on the person.

• Newton's Third Law: for every action there is an equal and opposite reaction / forces come in equal and opposite pairs that act on different objects (1m)
• Correct example: person pushes down on ground (weight/gravity), ground pushes up on person (normal contact force) - forces equal in size (1m)
• They do not cancel because they act on different objects (not the same object) (1m)

Action-reaction pairs are equal and opposite but act on DIFFERENT objects, so they cannot cancel. Cancellation requires forces to act on the SAME object. The person's weight acts on the Earth; the Earth's normal contact force acts on the person.

When the skydiver first jumps, their weight (downward force) is greater than air resistance (drag), so there is a resultant force downward and they accelerate. As their speed increases, the air resistance increases. Eventually the air resistance equals the weight - the forces are balanced and there is no resultant force. The skydiver then falls at a constant velocity called terminal velocity.

• Initially weight greater than air resistance/drag so there is a resultant force downwards and the skydiver accelerates (1m)
• As speed increases, air resistance/drag increases (1m)
• At terminal velocity, air resistance/drag equals weight - forces are balanced/resultant is zero - constant velocity (1m)

Terminal velocity occurs when weight = air resistance (balanced forces, no resultant). Before this, weight > drag so acceleration occurs. As speed increases, drag increases until balance is achieved.

• Correct substitution: F = 1200 x 15^2 / 50 = 1200 x 225 / 50 (1m)
• Correct intermediate step: F = 270000 / 50 (1m)
• Correct answer: F = 5400 N (1m)

F = mv^2 / r = 1200 x 15^2 / 50 = 1200 x 225 / 50 = 270000 / 50 = 5400 N. This inward (centripetal) force is provided by friction between the car tyres and the road surface.

A banked road is tilted inwards at an angle. The normal contact force from the road on the car acts perpendicular to the road surface. On a banked road this normal force is tilted, so it has a horizontal component that points towards the centre of the curve. This horizontal component provides the centripetal force needed for circular motion. Because the banking supplies centripetal force, the car can corner at higher speed without needing friction to provide it.

• The road is banked/tilted so the normal contact force acts at an angle (not straight upward) (1m)
• The normal force has a horizontal component directed towards the centre of the bend (1m)
• This horizontal component provides the centripetal force, reducing the need for friction and allowing higher speeds (1m)

On a flat road, friction provides the centripetal force for cornering. On a banked road, the tilted normal force has a horizontal inward component that provides some or all of the centripetal force instead. This reduces (or eliminates) the need for friction, allowing higher cornering speeds.

• Correct identification: v = 0 m/s, u = 15 m/s, a = -9.8 m/s^2 and rearrangement: s = (v^2 - u^2) / 2a (1m)
• Correct substitution: s = (0 - 225) / (2 x -9.8) = -225 / -19.6 (1m)
• Correct answer: s = 11.5 m (accept 11.4 to 11.5 m) (1m)

Taking upward as positive: initial velocity u = +15 m/s, acceleration a = -10 m/s² (gravity acts downward), final velocity v = 0 m/s (at maximum height). Using v² = u² + 2as: 0 = 15² + 2 × (-10) × s, so 0 = 225 - 20s, giving s = 225/20 = 11.25 m ≈ 11.3 m. Common mistake: forgetting that g acts downward (negative) when upward is positive.

The spaceship will continue to move at constant velocity - it will not slow down or change direction. This is because in deep space there is no resultant force acting on it (no friction, no air resistance, no gravity from nearby planets). According to Newton's First Law, an object with no resultant force acting on it continues at constant velocity.

• The spaceship continues at constant velocity / same speed and direction / does not slow down (1m)
• No resultant force acts on it (no friction / no air resistance in space) so Newton's First Law applies (1m)

In deep space there is no air resistance or friction, so no resultant force acts on the spaceship. By Newton's First Law, it continues at constant velocity indefinitely.

Centripetal force is given by F = mv^2 / r. If the radius r is smaller and the speed stays the same, the centripetal force must increase because force and radius are inversely proportional. A tighter bend requires a larger inward force to keep the car on the circular path.

• Centripetal force and radius are inversely proportional (from F = mv^2 / r) — as radius decreases, force increases (1m)
• A larger inward/centripetal force is needed to maintain circular motion on a tighter bend (1m)

The equation F = mv^2/r shows that centripetal force and radius are inversely proportional. At constant speed and mass, halving the radius doubles the centripetal force. A tighter bend means the direction of travel must change more sharply, requiring a greater inward force.

Distance is a scalar quantity — it has only magnitude (size) and no direction. Displacement is a vector — it has both magnitude and direction, describing how far an object has moved in a specific direction from its starting point. Speed is a scalar — it is the rate of change of distance. Velocity is a vector — it is the rate of change of displacement and includes direction.

• Displacement has direction, distance does not / displacement is a vector, distance is a scalar (1m)
• Velocity has direction, speed does not / velocity is a vector, speed is a scalar (1m)

Scalar quantities have magnitude only; vector quantities have magnitude AND direction. Distance and speed are scalars. Displacement and velocity are vectors — they specify both how much and in which direction.

Newton's First Law states that an object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted upon by a resultant (unbalanced) force. This is the principle of inertia.

Newton's Second Law states that the resultant force on an object equals its mass multiplied by its acceleration: F = ma. Force is measured in newtons (N), mass in kilograms (kg), and acceleration in m/s^2.

Newton's Third Law states that for every action there is an equal and opposite reaction. The swimmer pushes the water backwards (action), so the water pushes the swimmer forwards with an equal force of 200 N (reaction). The forces are equal in size, opposite in direction, and act on different objects.

Inertia is the tendency of an object to resist changes in its state of motion (whether at rest or moving). Objects with greater mass have greater inertia and require a larger resultant force to produce the same acceleration (F = ma). Kinetic energy is the energy of a moving object - this is a different concept.

The centripetal force always points towards the centre of the circle. This inward force is what keeps the object moving in a circular path rather than flying off in a straight line. Without it, the object would travel in a straight line (Newton's First Law). The force is centripetal (centre-seeking), not centrifugal.

The equation s = ut + ½at² links displacement (s), initial velocity (u), acceleration (a) and time (t). It is one of the kinematic (SUVAT) equations. Option A links v, u, a, t but not displacement. Option B links v, u, a, s but not time. F = ma is Newton's Second Law, not a kinematic equation.

When the string breaks, the centripetal force disappears. By Newton's First Law, the ball continues in a straight line at the velocity it had at the instant of breaking — which is tangential to the circle (at right angles to the radius). This is why mud flies off a spinning wheel tangentially, and why objects on a roundabout feel thrown outward.

Taking right as positive: momentum of 2 kg ball before = 2 x 3 = 6 kg m/s. Momentum of 1 kg ball before = 1 x (-2) = -2 kg m/s. Total momentum before = 6 + (-2) = 4 kg m/s. By conservation of momentum, total momentum after = 4 kg m/s. Momentum of 2 kg ball after = 2 x 0.33 = 0.66 kg m/s. Momentum of 1 kg ball after = 4 - 0.66 = 3.34 kg m/s. Velocity of 1 kg ball = 3.34 / 1 = 3.34 m/s to the right. KE before = 0.5 x 2 x 3 squared + 0.5 x 1 x 2 squared = 9 + 2 = 11 J. KE after = 0.5 x 2 x 0.33 squared + 0.5 x 1 x 3.34 squared = 0.11 + 5.58 = 5.69 J. Since KE after (5.69 J) is less than KE before (11 J), kinetic energy is not conserved. This is an inelastic collision.

• Total momentum before = (2 x 3) + (1 x -2) = 6 - 2 = 4 kg m/s (to the right) (1m)
• Momentum after: 2 x 0.33 = 0.66 kg m/s. Momentum of 1 kg ball = 4 - 0.66 = 3.34 kg m/s (1m)
• Velocity of 1 kg ball = 3.34 / 1 = 3.34 m/s to the right (accept answers consistent with conservation) (1m)
• KE before: 0.5 x 2 x 9 + 0.5 x 1 x 4 = 9 + 2 = 11 J (1m)
• KE after: 0.5 x 2 x 0.33^2 + 0.5 x 1 x 3.34^2 = 0.11 + 5.58 = 5.69 J (approximately) (1m)
• KE after < KE before, so kinetic energy is not conserved - this is an inelastic collision (1m)

Total p before = 4 kg m/s. After: 1 kg ball velocity ≈ 3.34 m/s right. KE before = 11 J, KE after ≈ 5.69 J. KE is not conserved, so this is an inelastic collision.

• States total momentum before = 0 (both stationary) (1m)
• Momentum of bullet after = 0.010 x 400 = 4 kg m/s (1m)
• Gun momentum = -4 kg m/s (opposite direction to bullet) (1m)
• Recoil velocity = 4 / 2.5 = 1.6 m/s (in opposite direction to bullet) (1m)

Total momentum before = 0. Bullet momentum = 0.010 x 400 = 4 kg m/s. Gun momentum = -4 kg m/s. Recoil velocity = 4 / 2.5 = 1.6 m/s in the opposite direction to the bullet.

Momentum before: ball A has p = 0.5 x 6 = 3 kg m/s; ball B has p = 0. Total = 3 kg m/s. After collision: ball A is stationary (p = 0), so ball B must have p = 3 kg m/s, giving velocity = 3/0.5 = 6 m/s. Total momentum after = 3 kg m/s. Momentum is conserved. Kinetic energy before: Ek = 0.5 x 0.5 x 36 = 9 J. After: Ek = 0.5 x 0.5 x 36 = 9 J. Kinetic energy is also conserved, so this is an elastic collision.

• Momentum before = 3 kg m/s; after collision ball B must have momentum 3 kg m/s (1m)
• Ball B velocity after = 6 m/s; total momentum after = 3 kg m/s, so momentum is conserved (1m)
• Kinetic energy before = 0.5 x 0.5 x 36 = 9 J; kinetic energy after = 9 J (1m)
• Kinetic energy is conserved; this is an elastic collision (1m)

Momentum: 3 kg m/s before and after (conserved). KE: 9 J before and after (conserved). This is an elastic collision.

• Correct total momentum before: 3 x 4 = 12 kg m/s (1m)
• Correct equation after: (3 + 1) x v = 12 (1m)
• Correct answer: v = 3 m/s (1m)

Total momentum before = 3 x 4 + 1 x 0 = 12 kg m/s. After: (3 + 1) x v = 12, so v = 12 / 4 = 3 m/s.

Before the explosion, the total momentum of the stationary rocket is zero. After the explosion, the two pieces fly apart in opposite directions. Their momenta are equal in magnitude but opposite in direction, so they cancel out. The total momentum after the explosion is also zero. This equals the total momentum before, so momentum is conserved.

• Total momentum before explosion is zero (rocket stationary) (1m)
• After explosion, pieces move in opposite directions with equal and opposite momenta (1m)
• Total momentum after = zero, which equals total momentum before, so momentum is conserved (1m)

Before: total momentum = 0. After: two pieces with equal and opposite momenta, total = 0. Momentum is conserved.

Change in momentum = m x change in velocity = m x (v - u). Dividing by time: F = m(v - u)/t. Since (v - u)/t is the definition of acceleration (a = change in velocity / time), this gives F = ma. So Newton's second law written using momentum is equivalent to F = ma when mass is constant.

• Change in momentum = mass x change in velocity (m x delta v or m(v-u)) (1m)
• Dividing by time: F = m(v-u)/t (1m)
• Identifies (v-u)/t as acceleration, giving F = ma (1m)

Change in momentum = m(v-u). F = m(v-u)/t = ma, since a = (v-u)/t.

• Correct substitution: p = 1500 x 20 (1m)
• Correct answer: 30,000 kg m/s (1m)

p = mv = 1500 x 20 = 30,000 kg m/s.

Momentum is a vector quantity because it has both magnitude and direction. The direction of the momentum is the same as the direction of the velocity. When calculating with momenta in opposite directions, one must be given a positive sign and one a negative sign.

• Momentum has both magnitude and direction (1m)
• Direction of momentum is the same as the direction of velocity / opposing momenta have opposite signs (1m)

Momentum is a vector - it has both magnitude (mass x speed) and direction (same as velocity). Objects moving in opposite directions have momenta with opposite signs.

Momentum (p) = mass (m) x velocity (v). The unit of momentum is kg m/s. Note that option D describes impulse, not momentum.

The unit of momentum is kg m/s (kilogram metres per second). This comes from the equation p = mv, where m is in kg and v is in m/s.

In a closed system, the total momentum before an event equals the total momentum after the event.

