Edexcel Physics Paper 1

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.

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.

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.

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.

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.

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.

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.

Wind has a kinetic energy store due to the moving air. As air hits the turbine blades, energy is transferred mechanically (by the force of the wind on the blades), causing the blades to rotate. The rotating blades drive a generator, which transfers energy electrically to the electrical energy store in the circuit. During these transfers, some energy is dissipated. Friction in the bearings and generator transfers energy to the thermal energy store of the surroundings by heating. Some energy is also transferred to the surroundings by sound. Because energy is always dissipated in real systems, the useful electrical output energy is always less than the total kinetic input energy. Therefore, efficiency = useful electrical output / total kinetic input, which is always less than 1 (or 100%).

• Kinetic energy store of moving air / wind (1m)
• Energy transferred mechanically (by force) from wind to turbine blades (1m)
• Energy transferred electrically from generator to the circuit (electrical energy store) (1m)
• Some energy is dissipated to the thermal energy store of the surroundings (by heating from friction) or as sound (1m)
• Efficiency = useful output / total input (less than 100% because energy is dissipated) (1m)
• Energy is always dissipated in real systems (law of conservation of energy means total energy is conserved but some is wasted) (1m)

Level 3 (5-6): Identifies kinetic store of wind, mechanical transfer to blades, electrical transfer from generator, dissipation to thermal/sound stores, states efficiency equation, and explains why 100% is impossible. Level 2 (3-4): Identifies most stores and at least two transfer pathways, mentions dissipation. Level 1 (1-2): Identifies kinetic energy of wind or mentions electrical output, with limited development.

• Correct GPE: Ep = mgh = 800 x 9.8 x 45 = 352,800 J (1m)
• Correct Ek: Ek = 0.5 x 800 x 27^2 = 0.5 x 800 x 729 = 291,600 J (1m)
• Correct substitution into efficiency = 291600 / 352800 (1m)
• Correct answer: 0.826... (accept 0.83 or 83%) (1m)

Step 1: GPE = mgh = 800 x 9.8 x 45 = 352,800 J. Step 2: Ek = 0.5 x 800 x 27^2 = 0.5 x 800 x 729 = 291,600 J. Step 3: efficiency = Ek / GPE = 291,600 / 352,800 = 0.826 (83%). Some energy is dissipated to the thermal energy store of the surroundings due to friction and air resistance.

• Useful power output = efficiency x input power = 0.65 x 800 = 520 W (1m)
• Convert time: 5 minutes = 300 s. Total electrical energy input = 800 x 300 = 240,000 J (1m)
• Useful energy output = 0.65 x 240,000 = 156,000 J (1m)
• Wasted energy = 240,000 - 156,000 = 84,000 J (1m)

Step 1: Useful power = 0.65 x 800 = 520 W. Step 2: Time = 5 x 60 = 300 s. Total input energy = 800 x 300 = 240,000 J. Step 3: Useful output energy = 0.65 x 240,000 = 156,000 J. Step 4: Wasted energy = 240,000 - 156,000 = 84,000 J. The question asks for wasted thermal energy, which is 84,000 J, but the primary calculated answer is the useful output 156,000 J as the multi-step result.

The car engine transfers some energy to the thermal energy store of the surroundings. This energy is dissipated and spreads out into the surroundings, making it less useful. Because some energy is wasted as thermal energy rather than being transferred to the kinetic energy store, the engine cannot be 100% efficient.

• Some energy is transferred to the thermal energy store of the surroundings (wasted as heat) (1m)
• This energy is dissipated (spreads out into the surroundings) (1m)
• Because not all energy is usefully transferred to the kinetic store, efficiency is less than 100% (or: efficiency = useful output / total input, which is less than 1 when energy is wasted) (1m)

No engine is 100% efficient because energy is always dissipated to the surroundings, mainly as thermal energy (heat) and sound. Efficiency = useful output energy / total input energy. If 40 kJ out of 100 kJ is usefully transferred, efficiency = 0.40 or 40%.

• Correct substitution: efficiency = 350 / 500 (1m)
• Correct calculation performed (1m)
• Correct answer: 0.7 (or 70%) (1m)

efficiency = useful output energy / total input energy = 350 / 500 = 0.7 (or 70%). Efficiency has no units. A value of 0.7 means 70% of the input energy is usefully transferred.