• Total momentum before = total momentum after (in a closed system / no external forces) (1m)

Conservation of momentum: total momentum before = total momentum after, provided the system is closed (no net external forces).

By conservation of momentum, the total momentum before (zero, as both were stationary) must equal the total momentum after. So the momenta of A and B must be equal and opposite, summing to zero.

Momentum is a vector. If east is positive, momentum before = +10 kg m/s, momentum after = -10 kg m/s (moving west). The change in momentum = -10 - (+10) = -20 kg m/s. The sign depends on which direction is defined as positive.

The principle of moments states that for a balanced system (in equilibrium), the sum of clockwise moments equals the sum of anticlockwise moments. The pivot is at the 50 cm mark. The 4 N weight is at the 20 cm mark, which is 30 cm from the pivot. The anticlockwise moment is 4 N times 30 cm equals 120 Ncm. For equilibrium, the clockwise moment from the 5 N weight must also equal 120 Ncm. So 5 N times d equals 120, giving d equals 24 cm from the pivot. The 5 N weight should be placed at the 74 cm mark. A source of error is that the metre rule may not be perfectly uniform in mass distribution, so the centre of mass may not be exactly at 50 cm. To reduce this error, check that the rule balances horizontally on the pivot before adding any weights, and repeat measurements to check consistency.

• States principle of moments: clockwise moment = anticlockwise moment (1m)
• Calculates anticlockwise moment: 4 N x 30 cm = 120 Ncm (or 4 x 0.30 = 1.2 Nm) (1m)
• Sets up equation for 5 N weight: 5 x d = 120 (or 1.2) (1m)
• Correct answer: d = 24 cm from pivot, so at 74 cm mark (or 26 cm mark on other side) (1m)
• Identifies a valid source of error (e.g. metre rule may not be exactly uniform/calibrated; pivot friction; difficulty reading exact position) (1m)
• Suggests a valid improvement (e.g. repeat and take mean; use a more precise ruler; ensure pivot is frictionless; check ruler is horizontal before starting) (1m)

Pivot is at 50 cm. The 4 N weight is at 20 cm, which is 30 cm from the pivot. Anticlockwise moment = 4 x 30 = 120 Ncm. For equilibrium: 5 x d = 120, so d = 24 cm from the pivot, placing the 5 N weight at the 74 cm mark.

• Correct anticlockwise moment: 60 x 2 = 120 Nm (1m)
• Correct equation: F x 3 = 120 (1m)
• Correct answer: F = 40 N (1m)

Anticlockwise moment = 60 x 2 = 120 Nm. For equilibrium: F x 3 = 120. F = 120 / 3 = 40 N.

A lever has a pivot (fulcrum), a load and an effort force. The effort is applied at a greater distance from the pivot than the load. Because moment = force x distance, the larger distance means that a smaller effort force can produce the same moment as a larger load force, so the heavy load can be lifted.

• Effort is applied at a greater distance from the pivot than the load (1m)
• Moment = force x distance stated or implied (1m)
• Larger distance allows smaller force to produce the same (or greater) moment (1m)

Levers allow a small effort to overcome a larger load by applying the effort at a greater distance from the pivot. Since M = F x d, the same moment can be produced with a smaller force if the distance is larger.

• Correct equation: moment = force x distance, so distance = moment / force (1m)
• Correct substitution: distance = 36 / 45 (1m)
• Correct answer: 0.8 m (1m)

Rearranging M = F x d gives d = M / F = 36 / 45 = 0.8 m.

A wide-based object is more stable because when tilted, the vertical line through the centre of gravity still falls within the base. The restoring moment from the weight brings the object back upright. A tall object with a high centre of gravity will topple when tilted because the vertical line through the centre of gravity quickly passes outside the base, causing the moment due to weight to rotate the object further away from upright.

• Wide base means vertical line through centre of gravity remains within the base when tilted (1m)
• Weight of object creates a restoring moment when centre of gravity is within base (1m)
• High centre of gravity means the line through it quickly falls outside the base, causing toppling moment (1m)

Stability depends on whether the weight of the object creates a restoring moment or a toppling moment. A low, wide-based object keeps the weight acting within the base through larger tilts, creating a restoring moment. A tall object's high centre of gravity quickly moves the weight line outside the base, creating a toppling moment.

Pushing at the edge (greatest distance from the pivot/hinges) maximises the perpendicular distance from the pivot, so for a given force the moment is largest. This makes it easiest to rotate the door. If the student pushed at half the distance, the perpendicular distance would be halved, so the moment would be halved for the same force. To produce the same moment to open the door, the student would need to apply twice the force.

• Larger distance from pivot (hinges) produces a larger moment for the same force (1m)
• At half the distance, the moment would be halved for the same force (1m)
• Twice the force would be needed to produce the same moment at half the distance (1m)

Moment = F x d. Greater distance from hinges = larger moment for same force. Halving distance halves the moment, so the force must double to produce the same moment.

• Correct substitution: moment = 20 x 0.5 (1m)
• Correct answer: 10 Nm (1m)

moment = F x d = 20 x 0.5 = 10 Nm

For a system in equilibrium, the total clockwise moment about a pivot equals the total anticlockwise moment about the same pivot.

• Clockwise moments equal anticlockwise moments (1m)
• Reference to equilibrium or balanced system (1m)

The principle of moments states that for equilibrium, the total clockwise moment equals the total anticlockwise moment about any pivot.

A moment is the turning effect of a force about a pivot point. It is calculated as moment = force x perpendicular distance from the pivot.

The equation for moment is M = F x d, where F is the force in Newtons and d is the perpendicular distance from the pivot in metres. The unit of moment is Newton-metres (Nm).

For a system to be in equilibrium (balanced), the principle of moments states that the sum of all clockwise moments must equal the sum of all anticlockwise moments about any pivot point.

The unit of moment is the Newton-metre (Nm).

• Newton-metre (Nm) stated as the unit (1m)

Moment = force (N) x distance (m), so the unit of moment is Newton-metres (Nm).

Moment = force x distance. Using a longer handle increases the distance from the pivot (the bolt), so a larger moment is created for the same applied force. This makes it easier to rotate the bolt.

Change in momentum = mass x change in velocity = 1500 x 15 = 22,500 kg m/s. This is the same in both cases. Force equals change in momentum divided by time. Without crumple zones: F = 22,500 divided by 0.12 = 187,500 N. With crumple zones: F = 22,500 divided by 0.8 = 28,125 N. The crumple zone reduces the force by a factor of 6.7 (187,500 divided by 28,125). The impulse (change in momentum) is the same in both cases because the car starts and ends with the same velocities. The crumple zone increases the collision time, which reduces the force for the same impulse. By Newton's Second Law, the smaller force produces a smaller acceleration on the passengers, greatly reducing the risk of serious injury.

• Change in momentum = 1500 x 15 = 22,500 kg m/s (in both cases) (1m)
• Force without crumple zone = 22,500 / 0.12 = 187,500 N (1m)
• Force with crumple zone = 22,500 / 0.8 = 28,125 N (1m)
• Crumple zone reduces force by a factor of approximately 6.7 (187,500 / 28,125) (1m)
• The impulse (change in momentum) is the same in both cases - the crumple zone does not reduce the change in momentum (1m)
• The much smaller force with crumple zones greatly reduces the risk of serious injury to passengers (by Newton's 2nd law, smaller force = smaller acceleration = less bodily harm) (1m)

Both cases: delta p = 22,500 kg m/s. Without crumple: F = 22,500/0.12 = 187,500 N. With crumple: F = 22,500/0.8 = 28,125 N. Crumple zone reduces force by factor ~6.7. Impulse same in both cases. Much smaller force with crumple zones greatly reduces injury risk.

• Change in momentum = 1200 x 30 = 36,000 kg m/s (1m)
• Correct equation: F = delta p / t (1m)
• Correct substitution: F = 36,000 / 0.06 (1m)
• Correct answer: 600,000 N (1m)

Delta p = 1200 x 30 = 36,000 kg m/s. F = 36,000 / 0.06 = 600,000 N.

The area under a force-time graph represents the impulse delivered to the object. Impulse equals change in momentum, so the area also represents the change in momentum of the object. If the same impulse is delivered over a longer time, the area must remain the same, but the graph becomes wider (longer time axis) and shorter (lower peak force). The peak force is reduced because the same total impulse is spread over a greater time interval.

• Area under force-time graph represents impulse (1m)
• Impulse equals change in momentum (1m)
• Same impulse over longer time: graph is wider (longer time) (1m)
• Peak force is lower (so that the area under the graph remains the same) (1m)

Area under F-t graph = impulse = change in momentum. Same impulse over longer time: graph widens and peak force decreases, maintaining the same area.

• Change in momentum = 0.4 x 15 = 6 kg m/s (1m)
• Correct equation: force = change in momentum / time (1m)
• Correct answer: force = 6 / 0.05 = 120 N (1m)

Delta p = 0.4 x 15 = 6 kg m/s. F = 6 / 0.05 = 120 N.

During a collision, a driver experiences a change in momentum (they decelerate from the car's speed to zero). An airbag increases the time taken for the driver to slow down. Since impulse = force x time = change in momentum, and the change in momentum is fixed, a longer time means the force acting on the driver is reduced. This smaller force means less injury to the driver.

• The airbag increases the time over which the driver decelerates / increases collision time (1m)
• Change in momentum is the same / impulse = F x t = change in momentum (constant) (1m)
• Longer time means smaller force (F = delta p / t), reducing risk of injury (1m)

Airbag increases collision time. Impulse = F x t = change in momentum (fixed). Longer time = smaller force = less injury.

Crumple zones are designed to deform and collapse during a collision, increasing the time over which the car (and its occupants) decelerates. The change in momentum (impulse) of the car is fixed. Since impulse = force x time, a longer collision time means the force acting on the car and its occupants is smaller. This reduces the forces on passengers, reducing the risk of serious injury.

• Crumple zone deforms during collision, increasing the time of the collision (1m)
• Change in momentum / impulse is the same regardless of crumple zone (1m)
• Longer time means smaller force (F = impulse / t), reducing injury to occupants (1m)

Crumple zones extend collision time. Same impulse over longer time = smaller force on passengers.

In both elastic and inelastic collisions, momentum is always conserved (total momentum before equals total momentum after). In an elastic collision, kinetic energy is also conserved - the total kinetic energy before equals the total kinetic energy after. In an inelastic collision, kinetic energy is not conserved; some kinetic energy is converted to thermal energy, sound, or deformation of materials.

• Both collisions conserve momentum (1m)
• Elastic collision: kinetic energy is also conserved (1m)
• Inelastic collision: kinetic energy is not conserved (converted to thermal energy / sound / deformation) (1m)

Both conserve momentum. Elastic collisions also conserve kinetic energy. Inelastic collisions convert some kinetic energy to thermal energy, sound or deformation.

• Correct substitution: impulse = 500 x 0.04 (1m)
• Correct answer: 20 Ns (1m)

Impulse = F x t = 500 x 0.04 = 20 Ns.

Impulse = force x time (F x t). Impulse is equal to the change in momentum of an object. Option D is momentum (p = mv), which is related but not the definition of impulse.

Impulse = F x t = change in momentum (delta p). This comes from Newton's second law: F = delta p / delta t, rearranged to F x delta t = delta p.

The unit of impulse is the Newton-second (Ns).

• Newton-second (Ns) or kg m/s (1m)

Impulse = F x t, so units = N x s = Ns. This is equivalent to kg m/s (the unit of momentum).

Impulse = F x t = change in momentum (constant). If t increases by a factor of 10 (from 0.02 to 0.2 s) and impulse is constant, the force must decrease by a factor of 10. F = impulse / t.

Crumple zones increase the time over which the collision occurs. Since impulse = F x t = change in momentum, and the change in momentum is fixed, a longer collision time means a smaller force. This reduces the force experienced by the passengers, reducing injury.

When the ball is thrown, work is done by the hand on the ball. Energy transfers from the chemical energy store of the thrower's muscles to the kinetic energy store of the ball. As the ball rises, kinetic energy decreases and gravitational potential energy increases. Air resistance acts on the ball throughout, transferring some energy from the kinetic energy store to the thermal energy store of the surroundings. At the top of the flight, almost all kinetic energy has been converted to gravitational potential energy. As the ball falls, gravitational potential energy decreases and kinetic energy increases. When the ball hits the ground, the kinetic energy is transferred to thermal energy (from the impact), sound energy, and elastic deformation energy of the ball and ground. Throughout all stages, the total energy remains constant — energy is conserved (it cannot be created or destroyed, only transferred between stores).

• Level 3 (5-6 marks): Logically sequenced answer including: (1) Throw: work done by hand on ball transfers chemical energy to kinetic energy. (2) Rising: kinetic energy decreases, gravitational potential energy increases as ball gains height. Work done against gravity. (3) If air resistance present: some kinetic energy continuously transferred to thermal energy. (4) Falling: gravitational potential energy decreases, kinetic energy increases. (5) Landing: kinetic energy transferred to other stores (thermal, sound, elastic deformation). (6) Overall: total energy is conserved throughout - energy transferred between stores but total remains constant. Must reference conservation of energy and work done. (6m)
• Level 2 (3-4 marks): Identifies most energy transfers correctly but may not fully explain work done or conservation, or may miss some stages. (4m)
• Level 1 (1-2 marks): Some relevant energy stores identified but transfers not clearly explained or linked incorrectly. (2m)

Throw: chemical energy (hand/muscles) transferred to kinetic energy of ball via work done by the throwing force. Rising: kinetic energy decreases, gravitational potential energy increases. Air resistance transfers some energy to thermal store throughout. Falling: gravitational PE converts back to kinetic energy. Landing: kinetic energy transferred to thermal energy, sound energy, and elastic deformation of ball/ground. Throughout: total energy is conserved (conservation of energy: energy cannot be created or destroyed, only transferred between stores).

• Correct time conversion: 5 minutes = 300 seconds (1m)
• Correct equation: W = Pt or rearrangement of P = W/t (1m)
• Correct substitution: W = 2400 x 300 (1m)
• Correct answer: 720000 J (or 720 kJ) (1m)

Time = 5 x 60 = 300 s. W = Pt = 2400 x 300 = 720000 J (720 kJ).

• Correct calculation of kinetic energy: Ek = 0.5 x 900 x 20^2 = 180000 J (1m)
• Statement that work done = kinetic energy gained = 180000 J (1m)
• Rearrangement: F = W / s = 180000 / 80 (1m)
• Correct answer: 2250 N (1m)

Ek = 0.5 x 900 x 400 = 180000 J. Work done = 180000 J. F = W/s = 180000 / 80 = 2250 N.

• Correct substitution: W = 75 x 8 (1m)
• Correct calculation: W = 600 (1m)
• Correct answer with unit: 600 J (1m)

W = Fs = 75 x 8 = 600 J.

• Correct weight: W = 500 x 9.8 = 4900 N (1m)
• Correct substitution: work done = 4900 x 12 (1m)
• Correct answer: 58800 J (or 58.8 kJ) (1m)

Weight = mg = 500 x 9.8 = 4900 N. Work done = F x d = 4900 x 12 = 58800 J.

• Correct rearrangement: F = W / s = 4500 / 15 (1m)
• Correct calculation: F = 300 (1m)
• Correct answer with unit: 300 N (1m)

Rearranging W = Fs: F = W / s = 4500 / 15 = 300 N.

When the brakes are applied, a friction force acts between the brake pads and the wheels. This friction force does work on the car. The kinetic energy of the car is transferred to thermal energy in the brakes and surroundings. The temperature of the brake pads and discs increases. The kinetic energy store of the car decreases to zero as the car stops.

• Friction force / braking force does work on the car (1m)
• Kinetic energy of the car is transferred / decreases as the car slows (1m)
• Energy is transferred to the thermal energy store of the brakes/surroundings / brakes heat up (1m)

Friction (braking force) does work on the car. The kinetic energy store decreases as work is done against friction. Energy is transferred to the thermal energy store of the brakes and surroundings - the brakes heat up.

Both students do the same amount of work because work done equals force multiplied by distance, and both students apply the same force (weight of the mass) over the same distance (0.5 m). The work done is equal. However, the first student has a greater power because power is the rate of energy transfer (or work done per unit time). The first student transfers the same energy in less time, so their power is greater.

• Both students do the same work done (because same force and same distance) (1m)
• Power is defined as rate of energy transfer / work done per unit time (P = W/t) (1m)
• First student has greater power because same work done in less time (1m)

Work done is the same (both do F x d = weight x 0.5 m). Power = W/t. Same work in less time = greater power for student 1.

Work done is a measure of the energy transferred when a force moves an object. Work is done when a force causes a displacement in the direction of the force. The amount of work done equals the force multiplied by the distance moved in the direction of the force. Work done is measured in joules.

• Work done is the energy transferred when a force causes movement / work done = force x distance (1m)
• Work is done only when there is movement in the direction of the force / force must cause displacement (1m)

Work is done when a force causes a displacement in the direction of the force. W = Fs. Work done equals the energy transferred. No movement = no work done.

Work done is calculated using: work done (W) = force (F) x distance (s). The unit of work done is the joule (J). Work is done when a force causes an object to move in the direction of the force.

Work done is a measure of energy transferred. When a force does work on an object, energy is transferred from one store to another (e.g. from chemical store in muscles to kinetic store of the object). Energy is always conserved - it cannot be created or destroyed, only transferred.

Work done = force x distance = 50 x 3 = 150 J. The unit is joules (J).

Work done = force x distance in the direction of the force. When lifting, the crane must exert a force upward to raise the beam, and the work done is the crane's lifting force multiplied by the vertical distance moved. (Note: if lifting at constant speed, the crane force equals the weight, so both options B and C give the same answer in that case - but the force applied by the crane is the force doing the work.)

• Level 3 (5-6 marks): Full description of all 4 phases: (1) Initial: weight > drag, accelerates downward. (2) As speed increases, drag increases, resultant force decreases, acceleration decreases. (3) First terminal velocity: drag = weight, constant speed. (4) Parachute opens: drag >> weight, decelerates. (5) New lower terminal velocity: drag = weight at lower speed. Logical, well-sequenced, correct physics throughout. (6m)
• Level 2 (3-4 marks): Most phases described but some missing. Physics mostly correct but some gaps. (4m)
• Level 1 (1-2 marks): Only basic description. Limited force/acceleration analysis. (2m)

Full skydive sequence: freefall acceleration -> decreasing acceleration -> first terminal velocity -> parachute opens (drag >> weight) -> deceleration -> new lower terminal velocity.

When the skydiver first jumps, weight acts downward and is greater than drag, so there is a resultant downward force. The skydiver accelerates downward. As speed increases, drag force increases. The resultant force decreases, so acceleration decreases. Eventually drag equals weight and the resultant force is zero. The skydiver no longer accelerates and falls at a constant speed called terminal velocity.

• Initially weight is greater than drag / resultant force is downward (1m)
• As speed increases, drag increases, so resultant force decreases and acceleration decreases (1m)
• When drag equals weight, resultant force is zero, acceleration is zero, constant terminal velocity reached (1m)

Key physics: drag increases with speed. Terminal velocity occurs when drag = weight (zero resultant force, zero acceleration).

During the curved section, weight is greater than drag. The resultant force is downward so the object accelerates. As speed increases, drag increases, so the resultant force decreases and the gradient of the graph decreases (acceleration decreases). During the horizontal section, drag equals weight so the resultant force is zero. Acceleration is zero and the object travels at constant speed - terminal velocity.

• Curved section: object accelerating, weight greater than drag, decreasing resultant force causes decreasing acceleration (gradient decreasing) (1m)
• Drag increases as speed increases, causing resultant force to reduce (further explanation of why gradient decreases) (1m)
• Horizontal section: drag equals weight, resultant force zero, acceleration zero, constant terminal velocity (1m)

Speed-time graph: curved (decreasing gradient) = decreasing acceleration due to increasing drag. Horizontal = zero acceleration = terminal velocity (drag = weight).

• Weight = 50 x 9.8 = 490 N downward (1m)
• Resultant force = 490 - 350 = 140 N downward (1m)
• Acceleration = F/m = 140/50 = 2.8 m/s^2 (downward) (1m)

W = 50 x 9.8 = 490 N. Resultant = 490 - 350 = 140 N down. a = F/m = 140/50 = 2.8 m/s^2 downward.

Without the parachute, the skydiver reaches terminal velocity when drag equals weight at a relatively high speed due to their small cross-sectional area. Opening the parachute greatly increases the cross-sectional area, which greatly increases the drag force at any given speed. The drag now exceeds weight, causing deceleration. The skydiver slows down until drag again equals weight, but now at a much lower speed. This new terminal velocity is lower because the larger parachute generates sufficient drag to balance weight at a lower speed.

• Parachute increases cross-sectional area (greatly increases drag force) (1m)
• Drag exceeds weight so there is a net upward force causing deceleration / slowing down (1m)
• New equilibrium (drag = weight) is reached at a lower speed (lower terminal velocity) (1m)

Parachute increases area -> increases drag. Drag > weight -> deceleration -> slower speed -> new equilibrium where drag = weight at lower terminal velocity.

• Weight = 80 x 9.8 = 784 N (1m)
• At terminal velocity, drag = weight = 784 N (1m)

W = 80 x 9.8 = 784 N. At terminal velocity drag = weight, so drag = 784 N.

A streamlined shape is designed to allow air to flow smoothly over the surface without creating turbulence. This reduces the drag force on the vehicle, which means it can travel at a higher speed for the same engine force, or reach a higher terminal velocity.

• Streamlining allows air to flow smoothly (reduces turbulence) (1m)
• Less turbulence means a smaller drag force on the vehicle (1m)

Streamlined shapes reduce turbulence, which reduces drag. Less drag means the vehicle can travel faster (higher terminal velocity) for the same driving force.

• Weight = 25 x 9.8 = 245 N (1m)
• Resultant force = 245 - 180 = 65 N downward (1m)

W = 25 x 9.8 = 245 N. Resultant = 245 - 180 = 65 N downward. The object is still accelerating downward.

At terminal velocity the downward weight force exactly equals the upward drag force. The resultant force is zero so there is no acceleration and the object falls at constant speed.

Drag force increases with speed. The faster an object moves through a fluid, the more air molecules it collides with per second, so the greater the resistive force.

Terminal velocity is the maximum constant speed reached by a falling object when the drag force equals its weight, so the resultant force is zero and there is no further acceleration.

• Maximum constant speed reached when drag equals weight (resultant force = 0 / no acceleration) (1m)

Terminal velocity: constant maximum falling speed when drag = weight (zero resultant force, zero acceleration).

Opening the parachute greatly increases the surface area, which greatly increases drag. The drag now exceeds weight so there is a net upward force, decelerating the skydiver (speed decreases).

Terminal velocity is reached when drag equals weight. Skydiver B has more weight so needs a higher speed to generate enough drag to balance it, giving a higher terminal velocity.

To demonstrate a transverse wave: hold one end of the slinky fixed and move the other end side to side (perpendicular to the length of the spring). This creates peaks and troughs that travel along the spring. To demonstrate a longitudinal wave: push and pull the end of the spring backwards and forwards (parallel to the spring). This creates compressions (coils pushed together) and rarefactions (coils spread apart) that travel along the spring. To change frequency: oscillate the end faster or slower - faster oscillation creates more waves and a shorter wavelength. To change amplitude: move the end through a larger or smaller distance - larger displacement produces a greater amplitude wave. The amplitude does not affect the wave speed or wavelength.

• Transverse wave: move one end of the spring side to side (perpendicular to the spring's length) (1m)
• Longitudinal wave: push and pull one end of the spring backwards and forwards (parallel to spring's length), creating compressions and rarefactions (1m)
• Frequency: increase the speed of oscillation to increase frequency; this decreases the wavelength (for same wave speed) (1m)
• Amplitude: oscillate the end through a larger distance to increase amplitude (1m)
• Observing wavelength: measure the distance between compressions (longitudinal) or between crests (transverse) (1m)
• Changing frequency increases the number of waves visible along the spring; changing amplitude does not affect the number of waves (1m)

This is an RPA-style question. Students should demonstrate understanding of both wave types and the independent effects of changing frequency and amplitude.

• Correct substitution: v = 440 x 0.77 (1m)
• Correct calculation (1m)
• Correct answer: 338.8 m/s (accept 339 or 340) (1m)

Using v = f x lambda: v = 440 x 0.77 = 338.8 m/s. This is close to the speed of sound in air (~340 m/s).

• Correct substitution: f = 1 / 0.025 (1m)
• Correct calculation (1m)
• Correct answer: 40 Hz (1m)

f = 1/T = 1/0.025 = 40 Hz. Frequency and period are reciprocals of each other.

• Correct rearrangement: lambda = v / f (1m)
• Correct substitution: lambda = 3 x 10^8 / 1 x 10^8 (1m)
• Correct answer: 3 m (1m)

Rearranging v = f x lambda gives lambda = v/f = (3 x 10^8) / (1 x 10^8) = 3 m.