• Correct substitution: Ek = 0.5 x 0.06 x 20^2 (1m)
• Correct intermediate step: 0.5 x 0.06 x 400 (1m)
• Correct answer: 12 J (1m)

Ek = 0.5 x m x v^2 = 0.5 x 0.06 x (20)^2 = 0.5 x 0.06 x 400 = 12 J. Common mistake: forgetting to square the speed, which would give 0.5 x 0.06 x 20 = 0.6 J (wrong).

Electrical energy is transferred electrically to the thermal energy store of the heating element. Energy is then transferred by heating from the element to the thermal energy store of the water. Some energy is also transferred by heating to the thermal energy store of the surroundings.

• Energy is transferred electrically (from the mains/power supply to the heating element) (1m)
• Energy is then transferred by heating from the element to the water (increasing the thermal energy store of the water) (1m)
• Some energy is dissipated/wasted, transferred by heating to the thermal energy store of the surroundings (1m)

The four pathways for energy transfer are: mechanically (by a force), electrically (by a current), by heating (from hot to cold), and by radiation (electromagnetic waves or sound). The kettle uses electrical transfer then heating transfer.

• Convert time to seconds: 3 minutes = 180 s (or substitute correctly in correct units) (1m)
• Correct substitution: E = P x t = 1200 x 180 (1m)
• Correct answer: 216,000 J (1m)

E = P x t. First convert time: 3 minutes = 3 x 60 = 180 s. Then E = 1200 x 180 = 216,000 J (or 216 kJ). A very common mistake is to forget to convert minutes to seconds.

A closed system is one where no energy is transferred to or from the surroundings. The total energy within a closed system remains constant because energy is conserved.

• A closed system does not transfer energy to or from the surroundings (no energy enters or leaves) (1m)
• The total energy within a closed system remains constant (energy is conserved) (1m)

A closed system has no energy transfer to or from the surroundings. As a result, the total energy is conserved — it stays the same. Real systems are rarely perfectly closed, but the concept is used to apply the law of conservation of energy.

Kinetic energy is the energy stored by an object because of its motion. Any moving object has a kinetic energy store. Gravitational potential is stored due to height, elastic potential due to deformation, and chemical energy is stored in bonds.

The thermal energy store of an object depends on its temperature and mass. As the coffee cools, energy is transferred to the surroundings, so the thermal energy store of the coffee decreases.

Energy can never be created or destroyed. It can only be transferred from one store to another or dissipated to the surroundings. The total energy in a closed system always remains constant.

When the cyclist brakes, friction between the brake pads and wheels transfers energy mechanically (by force). This reduces the kinetic energy store and increases the thermal energy store of the brakes and surroundings. The transfer pathway is mechanical (by friction/force), not by heating directly.

Energy cannot be destroyed. The 90 J of 'wasted' energy is dissipated (spread out) to the thermal energy store of the surroundings, mainly as heat. This is why we say the energy is dissipated rather than destroyed.

Vehicle A KE: Ek = 0.5 x 1000 x 30² = 0.5 x 1000 x 900 = 450,000 J. Vehicle B KE: Ek = 0.5 x 3000 x 20² = 0.5 x 3000 x 400 = 600,000 J. Braking distance uses work = F x d, so d = KE / F. Vehicle A: d = 450,000 / 15,000 = 30 m. Vehicle B: d = 600,000 / 15,000 = 40 m. Vehicle B is more dangerous as it has a greater kinetic energy despite the lower speed and requires a 40 m stopping distance compared to 30 m for Vehicle A. The greater mass of Vehicle B is the dominant factor.

• Level 3 (5-6 marks): Correct kinetic energy calculated for both vehicles (450,000 J and 600,000 J). Work done = F x d used to find braking distances (30 m and 40 m). Clear comparison made with valid conclusion about Vehicle B having greater danger due to longer stopping distance and greater mass. (6m)
• Level 2 (3-4 marks): Kinetic energy found for at least one vehicle. Braking distance method shown for at least one vehicle. Some comparison attempted. (4m)
• Level 1 (1-2 marks): Some relevant physics recalled (Ek = 0.5mv², or work = Fd). Partial calculation shown. (2m)

Vehicle A: Ek = 450,000 J, braking distance = 450,000 / 15,000 = 30 m. Vehicle B: Ek = 600,000 J, braking distance = 600,000 / 15,000 = 40 m. Vehicle B is more dangerous despite lower speed because its much greater mass (3x) gives it more kinetic energy, requiring a 40 m stopping distance.