The loudspeaker cone vibrates back and forth. When the cone moves forward it compresses the air in front of it, creating a region of higher pressure called a compression. When it moves back it creates a region of lower pressure called a rarefaction. These compressions and rarefactions travel through the air as a longitudinal wave.

• The loudspeaker cone vibrates back and forth (1m)
• Vibrations create compressions (higher pressure) and rarefactions (lower pressure) in the air (1m)
• Sound travels as a longitudinal wave with particles oscillating parallel to the direction of travel (1m)

Sound is produced by vibration. The cone pushes and pulls air, creating alternating high-pressure compressions and low-pressure rarefactions. These travel as a longitudinal wave.

To measure frequency: count the number of wave crests that pass a fixed point in one second, or use a stroboscope to freeze the wave pattern. To measure wavelength: measure the distance between two adjacent crests or troughs on the water surface. Then calculate wave speed using the equation v = f x lambda.

• Measure frequency by counting crests passing a fixed point per second (1m)
• Measure wavelength by measuring the distance between two adjacent crests (or troughs) (1m)
• Use v = f x lambda to calculate wave speed (1m)

In RPA8, frequency is measured by counting peaks per second, wavelength by measuring between consecutive crests, and wave speed is calculated from v = f x lambda.

• Correctly calculate frequency: f = 1/T = 1/0.5 = 2 Hz OR correctly state lambda = v x T (1m)
• Correct substitution: lambda = 8000 / 2 or lambda = 8000 x 0.5 (1m)
• Correct answer: 4000 m (4 km) (1m)

Step 1: f = 1/T = 1/0.5 = 2 Hz. Step 2: lambda = v/f = 8000/2 = 4000 m. Alternatively, lambda = v x T = 8000 x 0.5 = 4000 m.

The student is incorrect. The speed of a wave in a given medium is constant and does not depend on frequency. If frequency increases, the wavelength decreases proportionally so that wave speed stays the same. This is shown by the wave equation v = f x lambda - if v is constant and f increases, lambda must decrease.

• The student is incorrect - wave speed does not increase with frequency (1m)
• Wave speed is constant in a given medium (depends on the medium, not frequency) (1m)
• If frequency increases, wavelength decreases proportionally (v = f x lambda, v constant) (1m)

Wave speed is determined by the medium, not the frequency. The wave equation v = f x lambda means that if speed is constant and frequency increases, wavelength must decrease.

First, read the wavelength from the diagram by measuring from one crest to the next crest. Then use the wave speed equation: wave speed = frequency x wavelength (v = f x lambda). Multiply the frequency of 5 Hz by the wavelength read from the diagram to calculate the wave speed in metres per second.

• Identify wavelength from the diagram as the distance between two adjacent crests (or equivalent points) (1m)
• State the wave equation: wave speed = frequency x wavelength (v = f x lambda) (1m)
• Substitute frequency = 5 Hz and the wavelength from the diagram into the equation to calculate wave speed in m/s (1m)

Wave speed v = f x lambda. First read lambda from the diagram (crest to crest), then multiply by the given frequency of 5 Hz to get the speed in m/s.

The wavelength is the distance from one crest to the next crest (or one trough to the next trough) — one complete wave cycle. The amplitude is the maximum displacement of the wave from the equilibrium (rest) position, measured from the midline to the top of a crest.

• Wavelength: the distance from one crest to the next equivalent crest (or trough to trough / any two equivalent adjacent points) — one complete wave (1m)
• Amplitude: the maximum displacement from the equilibrium/rest position to the crest or trough (1m)

Wavelength (symbol lambda) is measured horizontally between two equivalent adjacent points (e.g. crest to crest). Amplitude is measured vertically from the equilibrium line to the peak (or trough). These are two of the four key wave properties.

In a transverse wave, the particles oscillate perpendicular to the direction of wave travel. In a longitudinal wave, the particles oscillate parallel to the direction of wave travel, creating compressions and rarefactions.

• In a transverse wave, particles oscillate perpendicular to the direction of wave travel (1m)
• In a longitudinal wave, particles oscillate parallel to the direction of wave travel (1m)

The key difference is the direction of oscillation relative to wave travel. Transverse = perpendicular; Longitudinal = parallel (with compressions and rarefactions).

A wave with a greater amplitude carries more energy. The amplitude is the maximum displacement of particles from their equilibrium position, and larger displacement requires more energy.

• Greater amplitude means more energy carried by the wave (1m)
• Amplitude is the maximum displacement from equilibrium / larger displacement requires more energy (1m)

Greater amplitude = more energy. This is because amplitude represents how far particles are displaced, and displacing them further requires more energy input.

Amplitude is the maximum displacement of a particle from the rest (equilibrium) position — it is measured vertically from the midline to the crest or trough. Wavelength is the horizontal distance from one point on the wave to the equivalent point on the next wave, such as from one crest to the next crest.

• Amplitude is the maximum displacement from the rest/equilibrium position (vertical measurement, from midline to crest or trough) (1m)
• Wavelength is the distance from one point to the equivalent point on the next wave / from crest to crest / one complete wave cycle (horizontal measurement) (1m)

Amplitude is a vertical measurement (from equilibrium to peak) telling you how much energy the wave carries. Wavelength is a horizontal measurement (from crest to crest) telling you the length of one complete wave cycle.

A crest is the highest point of a wave — the point of maximum positive displacement above the equilibrium position. A trough is the lowest point of a wave — the point of maximum negative displacement below the equilibrium position.

• Crest is the highest point / maximum positive displacement above the equilibrium position (1m)
• Trough is the lowest point / maximum negative displacement below the equilibrium position (1m)

Crest = maximum positive displacement (above equilibrium). Trough = maximum negative displacement (below equilibrium). Both are at the same distance from the equilibrium — that distance is the amplitude.

Waves transfer energy from one place to another without transferring matter. The medium through which the wave travels is not permanently displaced.

In a transverse wave, particles oscillate perpendicular (at right angles) to the direction of wave travel. Examples include light waves, water waves, and all electromagnetic waves.

Amplitude is the maximum displacement of a particle from its equilibrium (rest) position. A larger amplitude means the wave carries more energy.

Sound cannot travel through a vacuum because sound is a mechanical wave that needs a medium (such as air, water, or a solid) to travel through. In a vacuum there are no particles to vibrate and transfer the energy.

• Sound needs a medium (particles) to travel through / vacuum has no particles to vibrate (1m)

Sound is a mechanical wave requiring particles to transfer energy through compressions and rarefactions. A vacuum has no particles, so sound cannot travel through it.

Amplitude is the maximum displacement of a particle from its rest (equilibrium) position. It is measured from the midline to the crest (or trough), not from crest to trough.

Amplitude is the maximum displacement of a wave particle from the equilibrium (rest) position. This is measured vertically from the equilibrium line to the peak of a crest (or to the bottom of a trough). It is NOT the full crest-to-trough distance (that would be double the amplitude).

Period T = 1/f = 1/50 = 0.02 s. Frequency and period are reciprocals of each other.

Equipment needed: ray box with single slit, rectangular glass block, plain white paper, pencil, protractor, ruler. Method: (1) Place the glass block on the paper and draw its outline carefully. (2) Shine a ray from the ray box at the glass block at a chosen angle. Mark two points on the incident ray and two on the emergent ray. (3) Remove the block and join the incident ray points to the boundary. Draw the normal (perpendicular line) at the point of entry. (4) Join the emergent ray back to the block using the outline. The refracted ray inside the block can be constructed by connecting the entry and exit points. (5) Use a protractor to measure the angle of incidence (between the incident ray and the normal) and the angle of refraction (between the refracted ray and the normal). (6) Repeat for 5 or more different angles of incidence. (7) Calculate the refractive index using n = sin(i)/sin(r) for each set of readings. Calculate the mean value.

• Equipment: ray box (single slit), rectangular glass block, plain paper, pencil, protractor, ruler (1m)
• Place the glass block on paper and draw around its outline. Remove the block. (1m)
• Direct a ray of light at the glass block and mark the incident ray and the emergent ray on the paper (1m)
• Draw the normal at the point where the ray enters the glass block (perpendicular to the surface) (1m)
• Measure the angle of incidence and angle of refraction from the normal using a protractor. Repeat for several different angles of incidence. (1m)
• Calculate refractive index using n = sin(angle of incidence) / sin(angle of refraction). Calculate a mean value from multiple measurements for reliability. (1m)

This full 6-mark Level of Response question requires a complete, logical experimental method including equipment, procedure, measurements, and data analysis to find the refractive index.

The student places the glass block on paper and draws its outline. A ray of light from a ray box is directed at the glass block. The student marks the positions of the incident ray and the emergent ray. They draw the normal at the point where the ray enters the glass. They measure the angle of incidence and the angle of refraction from the normal using a protractor. By repeating this for different angles of incidence, they can investigate how the angle of refraction changes. The refractive index can be calculated from the results.

• Place glass block on paper, trace outline (1m)
• Direct a ray of light at the block and mark the incident and emergent rays (1m)
• Draw the normal at the point of entry and measure angles from the normal using a protractor (1m)
• Repeat for different angles of incidence to see how the angle of refraction changes (1m)

RPA9 method: trace block outline, direct ray at block, mark incident and emergent rays, draw normal, measure angles of incidence and refraction using protractor, repeat for different angles.

Light travels more slowly in water than in air because water is a denser medium. When the light ray enters the water at an angle, one side of the wavefront enters the water first and slows down before the other side. This causes the ray to bend towards the normal. The change in speed causes the change in direction called refraction.

• Light slows down when entering water (from air) because water is denser (1m)
• The change in speed causes the ray to change direction (1m)
• The ray bends towards the normal when entering a denser medium (1m)

Refraction is caused by the change in speed of light at the boundary. Denser medium = slower speed = bends towards normal.

• Correct substitution: n = sin 30 / sin 19 = 0.50 / 0.33 (1m)
• Correct calculation (1m)
• Correct answer: n = 1.52 (no units, refractive index is dimensionless) (1m)

n = sin(30) / sin(19) = 0.50 / 0.33 = 1.52. This is close to the actual refractive index of glass (~1.5). Refractive index has no units.

Optical fibres are made of glass or plastic. Light travels along the fibre and hits the boundary between the glass and the surrounding material at an angle greater than the critical angle. This causes total internal reflection, where all the light is reflected back into the glass and none escapes. The light bounces along the fibre from one end to the other, allowing signals to be transmitted over long distances with very little energy loss.

• Light hits the glass-air boundary at an angle greater than the critical angle (1m)
• Total internal reflection occurs - all light is reflected back into the glass/fibre (1m)
• Light bounces along the fibre and signals are transmitted with little energy loss over long distances (1m)

TIR in optical fibres: light hits boundary at angle > critical angle, reflects completely, bounces along fibre, transmitting signals with minimal loss.

• Correct rearrangement: v = c/n (1m)
• Correct substitution: v = 3.0 x 10^8 / 1.5 (1m)
• Correct answer: 2.0 x 10^8 m/s (1m)

v = c/n = (3.0 x 10^8) / 1.5 = 2.0 x 10^8 m/s. Light travels more slowly in glass than in air, which is why it bends towards the normal when entering glass.

A higher refractive index means the critical angle for total internal reflection is smaller. In diamond, the critical angle is about 24 degrees, much smaller than for glass (about 42 degrees). This means that light entering a diamond is more likely to undergo total internal reflection inside it. When a diamond is cut with many flat faces at precise angles, light bounces around inside many times through total internal reflection before emerging, creating the sparkling appearance.

• Higher refractive index means smaller critical angle for total internal reflection (1m)
• Light is more likely to undergo total internal reflection inside the diamond (1m)
• Light bounces around inside the many cut faces before emerging, creating the sparkling appearance (1m)

Higher refractive index = smaller critical angle = more TIR inside the diamond = light bounces around many faces before emerging = sparkling effect.

The angle of incidence equals the angle of reflection. Both angles are measured from the normal to the surface at the point of incidence.

• The angle of incidence equals the angle of reflection (1m)
• Both angles are measured from the normal (to the surface) (1m)

The law of reflection: angle of incidence = angle of reflection. Both angles measured from the normal (perpendicular to surface).

• Correct calculation: angle of incidence = 90 - 35 = 55 degrees (1m)
• Angle of reflection = 55 degrees (by law of reflection) (1m)

Angles in reflection are measured from the NORMAL (perpendicular to the surface). If the ray makes 35 degrees with the surface, it makes 90 - 35 = 55 degrees with the normal. By the law of reflection, angle of reflection = 55 degrees.

The image formed by a plane mirror is virtual (cannot be projected on a screen), upright, and the same size as the object. The image appears to be the same distance behind the mirror as the object is in front of it.

• Image is virtual and appears behind the mirror / same distance behind as object is in front (1m)
• Image is the same size as the object and upright (1m)

Plane mirror image properties: virtual (behind mirror), same distance behind as object is in front, same size, upright, laterally inverted.

Specular reflection occurs from a smooth surface such as a mirror. The incident rays are parallel, and all the reflected rays are also parallel, producing a clear image. Diffuse reflection occurs from a rough surface. The surface has many tiny irregularities at different angles, so parallel incident rays are reflected in many different directions and no clear image is formed.

• Specular reflection: smooth surface, reflected rays are parallel / regular / produce a clear image (1m)
• Diffuse reflection: rough surface, reflected rays scatter in many directions / no clear image / surface at different angles (1m)

Specular = smooth surface, parallel reflected rays, clear image. Diffuse = rough surface, scattered reflected rays, no clear image.

The law of reflection states that the angle of incidence equals the angle of reflection. Both angles are measured from the normal (a line perpendicular to the surface at the point of incidence).

When light enters a denser medium (like glass from air), it slows down and bends towards the normal. The angle of refraction is smaller than the angle of incidence.

Diffuse reflection occurs because a rough surface has many tiny facets at different angles. Each facet obeys the law of reflection, but since they point in different directions, light is reflected in many different directions overall. This is why rough surfaces appear matt.

Total internal reflection occurs when light is travelling in a denser medium (e.g. glass) and hits the boundary with a less dense medium (e.g. air) at an angle greater than the critical angle. The light is completely reflected back into the denser medium.

Ultrasound is sound with a frequency above 20,000 Hz, which is above the range of human hearing. A transducer (probe) is placed on the surface of the skin and emits pulses of ultrasound into the body. At each boundary between tissues of different densities, some of the ultrasound is reflected back as an echo. The time taken for each echo to return to the transducer is measured. The depth of each boundary is calculated using the equation distance = wave speed x time, and this value is divided by two because the wave travels to the boundary and back. By repeating this process across the organ, the computer builds up a two-dimensional cross-sectional image which is displayed on a monitor.

• Ultrasound is sound with frequency above 20,000 Hz / above human hearing range (1m)
• A transducer (probe) is placed on the skin to emit pulses of ultrasound into the body (1m)
• Pulses are reflected (echoes) at boundaries between different types of tissue (1m)
• The time for each echo to return to the transducer is measured (1m)
• Depth of boundary calculated using distance = wave speed x time, divided by 2 (wave travels to boundary and back) (1m)
• Multiple echoes build up a 2D cross-sectional image displayed on a computer/monitor (1m)

Full Level 3 answer: Ultrasound (f > 20 kHz) is emitted from a transducer placed on the skin. Pulses travel into the body. At each tissue boundary, some sound is reflected (echo). The time for each echo to return is recorded. Using distance = v x t and dividing by 2 (pulse goes to boundary and back), the depth of each boundary is calculated. Repeating across the body builds a 2D cross-sectional image displayed on a monitor.

• Correct substitution: v = 400 x 0.85 (1m)
• Correct calculation performed (1m)
• Correct answer: 340 m/s (1m)

v = f x lambda = 400 x 0.85 = 340 m/s. This is approximately the speed of sound in air at room temperature.

• Correct rearrangement: wavelength = wave speed / frequency (1m)
• Correct substitution: wavelength = 1500 / 2,000,000 (1m)
• Correct answer: 0.00075 m (or 7.5 x 10^-4 m or 0.75 mm) (1m)

wavelength = v / f = 1500 / 2,000,000 = 0.00075 m = 7.5 x 10^-4 m. Short wavelengths are needed in medical imaging to detect small features.

Ultrasound is sound with a frequency above the human hearing range (above 20,000 Hz). In medical imaging, pulses of ultrasound are transmitted into the body. When the ultrasound reaches a boundary between different tissues, some of the wave is reflected back. The time taken for each reflected pulse to return is measured. Since the speed of ultrasound in tissue is known, the distance to each boundary can be calculated. A computer uses these distances and reflected signal strengths to build up an image of the internal structures.

• Ultrasound is sound with a frequency above 20,000 Hz (above human audible range) (1m)
• Ultrasound pulses are reflected at boundaries between different tissues; the time for each reflection (echo) to return is measured (1m)
• The distance to each boundary is calculated using the speed of ultrasound in tissue; a computer builds up an image from the data (1m)

Ultrasound (f > 20 kHz) is used medically because it can penetrate the body and reflects at tissue boundaries. Timing the echoes and using distance = speed x time allows the depth of each boundary to be found. Multiple reflections are combined into an image.

• Identifies that total distance = speed x time = 1500 x 0.06 = 90 m (1m)
• Divides by 2 to get one-way distance: 90 / 2 = 45 m (1m)
• Correct answer: 45 m (1m)

Total distance = 1500 x 0.06 = 90 m. But this is the distance there and back, so depth = 90 / 2 = 45 m. The factor of 2 is a very common exam trap.

Sound travels faster in steel than in air because the particles in steel are much closer together (higher density solid) and are more strongly bonded, so they can transfer vibrations to neighbouring particles more quickly. Sound cannot travel through a vacuum because there are no particles to vibrate and transfer the energy from particle to particle.

• Particles are closer together in steel (denser medium) so vibrations are transmitted more quickly (1m)
• Particles in steel are more strongly bonded so the restoring forces are greater, allowing faster transfer (1m)
• A vacuum has no particles, so there is nothing to vibrate and energy cannot be transferred (1m)

Speed of sound depends on medium: (1) particle spacing - closer particles in solids allow faster transfer, (2) bond strength - stiffer bonds restore position faster. Vacuum = no particles = no medium = no sound transmission.

Infrasound is sound with a frequency below 20 Hz, which is below the lower limit of human hearing. Some animals such as elephants and whales use infrasound for long-distance communication because low-frequency sounds travel further with less energy loss. Infrasound is also used to detect natural events such as earthquakes and volcanic eruptions, as these events generate infrasound waves that travel great distances.

• Infrasound has a frequency below 20 Hz (below human hearing range) (1m)
• Used by animals (e.g. elephants/whales) for long-distance communication (1m)
• Used to detect earthquakes, volcanic eruptions or other natural events (1m)

Infrasound: f < 20 Hz. Uses: (1) animal communication - elephants, whales, tigers use infrasound to communicate over km; (2) geophysical monitoring - infrasound detectors can warn of approaching tsunamis, earthquakes, and volcanic eruptions.

A sound wave is produced when an object vibrates. The vibrations cause particles in the surrounding medium to oscillate parallel to the direction of wave travel, creating compressions and rarefactions. Energy is transferred from particle to particle through the medium without the particles themselves moving overall.

• Object/source vibrates, causing particles to oscillate parallel to direction of travel (compressions and rarefactions) (1m)
• Energy is transferred through the medium particle to particle; particles do not travel with the wave (1m)

Sound waves are longitudinal: particles vibrate parallel to the wave's direction of travel. This creates alternating compressions (high pressure, particles close together) and rarefactions (low pressure, particles spread out). Energy passes from particle to particle; the particles themselves do not travel with the wave.

The amplitude of a sound wave determines its loudness: a larger amplitude means more energy is transferred, so the sound is louder. The frequency of a sound wave determines the pitch: a higher frequency means the particles vibrate more times per second, which the ear detects as a higher pitched sound.

• Amplitude determines loudness: larger amplitude = louder sound (more energy transferred) (1m)
• Frequency determines pitch: higher frequency = higher pitched sound (more vibrations per second) (1m)

Amplitude and loudness: larger amplitude = more energy = louder sound. Frequency and pitch: higher frequency = more vibrations per second = higher pitch. These are independent properties -- you can have a loud high-pitched sound or a quiet low-pitched sound.

Sound is a longitudinal wave. The particles vibrate parallel to the direction of wave travel, creating compressions (regions of high pressure) and rarefactions (regions of low pressure).

Sound is a mechanical wave and requires a medium (particles) to travel through. A vacuum contains no particles, so sound cannot travel through it. This is why space is silent.

The normal human hearing range is approximately 20 Hz to 20,000 Hz (20 kHz). Frequencies below 20 Hz are called infrasound; frequencies above 20,000 Hz are called ultrasound. Both are inaudible to humans.

Frequency determines the pitch of a sound. A higher frequency means more vibrations per second, which the ear perceives as a higher pitched sound. Amplitude determines loudness, not frequency. The speed of sound depends on the medium, not frequency.

X-rays have short wavelengths and high frequencies, giving them high energy. They are ionising radiation. Benefits: X-rays can penetrate soft tissue but are absorbed by bone and dense materials, so they produce images of bones and internal structures. CT scans use multiple X-ray images to build 3D images of organs. Risks: because they are ionising, X-rays can damage DNA in cells, potentially causing mutations and cancer. Precautions: patients wear lead aprons to protect parts of the body not being imaged, because lead absorbs X-rays effectively. Medical staff leave the room or stand behind lead screens. The dose is kept as low as possible, and the number of X-rays a patient receives is limited.

• X-rays have short wavelength and high frequency/energy, making them highly penetrating (1m)
• Benefit: X-rays can pass through soft tissue but are absorbed by dense material like bone, allowing bones and internal structures to be imaged (1m)
• Benefit: used in CT scans for detailed 3D imaging of internal organs (1m)
• Risk: X-rays are ionising radiation and can damage DNA in cells, potentially causing mutations and cancer (1m)
• Precaution: lead shielding / lead aprons used to absorb X-rays and protect tissue not being imaged (1m)
• Precaution: exposure time minimised / doses kept as low as reasonably possible (ALARP principle) / medical staff leave the room (1m)

This Level of Response question requires balanced discussion of benefits (imaging, CT scans) and risks (ionising, DNA damage, cancer) with specific precautions (lead shielding, minimal exposure).

All objects above absolute zero emit infrared radiation. Hotter objects emit more infrared radiation and at shorter wavelengths. A thermal imaging camera detects the infrared radiation emitted by objects. The camera converts the detected infrared radiation into a visible image. Brighter areas on the image represent regions that are emitting more infrared radiation, which means they are at a higher temperature.

• All objects above absolute zero emit infrared radiation (1m)
• The camera detects infrared radiation emitted by objects (1m)
• Hotter objects emit more infrared radiation (higher intensity) (1m)
• Brighter areas on the image correspond to higher temperature / more infrared emitted (1m)

Thermal imaging works because all objects emit infrared. Hotter = more IR emitted = brighter in image. Camera detects IR and converts to visible image.

The student's claim is incorrect. Gamma rays are far more dangerous than microwaves because they are ionising radiation — they have enough photon energy to remove electrons from atoms and cause mutations in DNA, leading to cancer. Microwaves are non-ionising; they heat water molecules by causing them to vibrate but cannot ionise atoms or damage DNA directly. The reason a microwave oven heats food more visibly is because it is designed to concentrate microwave energy on food at high power. A sealed gamma source in a lab emits ionising radiation continuously but cannot be felt or seen — this absence of visible effect does not mean it is safe. Ionising radiation is dangerous precisely because it causes invisible cumulative DNA damage.

• Gamma rays are ionising radiation — they have enough energy to remove electrons from atoms / ionise atoms (1m)
• Ionising radiation can damage/mutate DNA, leading to cancer / cell death (1m)
• Microwaves are non-ionising — they only cause heating by making molecules (water) vibrate; cannot ionise atoms or damage DNA directly (1m)
• The visible effect of a microwave oven is due to high concentrated power, not intrinsic danger of microwaves / gamma sources emit invisible ionising radiation continuously (absence of visible effect does not mean it is safe) (1m)

The key distinction in the electromagnetic spectrum is between ionising and non-ionising radiation. Gamma rays (and X-rays, UV) are ionising: they carry enough energy per photon to remove electrons from atoms, which can break chemical bonds in DNA and cause mutations or cancer. Microwaves, infrared, and radio waves are non-ionising — they deposit energy as heat (molecular vibrations) but cannot ionise atoms. A microwave oven looks dangerous because it is a high-powered device concentrating energy, but the radiation type is far less intrinsically hazardous than gamma, which invisibly damages DNA at much lower power levels.

UV radiation is ionising. It has enough energy to ionise atoms in skin cells, causing mutations in DNA. These mutations can lead to skin cancer. The ozone layer in the atmosphere absorbs most of the UV radiation from the Sun, preventing it from reaching Earth's surface. Sunscreen absorbs or reflects UV radiation before it reaches the skin, reducing the intensity of UV that penetrates to the deeper skin cells where DNA damage can occur.