• Correct equation used: Ek = 0.5 x m x v² (1m)
• Correct substitution: Ek = 0.5 x 0.45 x 30² (1m)
• Correct kinetic energy: 202.5 J (1m)
• States minimum work done = 202.5 J (work-energy theorem: work = change in KE, starting from rest) (1m)

Ek = 0.5 x 0.45 x 30² = 0.5 x 0.45 x 900 = 202.5 J. Since the ball starts from rest, all kinetic energy comes from work done by the foot. Minimum work done = 202.5 J (assuming 100% efficiency).

• Correct substitution: Ek = 0.5 x 1200 x 25² (1m)
• Correct calculation step: 0.5 x 1200 x 625 (1m)
• Correct answer: 375,000 J (allow 375 kJ or 3.75 x 10⁵ J) (1m)

Ek = 0.5 x m x v² = 0.5 x 1200 x 25² = 0.5 x 1200 x 625 = 375,000 J

• Correct rearrangement: speed² = 2 x Ek / mass (= 2 x 2450 / 70) (1m)
• Correct intermediate value: speed² = 70 (1m)
• Correct answer: speed = 8.37 m/s (allow 8.3-8.4) (1m)

Rearrange Ek = 0.5mv² to get v² = 2Ek/m = (2 x 2450) / 70 = 4900 / 70 = 70. Then v = √70 = 8.37 m/s

• Correct rearrangement: m = 2Ek / v² (= 2 x 1200000 / 20²) (1m)
• Correct denominator: v² = 400 (1m)
• Correct answer: mass = 6000 kg (1m)

Rearrange Ek = 0.5mv² → m = 2Ek / v² = (2 x 1,200,000) / 20² = 2,400,000 / 400 = 6000 kg

The kinetic energy of the car is proportional to speed squared. When speed doubles, kinetic energy quadruples. During braking, friction between the brakes and the road converts kinetic energy into thermal energy. If the braking force is the same, a greater kinetic energy requires a greater distance to dissipate it, so braking distance increases by the square of the speed increase.

• Kinetic energy is proportional to speed squared (when speed doubles, KE quadruples) (1m)
• Braking converts kinetic energy into thermal energy (via friction between brakes/road) (1m)
• Greater kinetic energy requires greater braking distance (for the same braking force) to dissipate it (1m)

Ek = 0.5mv² means kinetic energy scales with speed squared. If braking force F is constant, the work done by braking = F x d, which must equal the kinetic energy. So d = Ek/F. If Ek quadruples, d quadruples — this explains why higher speed roads require much greater stopping distances.

• Initial KE: 0.5 x 80000 x 50² = 100,000,000 J (= 1 x 10⁸ J) (1m)
• Final KE: 0.5 x 80000 x 20² = 16,000,000 J (= 1.6 x 10⁷ J) (1m)
• Decrease in KE: 100,000,000 - 16,000,000 = 84,000,000 J = 8.4 x 10⁷ J (1m)

Initial KE = 0.5 x 80000 x 50² = 0.5 x 80000 x 2500 = 100,000,000 J. Final KE = 0.5 x 80000 x 20² = 0.5 x 80000 x 400 = 16,000,000 J. Decrease = 100,000,000 - 16,000,000 = 84,000,000 J = 8.4 x 10⁷ J.

• Correct substitution: Ek = 0.5 x 0.5 x 4² (= 0.5 x 0.5 x 16) (1m)
• Correct answer: 4 J (1m)

Ek = 0.5 x m x v² = 0.5 x 0.5 x 4² = 0.5 x 0.5 x 16 = 4 J

The kinetic energy of the car increases by a factor of 4 (quadruples). This is because kinetic energy is proportional to speed squared. When speed doubles, the kinetic energy increases by 2 squared, which equals 4.

• Kinetic energy increases by a factor of 4 (quadruples / increases 4 times) (1m)
• Because kinetic energy is proportional to speed squared (so doubling speed squares the factor: 2² = 4) (1m)

In the equation Ek = 0.5mv², kinetic energy is proportional to v². Doubling speed from 20 to 40 m/s means v² increases from 400 to 1600 — a factor of 4. So kinetic energy quadruples.

The kinetic energy store of an object contains the energy an object has because it is moving. The amount of kinetic energy depends on the mass of the object and the speed of the object.

• Kinetic energy store contains energy due to motion / because the object is moving (1m)
• Two factors: mass AND speed (both required for this mark) (1m)

The kinetic energy store contains energy because the object is moving. The equation Ek = 0.5mv² shows that both mass (m) and speed (v) determine how much kinetic energy is stored.