• UV radiation is ionising — it has enough energy to ionise atoms / damage/mutate DNA in skin cells (1m)
• DNA mutations caused by UV can lead to skin cancer (1m)
• The ozone layer in the atmosphere absorbs UV radiation from the Sun, preventing most of it reaching Earth's surface (1m)
• Sunscreen absorbs or reflects UV radiation, reducing the amount that reaches skin cells (1m)

UV radiation is ionising — its photons carry enough energy to remove electrons from atoms in biological tissue, breaking chemical bonds in DNA. Accumulated DNA mutations are a major cause of skin cancer. Two protection mechanisms exist: (1) the stratospheric ozone layer absorbs most incoming UV before it reaches the surface, which is why ozone depletion increases cancer risk; (2) sunscreen contains compounds that absorb UV photons and re-emit the energy harmlessly as heat, preventing UV from penetrating to the basal layer of the skin where DNA damage is most harmful.

Ultraviolet radiation is ionising radiation. It can damage DNA in skin cells, which can lead to mutations and skin cancer. High doses can also cause sunburn and damage to the eyes. The ozone layer in the atmosphere absorbs most of the ultraviolet radiation from the Sun, protecting life on Earth.

• UV is ionising radiation / can damage DNA (1m)
• Can cause skin cancer / mutations in cells (1m)
• Also causes sunburn and/or eye damage / ozone layer provides protection (1m)

UV radiation is dangerous because it is ionising - it can damage DNA, cause mutations, lead to skin cancer, and cause sunburn and eye damage.

• Correct rearrangement: lambda = v/f (1m)
• Correct substitution: lambda = 3 x 10^8 / 6 x 10^5 (1m)
• Correct answer: 500 m (1m)

lambda = v/f = (3 x 10^8) / (6 x 10^5) = 500 m. Radio waves have long wavelengths - this is typical for AM radio.

• Correct rearrangement: f = v/lambda (1m)
• Correct substitution: f = 3 x 10^8 / 5 x 10^-7 (1m)
• Correct answer: 6 x 10^14 Hz (1m)

f = v/lambda = (3 x 10^8) / (5 x 10^-7) = 6 x 10^14 Hz. This is in the visible light frequency range.

• Correct rearrangement: f = v/lambda (1m)
• Correct substitution: f = 3 x 10^8 / 1 x 10^-10 (1m)
• Correct answer: 3 x 10^18 Hz (1m)

f = v/lambda = (3 x 10^8) / (1 x 10^-10) = 3 x 10^(8-(-10)) = 3 x 10^18 Hz. X-rays have very high frequencies.

Gamma rays are ionising radiation with very high energy. They can kill cancer cells by ionising the DNA within them, preventing cell division. For sterilisation, gamma rays kill bacteria and viruses on medical equipment by ionising and destroying their DNA, without leaving any harmful residue.

• Gamma rays are ionising / have high energy (1m)
• Kill cancer cells by ionising/damaging DNA, preventing cell division (1m)
• Kill bacteria/viruses on equipment by destroying their DNA (sterilisation without chemical residue) (1m)

Gamma rays work in both applications because they are ionising - they damage and destroy DNA. In cancer treatment this kills tumour cells; in sterilisation this kills pathogens.

Gamma rays have a much shorter wavelength and much higher frequency than radio waves. Higher frequency means higher energy. Because gamma rays have very high energy they are ionising radiation - they can knock electrons off atoms and damage DNA in cells, which can cause cancer. Radio waves have very low frequency and energy and are non-ionising, so they do not damage cells in the same way.

• Gamma rays have shorter wavelength / higher frequency than radio waves (1m)
• Higher frequency means higher energy; gamma rays are ionising radiation (1m)
• Gamma rays can damage DNA / cause mutations / cause cancer; radio waves are non-ionising and do not damage DNA (1m)

Higher frequency = higher energy. Gamma is ionising (damages DNA, causes cancer). Radio waves are non-ionising and do not damage cells the same way.

In order of increasing frequency: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays.

• Radio waves and microwaves at the low-frequency end, gamma rays at the high-frequency end (correct relative ordering) (1m)
• All seven types named in the correct sequence: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma (1m)

Mnemonic: 'Raging Martians Invaded Venus Using X-ray Guns'. Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma.

Microwaves are used in satellite communication and mobile phone networks. They are also used in microwave ovens to cook food because water molecules in food absorb microwave radiation and heat up.

• Communication: satellite communication / mobile phones / wireless networks (1 mark) (1m)
• Cooking: microwave ovens / heating food (1 mark) (1m)

Microwaves have two main uses: (1) communication (satellites, mobile phones, Wi-Fi) and (2) cooking food (microwave ovens heat water molecules in food).

Radio waves are used for broadcasting television and radio programmes. They are also used for communication in mobile phones, emergency services, and long-distance communication with satellites and spacecraft.

• Broadcast radio and/or television programmes (1 mark) (1m)
• Mobile phone communication / emergency services / satellite communication (1 mark - any one) (1m)

Radio waves are used in broadcasting (radio, TV) and communication (mobile phones, emergency services, satellites).

All electromagnetic waves travel at the speed of light in a vacuum: 3 x 10^8 m/s. This is one of the fundamental constants of physics.

Gamma rays have the highest frequency (and shortest wavelength) in the electromagnetic spectrum. The order from lowest to highest frequency is: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma.

All electromagnetic waves are transverse waves. The electric and magnetic fields oscillate perpendicular to the direction of wave travel. Unlike sound, they do not need a medium to travel.

All objects above absolute zero emit infrared radiation. Hotter objects emit more infrared radiation and at shorter wavelengths. This is the basis of thermal imaging.

• Level 3 (5-6 marks): Detailed coherent account. Includes all three ray construction rules (parallel ray through far F; ray through centre straight; ray through near F parallel). States how the intersection point gives the image. Correctly describes how image type, orientation, and size change with object distance (beyond 2F: real, inverted, diminished; at 2F: same size; F to 2F: real, inverted, magnified; inside F: virtual, upright, enlarged). Mentions that image distance also changes. (6m)
• Level 2 (3-4 marks): Describes at least two correct ray rules and some correct image properties for different object distances, but the account is incomplete or has minor inaccuracies. (4m)
• Level 1 (1-2 marks): Knows some basic facts (e.g. parallel ray through focal point) and one or two image properties but account is limited. (2m)

Ray construction: (1) ray // axis -> through far F; (2) ray through centre -> straight; (3) ray through near F -> // axis. Image at intersection. Object distances: beyond 2F: real, inverted, diminished; at 2F: real, inverted, same size; between F and 2F: real, inverted, magnified; inside F: virtual, upright, enlarged (magnifying glass).

A real image is formed when light rays actually converge at a point after passing through the lens. A real image can be projected onto a screen. A converging lens produces a real image when the object is beyond the focal point. A virtual image is formed when light rays appear to diverge from a point but do not actually pass through it. A virtual image cannot be projected onto a screen. A converging lens produces a virtual image when the object is between the lens and the focal point (as in a magnifying glass).

• Real image: rays actually converge at a point; can be projected onto a screen (1m)
• Virtual image: rays appear to diverge from a point but do not actually pass through it; cannot be projected (1m)
• Correct situation for each: real image when object is beyond focal point; virtual image when object is between lens and focal point (1m)

Real image: rays actually pass through the image point; can be caught on a screen; formed when object is beyond the focal point of a converging lens. Virtual image: rays only appear to come from the image point; cannot be projected; formed when object is inside the focal length of a converging lens (magnifying glass effect).

To draw a ray diagram for a converging lens with the object beyond the focal point: a ray parallel to the principal axis refracts through the lens and passes through the far focal point; a ray passing through the optical centre of the lens continues in a straight line without changing direction; a ray passing through the near focal point refracts through the lens and emerges parallel to the principal axis. Where these refracted rays cross on the far side of the lens, the real image is formed. Because the rays cross below the axis, the image is inverted. Because the image is on the far side of the lens from the object, it is real and can be projected.

• Ray parallel to principal axis refracts and passes through the far focal point (1m)
• Ray through optical centre passes straight through (undeviated) (1m)
• Rays cross on the far side forming a real, inverted image (real because rays actually converge; inverted because they cross the axis) (1m)

Three standard rays: (1) parallel to axis -> through far focal point; (2) through optical centre -> straight; (3) through near focal point -> parallel to axis. Image where any two cross on far side = real + inverted. Object beyond F gives real, inverted image (size depends on how far beyond F).

When the object is placed inside the focal length of a converging lens, the image formed is virtual, upright, and enlarged. The image is virtual because the diverging refracted rays do not actually converge - they only appear to come from a point further away when traced back. The image is on the same side as the object. The object must be inside the focal length because if it is placed beyond the focal point, the rays converge on the far side producing a real inverted image, which cannot be used as a simple magnifier that you look through.

• Image is virtual, upright, and enlarged (all three properties) (1m)
• The image is virtual because the refracted rays diverge (do not actually converge); it appears on the same side as the object (1m)
• Object must be inside focal length because beyond the focal point, rays converge producing a real, inverted image that cannot be used as a magnifier (1m)

Magnifying glass: object inside F -> rays refract but still diverge. The eye traces them back to find a virtual, upright, enlarged image on the same side as the object. Beyond F: rays converge on far side -> real, inverted image (not useful as magnifier).

A converging (convex) lens refracts parallel rays of light so that they come together at a point called the focal point. A diverging (concave) lens refracts parallel rays of light so that they spread apart as if coming from a single point on the same side as the incoming light. The focal point of a diverging lens is virtual.

• Converging (convex) lens refracts parallel rays to meet at a real focal point (on the opposite side) (1m)
• Diverging (concave) lens refracts parallel rays so they spread apart and appear to come from a virtual focal point (on the same side as incoming light) (1m)

Converging (convex) lens: parallel rays converge at the real focal point on the far side. Diverging (concave) lens: parallel rays spread out and appear to originate from a virtual focal point on the near side. This is why concave lenses always produce virtual images.

• Correct rearrangement: image height = magnification x object height (1m)
• Correct answer: image height = 6 x 2.5 = 15 cm (1m)

Image height = magnification x object height = 6 x 2.5 = 15 cm. The magnification of 6 means the image is 6 times taller than the object.

• Correct substitution: magnification = 15 / 0.05 (1m)
• Correct answer: 300 (no units, it is a ratio) (1m)

Magnification = image height / object height = 15 mm / 0.05 mm = 300. Both values are in mm so no unit conversion needed. Magnification has no units.

• Converts units so both are the same: 300 m = 300,000 mm (or 6 mm = 0.006 m) (1m)
• Correct answer: magnification = 6 / 300,000 = 0.000020 (or 2.0 x 10^-5) (1m)

Convert 300 m to mm: 300 x 1000 = 300,000 mm. Magnification = 6 / 300,000 = 0.000020 = 2.0 x 10^-5. This very small magnification makes sense -- a huge building produces a tiny image on a camera sensor.

A person with long-sightedness needs a converging (convex) lens. Long-sightedness occurs because the eye forms the image of near objects behind the retina rather than on it. A converging lens refracts the light rays together before they enter the eye, so the eye can then focus them correctly onto the retina. The converging lens effectively moves the near point of the eye to a closer distance.

• A converging (convex) lens is needed for long-sightedness (1m)
• The converging lens refracts light rays together before entering the eye so the image falls on the retina (rather than behind it) (1m)

Long-sight: image forms behind retina. Fix: converging lens refracts rays together before entering eye, shifting focus point forward onto the retina. Short-sight (myopia): image forms in front of retina; fix with diverging (concave) lens.

A convex (converging) lens refracts parallel rays of light so that they converge at a single point called the focal point. The distance from the centre of the lens to this point is the focal length.

A virtual image cannot be projected onto a screen because the light rays do not actually converge - they only appear to diverge from that point. A virtual image appears to be on the same side as the object (behind the lens for a converging lens when the object is within the focal length).

A concave (diverging) lens always produces a virtual, upright, diminished image regardless of where the object is placed. The image appears on the same side as the object and is always smaller.

Magnification = image height / object height = 12 / 4 = 3. A magnification greater than 1 means the image is larger than the object. Magnification has no units as it is a ratio.

P-waves are longitudinal waves that can travel through both solids and liquids. S-waves are transverse waves that can only travel through solids. When seismologists detected no S-waves on the far side of the Earth from an earthquake, they concluded that the outer core must be liquid, because S-waves cannot pass through it. P-waves can travel through the outer core but are refracted (bent) at the boundary because their speed changes as they enter material of different density. This refraction creates a P-wave shadow zone where P-waves are not detected. However, P-waves are detected beyond the shadow zone, showing they have been refracted through the inner core. The fact that P-waves can travel through the inner core indicates it is solid. This evidence led to the model of Earth as a solid crust and mantle, a liquid outer core, and a solid inner core.

• P-waves are longitudinal and travel through solids and liquids; S-waves are transverse and travel only through solids (1m)
• S-waves do not reach the far side of the Earth — they are blocked by the outer core, which must be liquid (1m)
• P-waves are refracted (change direction) at the core boundary because their speed changes (1m)
• P-wave shadow zone (region where P-waves are not detected) also provides evidence of a distinct core (1m)
• P-waves are detected on the far side despite the shadow zone, showing they refract through the inner core — the inner core must be solid (1m)
• Overall model: solid crust, solid mantle, liquid outer core, solid inner core (1m)

Full model answer: P-waves = longitudinal, travel through all materials. S-waves = transverse, only through solids. S-wave shadow zone (no S-waves on far side) = liquid outer core. P-wave shadow zone + refraction at core boundary = core has different properties from mantle. P-waves refracted through core but still detected on far side = solid inner core. Earth model: crust (solid), mantle (solid, convects slowly), outer core (liquid iron/nickel), inner core (solid iron/nickel under extreme pressure).

• Correct rearrangement: wavelength = wave speed / frequency (1m)
• Correct substitution: wavelength = 8000 / 2.5 (1m)
• Correct answer: 3200 m (or 3.2 km) (1m)

wavelength = v / f = 8000 / 2.5 = 3200 m. Seismic waves have very long wavelengths because they travel at high speeds but at low frequencies.

• Correct rearrangement: frequency = wave speed / wavelength (1m)
• Correct substitution: frequency = 4000 / 400 (1m)
• Correct answer: 10 Hz (1m)

frequency = v / wavelength = 4000 / 400 = 10 Hz. Seismic waves typically have very low frequencies (below 20 Hz) which is why they are inaudible to humans.

After an earthquake, seismometers around the world detect seismic waves. S-waves, which are transverse, are not detected on the side of the Earth opposite the earthquake because they cannot travel through the liquid outer core. This creates an S-wave shadow zone and provides evidence that the outer core is liquid. P-waves, which are longitudinal, are refracted (bent) as they pass through the core because their speed changes with rock density. This refraction creates a P-wave shadow zone at certain angles, confirming that the outer core has a different composition from the mantle. The detection of P-waves (but not S-waves) on the opposite side of the Earth provides evidence that the inner core is solid.

• S-waves cannot travel through the liquid outer core; S-wave shadow zone provides evidence the outer core is liquid (1m)
• P-waves are refracted (change direction) as they pass through the Earth because wave speed changes with density/material; this creates a P-wave shadow zone (1m)
• The detection of P-waves (refracted through core) but absence of S-waves on far side provides evidence for a solid inner core and liquid outer core structure (1m)

S-wave shadow zone -> outer core is liquid (S-waves = transverse, cannot go through liquid). P-wave refraction + shadow zone -> core has different density/properties from mantle. P-waves still arriving on far side (just refracted) -> inner core is solid (P-waves can pass through).

Seismic waves travel in curved paths through the Earth because the speed of the waves increases with depth as density and pressure increase. Since the wave speed gradually increases with depth, the waves are continuously refracted (bent) at each rock layer boundary. This continuous refraction causes the waves to follow a curved path, bending away from the vertical as they travel deeper, and then curving back upwards towards the surface on the far side.

• Wave speed increases with depth because density and pressure increase with depth (1m)
• Waves are continuously refracted (bent) at each layer boundary as speed changes (1m)
• The continuous refraction bends the waves gradually, creating a curved (not straight) path through the Earth (1m)

Speed increases with depth (greater pressure/density). Refraction at each layer bends the wave slightly. Thousands of layers = smooth curved path. Waves curve away from vertical while going deeper, then curve back to surface on far side.

• Uses correct equation: distance = speed x time (1m)
• Correct substitution: distance = 7200 x 1800 (1m)
• Correct answer: 12,960,000 m (or 12,960 km) (1m)

distance = speed x time = 7200 x 1800 = 12,960,000 m = 12,960 km. This is slightly larger than the Earth's radius (~6,371 km), which makes sense for a wave that has travelled a curved path through the mantle.

Each seismometer records the time at which P-waves and S-waves arrive. Since P-waves travel faster than S-waves, the time difference between the arrival of P-waves and S-waves at a seismometer can be used to calculate the distance from that seismometer to the earthquake. Each seismometer gives a different distance, which is drawn as a circle around that seismometer. Data from three seismometers gives three circles. The epicentre of the earthquake is at the single point where all three circles intersect.

• Time difference between P-wave and S-wave arrival at each seismometer gives the distance from seismometer to earthquake (P-waves travel faster) (1m)
• The distance from each seismometer is used to draw a circle centred on that seismometer (1m)
• Three circles from three seismometers intersect at a single point which is the epicentre (1m)

P-waves faster than S-waves -> arrival time gap = distance indicator. Each seismometer: time gap -> distance -> circle on map. Three circles from three stations intersect at one point = epicentre. Two circles give two possible points; three circles give one unique point.

P-waves are longitudinal waves: particles vibrate parallel to the direction of wave travel. S-waves are transverse waves: particles vibrate perpendicular to the direction of wave travel. P-waves can travel through solids and liquids. S-waves can only travel through solids and cannot pass through liquids.

• P-waves are longitudinal (particles vibrate parallel); S-waves are transverse (particles vibrate perpendicular to wave direction) (1m)
• P-waves travel through solids and liquids; S-waves can only travel through solids (1m)

P-wave = longitudinal (particles parallel to travel); S-wave = transverse (particles perpendicular to travel). P-waves travel through solids AND liquids; S-waves only through solids (transverse waves cannot propagate through liquids).

P-waves (primary waves) travel faster than S-waves (secondary waves) through the same material. Because they travel faster, P-waves cover the distance from the earthquake to the seismometer in less time and arrive first. S-waves are slower and therefore arrive at the seismometer later. The greater the distance from the earthquake, the larger the time gap between the P-wave and S-wave arrivals.

• P-waves travel faster than S-waves (in the same material) (1m)
• Faster waves cover the same distance in less time, so P-waves arrive first (1m)

P-waves travel faster through rock than S-waves. Same distance, faster speed = less time to arrive. The primary (P) and secondary (S) naming reflects the order of arrival at seismometers.

P-waves (primary waves) are longitudinal waves. Particles vibrate parallel to the direction of wave travel, forming compressions and rarefactions. P-waves can travel through solids, liquids, and gases.

S-waves are transverse waves and cannot travel through liquids. The Earth's outer core is liquid, so S-waves are blocked there and cannot continue to the other side of the Earth. This is how scientists deduced that the outer core is liquid.

Both P and S-waves create shadow zones but in different ways. S-waves cannot pass through the liquid outer core at all, creating a large S-wave shadow zone. P-waves are refracted (bent) as they pass through the liquid outer core, creating a P-wave shadow zone at specific angles. Together these provide evidence for the structure of Earth's interior.

When a seismic wave enters a denser rock layer, it speeds up (denser rock is stiffer, transmitting vibrations faster). When a wave speeds up as it crosses a boundary at an angle, it bends away from the normal. This is refraction of seismic waves.

• Place the bar magnet on a piece of paper and draw around it (1m)
• Place the plotting compass at one end of the magnet (near the north pole) (1m)
• Mark the position of the compass needle tip (or tail) with a dot (1m)
• Move the compass so its tail is at the previous dot; mark the new needle tip; repeat to trace a field line (1m)
• Lift the compass and start a new line at a different position around the magnet (1m)
• Arrows on field lines show direction from north to south pole outside the magnet (1m)

Level 3 (5-6 marks): Valid field pattern obtained; steps clearly sequenced including: draw magnet outline, start near pole, mark dots, move compass tail to previous dot tip, repeat, add arrows. Level 2 (3-4 marks): Most steps identified but not fully sequenced. Level 1 (1-2 marks): Some relevant equipment or steps mentioned.

An electromagnet works by passing an electric current through a coil of wire. The current creates a magnetic field around the coil. Adding an iron core inside the coil increases the strength of the magnetic field. The strength can be increased by increasing the current flowing through the coil or by increasing the number of turns in the coil.

• Current flowing through the coil produces a magnetic field (1m)
• Increasing the current increases the strength of the field (1m)
• Increasing the number of turns (coils) of wire increases the strength (1m)

An electromagnet relies on the magnetic effect of a current. A coil of wire carrying current acts like a bar magnet. Strength increases with more current (higher current = stronger field) or more turns of wire (each turn adds to the total field).

• Correct rearrangement: I = F / (B x l) (1m)
• Correct substitution: I = 0.60 / (0.50 x 0.40) (1m)
• Correct answer: I = 3 A (1m)

Rearranging F = BIl to find current: I = F / (B x l) = 0.60 / (0.50 x 0.40) = 0.60 / 0.20 = 3 A.

Magnetic field lines show the direction of the magnetic field at each point — the direction in which a free north pole would move. Where the field lines are closer together the field is stronger. Where the field lines are further apart the field is weaker.

• Field lines show the direction of the magnetic field (direction a north pole would move) (1m)
• Closer field lines indicate a stronger magnetic field (1m)
• Further apart field lines indicate a weaker magnetic field (1m)

Field lines are a visual tool. Their direction shows which way the field acts (defined as the direction a free north pole would move). Their spacing shows the field strength — closely-spaced lines mean the field is strong, widely-spaced lines mean the field is weak.

A permanent magnet always produces a magnetic field and cannot be switched off. An electromagnet only produces a field when current flows through it, so it can be switched on and off. This is an advantage in industrial applications such as cranes used to lift and drop metal objects — the electromagnet can be turned off to release the metal.

• An electromagnet can be switched on and off (by controlling the current) (1m)
• A permanent magnet cannot be switched off (always has a magnetic field) (1m)
• Valid industrial advantage stated (e.g. electromagnet crane can release metal by turning off current) (1m)

The key difference is control. An electromagnet can be switched on and off — useful when you need to pick up and drop metal objects (e.g. in a scrap metal yard). A permanent magnet is always magnetised, which is useful when you always need the field but inconvenient when you need to release objects.

• Correct substitution: F = 0.30 x 4.0 x 0.25 (1m)
• Correct calculation: F = 0.30 N (1m)
• Correct unit: N (newtons) (1m)

Using F = BIl: F = 0.30 x 4.0 x 0.25 = 0.30 N. The equation F = BIl gives the force on a current-carrying conductor in a magnetic field when the wire is perpendicular to the field.

Like poles repel each other. Unlike poles attract each other.

• Like poles repel (same poles push away from each other) (1m)
• Unlike poles attract (opposite poles pull toward each other) (1m)

Like poles (N-N or S-S) repel because their magnetic field lines push against each other. Unlike poles (N-S) attract because their field lines connect and pull the magnets together.

• Field strength doubles (increases by a factor of 2) (1m)
• Correct reasoning: each turn contributes to the field, so more turns = proportionally stronger field (1m)

The magnetic field strength of a solenoid is directly proportional to the number of turns. If the number of turns doubles (with current constant), the field strength doubles (factor of 2).

Soft iron is used because it is easily magnetised when current flows and easily loses its magnetism when the current is switched off. Steel is a hard magnetic material that retains its magnetism — this is not useful when you need the electromagnet to switch off cleanly.

• Soft iron loses its magnetism when the current is switched off (easily demagnetised) (1m)
• Steel retains its magnetism (stays magnetised even when current is off), which is undesirable for an electromagnet core (1m)

Soft iron is a magnetically soft material — it magnetises and demagnetises very easily. This is ideal for an electromagnet core because you want the magnetism to disappear when the current is switched off. Steel (hard magnetic material) keeps its magnetism, so it would not switch off cleanly.

Like poles repel. When two north poles (or two south poles) are brought together, the magnetic fields push them apart. Unlike poles (north and south) attract.

Magnetic field lines run from north to south outside the magnet (and from south to north inside the magnet). The direction of the field line is the direction a free north pole would move.

A solenoid is a coil of wire that produces a magnetic field when an electric current flows through it. The field inside the solenoid is uniform and resembles the field of a bar magnet.

A compass needle aligns with the local magnetic field. The north end of the compass points in the direction of the field. If the compass points toward the south pole of the magnet, the field direction at that point is toward the south pole — which is consistent with field lines entering a south pole.