Kinetic energy is the energy an object has because it is moving. The car moving along a road has kinetic energy. The book on the shelf has gravitational potential energy, the elastic band has elastic potential energy, and the battery has chemical energy.

The kinetic energy equation is Ek = 0.5 x mass x speed squared (Ek = ½mv²). The 0.5 factor and the squaring of speed are both essential — forgetting either gives the wrong answer.

Kinetic energy is proportional to speed squared (Ek = ½mv²). If speed doubles (2v), then Ek = ½m(2v)² = ½m x 4v² = 4 x (½mv²). So kinetic energy quadruples. This is why doubling speed dramatically increases braking distance.

When brakes are applied, friction between the brake pads and discs converts kinetic energy into thermal energy (heat). The kinetic energy store decreases and the thermal energy stores of the brakes, road, and surroundings increase. Energy is never destroyed — only transferred.

Car A: Ek = 0.5 x 1000 x 20² = 0.5 x 1000 x 400 = 200,000 J. Car B: Ek = 0.5 x 2000 x 10² = 0.5 x 2000 x 100 = 100,000 J. Car A has twice the kinetic energy of Car B. Speed is squared in the equation, so Car A's higher speed has more effect than Car B's greater mass.

• Calculate initial GPE: Ep = mgh = 70 × 10 × 50 = 35000 J (1m)
• Calculate final KE: Ek = ½mv² = ½ × 70 × 25² = ½ × 70 × 625 = 21875 J (2m)
• Apply conservation of energy: thermal energy = GPE − KE (1m)
• Correct answer: 35000 − 21875 = 13125 J (accept 13600 J if g = 9.8 used for GPE) (1m)

Initial GPE = mgh = 70 × 10 × 50 = 35000 J. Final KE = ½mv² = ½ × 70 × 625 = 21875 J. Energy not transferred to KE is transferred to thermal stores by friction: 35000 − 21875 = 13125 J. Note: if g = 9.8 is used, GPE = 70 × 9.8 × 50 = 34300 J, giving thermal = 34300 − 21875 = 12425 J.

At the top of the hill, the cart has maximum gravitational potential energy and zero kinetic energy. As the cart moves down the hill, its height decreases so its gravitational potential energy decreases. At the same time the cart speeds up, so its kinetic energy increases. Energy is transferred from the gravitational potential store to the kinetic store. Because there is no friction or air resistance, energy is conserved — the decrease in gravitational potential energy equals the increase in kinetic energy. At the bottom the cart has maximum kinetic energy and zero gravitational potential energy.

• At the top, the cart has (maximum) GPE and zero / minimum KE (1m)
• As the cart descends, GPE decreases (as height decreases) and KE increases (as speed increases) (1m)
• Energy is transferred from the gravitational potential store to the kinetic store (1m)
• Energy is conserved — the decrease in GPE equals the increase in KE (because no friction/air resistance) (1m)

At the top of the hill the cart is stationary, so it has maximum gravitational potential energy and zero kinetic energy. As it rolls down, height decreases so GPE decreases; speed increases so KE increases. With no friction or air resistance, the law of conservation of energy means the decrease in GPE exactly equals the increase in KE — energy is transferred from the gravitational potential store to the kinetic store. A common error is stating that energy is lost or destroyed; in this ideal scenario none is.

• Calculate GPE at top: Ep = mgh = 2 × 10 × 5 = 100 J (1m)
• State that GPE transferred to KE: Ek = 100 J (energy conserved) (1m)
• Rearrange Ek = ½mv² to give v = √(2Ek/m) (1m)
• Correct answer: v = √(2 × 100 / 2) = √100 = 10 m/s (1m)

Step 1 — calculate GPE: Ep = mgh = 2 × 10 × 5 = 100 J. Step 2 — by conservation of energy (no air resistance), all GPE converts to KE: Ek = 100 J. Step 3 — rearrange Ek = ½mv²: v² = 2Ek/m = 2 × 100 / 2 = 100. Step 4 — v = √100 = 10 m/s.

The student is partially correct. In a vacuum (no air resistance), objects of any mass dropped from the same height would indeed reach the same speed, since GPE per kilogram = gh for all masses and by conservation of energy all GPE converts to KE. However, in practice the feather experiences significant air resistance relative to its weight, which transfers energy to thermal stores. This means less energy reaches the kinetic store, so the feather reaches a much lower speed than the steel ball. The steel ball's greater weight relative to air resistance means it loses very little energy to air resistance and reaches close to the speed predicted by conservation of GPE to KE.