An electric motor converts electrical energy into kinetic energy. A coil carrying current is placed in a magnetic field. Each side of the coil experiences a force given by F = BIl (the motor effect). The forces on opposite sides of the coil act in opposite directions, creating a turning effect (torque) that makes the coil rotate. A commutator reverses the direction of the current every half turn so that the forces continue to turn the coil in the same direction throughout its rotation. To increase the torque: (1) Increase the current — since F = BIl, a larger current produces a larger force on each side, giving a greater turning effect. (2) Increase the magnetic flux density — a stronger magnetic field also increases the force on each side via F = BIl. Alternatively, increasing the number of turns in the coil increases the total force as each turn contributes additional force.

• The motor effect: current in a coil inside a magnetic field experiences a force (F = BIl) (1m)
• Forces on opposite sides of the coil act in opposite directions, creating a turning effect (torque) (1m)
• The commutator reverses current every half turn to maintain rotation in the same direction (1m)
• Increasing the current through the coil increases the force (F = BIl, so larger I → larger F) (1m)
• Increasing the magnetic flux density increases the force (larger B → larger F) (1m)
• Increasing the number of turns in the coil increases the total force on the coil (1m)

Level 3 (5-6 marks): Clear logical sequence covering the motor effect, the creation of torque from opposing forces on the coil sides, the role of the commutator, and two specific, well-reasoned ways to increase torque. Level 2 (3-4 marks): Motor effect described and at least one way to increase torque identified but reasoning incomplete. Level 1 (1-2 marks): Some relevant physics mentioned but not clearly connected.

An alternating current passes through the coil attached to the speaker cone. The coil is inside a permanent magnetic field. The motor effect produces a force on the coil. Because the current is alternating, the force on the coil repeatedly reverses direction, causing the coil and attached cone to vibrate back and forth. These vibrations of the cone create pressure waves in the air, which we hear as sound.

• Alternating current flows through the coil (which is in a magnetic field) (1m)
• The motor effect produces a force on the coil (1m)
• The alternating current causes the force to repeatedly reverse direction, making the coil vibrate (1m)
• The vibrating cone creates pressure waves (compressions and rarefactions) in the air, which we hear as sound (1m)

Key chain: AC current → alternating force (motor effect) → coil vibrates → cone vibrates → pressure waves in air → sound. The frequency of the AC matches the frequency of vibration and hence the frequency (pitch) of the sound produced.

• Correct substitution: F = 0.40 x 6.0 x 0.15 (1m)
• Correct calculation: F = 0.36 N (1m)
• Correct unit: N (newtons) (1m)

F = BIl = 0.40 x 6.0 x 0.15 = 0.36 N. The force is 0.36 newtons. Remember the wire must be perpendicular to the field for this equation to apply directly.

• Correct rearrangement: B = F / (I x l) (1m)
• Correct substitution: B = 0.80 / (5.0 x 0.20) (1m)
• Correct answer: B = 0.80 T (1m)

Rearranging F = BIl: B = F / (I x l) = 0.80 / (5.0 x 0.20) = 0.80 / 1.0 = 0.80 T.

A current-carrying coil is placed inside a magnetic field. The motor effect produces a force on each side of the coil. Because the forces act on opposite sides of the coil in opposite directions, this creates a turning effect (torque) which causes the coil to rotate. The commutator reverses the direction of current in the coil every half turn, keeping the coil rotating in the same direction.

• Force on the coil is due to the motor effect (current in magnetic field) (1m)
• Forces on opposite sides of the coil are in opposite directions, creating a turning effect (rotation) (1m)
• The commutator reverses the current direction every half turn, maintaining continuous rotation in the same direction (1m)

In an electric motor: 1) Current flows through a coil inside a magnetic field; 2) Motor effect creates forces on the sides of the coil; 3) The forces on opposite sides act in opposite directions, creating a turning effect; 4) The commutator swaps the current direction every half turn so the coil always turns the same way rather than oscillating.

• Correct calculation of force on one turn: F = 0.20 x 2.5 x 0.05 = 0.025 N (1m)
• Correct multiplication by number of turns: 0.025 x 80 (1m)
• Correct total force: 2.0 N (1m)

Force on one turn = BIl = 0.20 x 2.5 x 0.05 = 0.025 N. Total force on one side = 0.025 x 80 turns = 2.0 N. Each turn of the coil contributes to the total force.

You can reverse the direction of the force by reversing the direction of the current through the conductor, or by reversing the direction of the magnetic field. However, if both the current direction and the magnetic field direction are reversed at the same time, the force direction stays the same.

• Reversing the direction of the current through the conductor reverses the force direction (1m)
• Reversing the direction of the magnetic field also reverses the force direction (1m)
• Reversing both the current and the field simultaneously does not reverse the force (1m)

Using Fleming's left-hand rule: reversing either the current or the field alone reverses the force. But reversing both inputs together gives the same output direction — the two reversals cancel each other out.

When a current flows through a wire, moving charges (electrons) create their own magnetic field around the wire. When this wire is placed inside an external magnetic field, the two magnetic fields interact. The interaction between the wire's magnetic field and the external magnetic field produces a resultant force on the wire, acting perpendicular to both the current direction and the field direction.

• Current in the wire creates its own magnetic field (due to moving charges / electrons) (1m)
• The wire's magnetic field interacts with the external magnetic field (1m)
• This interaction produces a force on the wire acting perpendicular to both the current and the field (1m)

Current = moving charges. Moving charges create a magnetic field around the wire. This field interacts with the external field. The interaction produces a force perpendicular to both current direction and external field direction — this is the motor effect.

In Fleming's left-hand rule: the First finger points in the direction of the magnetic Field; the seCond finger points in the direction of the Current; and the thuMb points in the direction of the Motion (force).

• First finger = magnetic field direction AND second finger = current direction (1m)
• Thumb = direction of the force / motion / thrust (1m)

Fleming's left-hand rule mnemonic: thuMb = Motion (force); First finger = Field; seCond finger = Current. Hold the left hand so all three are at right angles to each other.

Hold the left hand so the first finger points in the direction of the magnetic field, the second finger points in the direction of the conventional current. The thumb then points in the direction of the force on the conductor.

• First finger represents the direction of the magnetic field; second finger represents the direction of conventional current (1m)
• The thumb points in the direction of the force (motion) on the conductor (1m)

Fleming's left-hand rule is a mnemonic for the motor effect. Left hand: First finger = Field, seCond finger = Current, thuMb = Motion (force). Point first two fingers and thumb at right angles to each other — they indicate the three perpendicular directions.

If the direction of the current is reversed, the direction of the force on the conductor is also reversed. The force acts in the opposite direction to before.

• The direction of the force is reversed when the current is reversed (1m)
• The force acts in the opposite direction (accept: the conductor moves in the opposite direction) (1m)

From F = BIl: the force magnitude stays the same, but reversing the current direction means Fleming's left-hand rule gives the opposite direction for the thumb (force). The force direction reverses.

If the current is increased, the force on the conductor increases. The force is directly proportional to the current, so doubling the current doubles the force.

• The force increases when the current is increased (1m)
• The force is directly proportional to the current (accept: reference to F = BIl showing F increases with I) (1m)

From F = BIl, if B and l are constant, F is directly proportional to I. Increasing current increases the force by the same factor.

The motor effect is the force experienced by a current-carrying conductor in a magnetic field. This force is what drives an electric motor. Option B describes electromagnetic induction (the generator effect), not the motor effect.

In Fleming's left-hand rule: the First finger = magnetic Field direction; the seCond finger = conventional Current direction; the thuMb = direction of Motion (force on the conductor). A common memory trick: FBI — Field, Current (I), force (motion).

Fleming's left-hand rule gives the direction of the force (motion) on a current-carrying conductor in a magnetic field. The First finger = Field direction; the seCond finger = Current direction; the thuMb = Motion (force) direction.

Using Fleming's left-hand rule: the First finger points in the direction of the magnetic Field (upward), the seCond finger points in the direction of conventional Current (east), and the thuMb points in the direction of the Motion (force). With field up and current east, the thumb points out of the page toward the viewer.

The force on a conductor is given by F = BIl. The force increases with greater current (I), greater magnetic flux density (B), or greater length of wire (l). Reversing the field direction reverses the force direction but does not change its magnitude — so the size of the force is unchanged.

From F = BIl, the force F is directly proportional to the magnetic field strength B (when current I and wire length l stay constant). If B is doubled, F doubles. Reversing the field direction would reverse the force, but simply increasing it doubles the force.

At the power station, a step-up transformer increases the voltage from around 25 kV to 400 kV for transmission through the National Grid. Because power = voltage x current, increasing the voltage means the current is lower for the same power. Lower current means less power is wasted as heat in the cables, since the power lost in a cable is given by P = I²R — halving the current reduces the power loss to one quarter. Near homes and businesses, step-down transformers reduce the voltage to a safe level of 230 V. This makes the system highly efficient because very little energy is wasted during transmission over long distances. Transformers only work with alternating current (AC) because the continuously changing current produces a changing magnetic field in the primary coil, which induces a voltage in the secondary coil by electromagnetic induction.

• Step-up transformer at the power station increases voltage (e.g. from 25 kV to 400 kV) for transmission (1m)
• High voltage means low current in transmission cables (for the same power) (1m)
• Low current means less power is lost as heat in the cables (P = I²R), so transmission is more efficient (1m)
• Step-down transformers reduce voltage to safe levels (e.g. 230 V) for use in homes and businesses (1m)
• Energy is transferred efficiently across long distances because the cables waste less power (1m)
• Transformers only work with AC — the changing current creates a changing magnetic field for electromagnetic induction (1m)

Level 3 (5-6 marks): Logically sequenced account covering step-up at power station, high voltage / low current argument with P = I²R reasoning, step-down before homes, efficiency benefit, and why AC is essential. Level 2 (3-4 marks): Most of the key ideas present but not fully linked or sequenced. Level 1 (1-2 marks): Some relevant terms used (step-up, step-down, voltage, current) but not connected into a coherent explanation.

In an AC generator, a coil of wire is rotated inside a magnetic field. As the coil rotates, it cuts through the magnetic field lines, so the magnetic field through the coil is continuously changing. By electromagnetic induction, this induces an alternating voltage in the coil. As the coil rotates the direction of the induced voltage reverses every half turn, producing alternating current. The slip rings maintain electrical contact with the external circuit while allowing the coil to rotate freely.

• The coil rotates inside a magnetic field (1m)
• The rotation causes the magnetic field through the coil to change, inducing a voltage by electromagnetic induction (1m)
• The voltage is alternating because the direction reverses every half turn of the coil (1m)
• Slip rings maintain electrical contact between the rotating coil and the external circuit (1m)

An AC generator (alternator) converts kinetic energy to electrical energy using electromagnetic induction. Key features: rotating coil + permanent magnetic field → changing flux → induced AC voltage. Slip rings (not commutator) allow AC output by maintaining contact without reversing current. A DC generator (dynamo) uses a commutator instead, which reverses connections every half turn to give DC output.

When the magnet moves into the coil, the magnetic flux through the coil increases. This changing flux induces an EMF and a current, causing the galvanometer to deflect in one direction (say, to the right). By Lenz's law, the induced current flows to oppose the entry of the magnet — the coil acts as if it is trying to repel the incoming magnet. When the magnet is stationary inside the coil, the flux is not changing, so no EMF is induced and the galvanometer returns to zero. When the magnet is pulled out, the flux decreases. An EMF is induced again, but this time in the opposite direction — the galvanometer deflects in the other direction (to the left). By Lenz's law, the induced current now opposes the withdrawal of the magnet — the coil acts as if it is trying to attract and retain the magnet. The size of the deflection is larger if the magnet moves faster (greater rate of change of flux).

• Moving in: galvanometer deflects in one direction (e.g. to the right) — an EMF is induced because the magnetic flux through the coil is changing (1m)
• Stationary: galvanometer reads zero — no relative motion means no change in flux, so no induced EMF or current (1m)
• Moving out: galvanometer deflects in the opposite direction — the flux is decreasing, so the induced current reverses direction to oppose the withdrawal (Lenz's law) (1m)
• Lenz's law applied: when moving in, the induced current opposes entry (coil repels magnet); when moving out, the induced current opposes withdrawal (coil attracts magnet). Faster movement gives a larger deflection. (1m)

Three phases: (1) Moving in — flux increases, EMF induced, galvanometer deflects. By Lenz's law, induced current opposes entry (coil repels magnet). (2) Stationary — flux constant, no EMF, galvanometer at zero. (3) Moving out — flux decreases, EMF induced in opposite direction, galvanometer deflects the other way. By Lenz's law, induced current opposes withdrawal (coil attracts magnet). Speed of movement affects the size (not direction) of deflection.

• Correct substitution: 240 / Vs = 200 / 50 (1m)
• Correct rearrangement: Vs = 240 x 50 / 200 (1m)
• Correct answer: Vs = 60 V (1m)