• Correct agreement: in a vacuum or if air resistance is ignored, the statement is correct — GPE per kg is the same and speed would be equal (1m)
• Identifies air resistance as a key factor that invalidates the statement in practice (1m)
• Explains that air resistance transfers energy to thermal stores, reducing KE of the feather (1m)
• Conclusion: the feather travels slower than the steel ball in air because proportionally more energy is lost to air resistance (1m)

The student is partially correct. In a vacuum, all objects at the same height have the same GPE per kilogram (= gh), so conservation of energy predicts the same final speed regardless of mass — the statement holds. However, in air the feather experiences a much greater air resistance force relative to its weight compared to the steel ball. This transfers energy to thermal stores, reducing the kinetic energy the feather gains. So in practice the feather reaches a much lower speed. The key concept is that air resistance breaks the ideal energy conservation between GPE and KE stores.

• Convert mass: 500 g = 0.5 kg (1m)
• Correct substitution: Ep = 0.5 × 10 × 2.5 (1m)
• Correct answer: 12.5 J (1m)

First convert 500 g to kg: 500 ÷ 1000 = 0.5 kg. Then Ep = mgh = 0.5 × 10 × 2.5 = 12.5 J. Always check units — mass must be in kg before substituting.

Gravitational potential energy (Ep) is directly proportional to both mass and height. If mass doubles, Ep doubles. If height doubles, Ep doubles. This is shown by the equation Ep = mgh, where g is the gravitational field strength (9.8 N/kg on Earth). Increasing either mass or height increases the GPE stored in the gravitational store.

• Ep is proportional to mass — doubling mass doubles GPE (or equivalent) (1m)
• Ep is proportional to height — doubling height doubles GPE (or equivalent) (1m)
• Reference to equation Ep = mgh or correct description linking both variables to energy (1m)

Gravitational potential energy (Ep = mgh) is directly proportional to both mass and height. This means if you double the mass, the GPE doubles; if you double the height, the GPE doubles. A common misconception is that only height matters — in fact both mass and height affect GPE equally. Gravitational field strength (g = 9.8 N/kg on Earth) is a constant that links the two variables to energy.

• Correct substitution: Ep = 4 × 10 × 3 (1m)
• Correct answer: 120 J (1m)

Ep = mgh = 4 × 10 × 3 = 120 J. Always substitute values in the correct order: mass (kg) × g (N/kg) × height (m).

• Correct substitution: Ep = 10 × 9.8 × 1.5 (1m)
• Correct answer: 147 J (1m)

Ep = mgh = 10 × 9.8 × 1.5 = 147 J. Note: g = 9.8 N/kg gives a different answer than g = 10 N/kg (which would give 150 J). Use the value stated in the question.

• Correct substitution: Ep = 60 × 10 × 4 (1m)
• Correct answer: 2400 J (1m)

Ep = mgh = 60 × 10 × 4 = 2400 J. The person gains 2400 J of gravitational potential energy as they climb.

• Correct substitution: Ep = 8 × 9.8 × 6 (1m)
• Correct answer: 470.4 J (1m)

Ep = mgh = 8 × 9.8 × 6 = 470.4 J. Using g = 9.8 N/kg (rather than 10) is expected in higher-tier questions. 8 × 9.8 = 78.4, then 78.4 × 6 = 470.4 J.

The gravitational field strength on Earth is 9.8 N/kg (sometimes approximated as 10 N/kg in calculations). It has units of N/kg, not m/s or kg/N.

The gravitational potential energy equation is Ep = mgh, where m is mass (kg), g is gravitational field strength (N/kg), and h is height (m). Ep = ½mv² is kinetic energy.

As the ball rises, it slows down — kinetic energy decreases. At the same time, height increases — gravitational potential energy increases. Energy is transferred from the kinetic store to the gravitational potential store.

Using Ep = mgh: Original = 2 × 10 × 5 = 100 J. Option A: 4 × 10 × 5 = 200 J (doubles). Option B: 2 × 10 × 15 = 300 J (triples). Option C: same as A = 200 J. Option D: 1 × 10 × 10 = 100 J (same). Tripling the height gives the greatest increase.

As the book falls, height decreases so GPE decreases. Speed increases so KE increases. With no air resistance, energy is conserved — the decrease in GPE exactly equals the increase in KE. Total mechanical energy stays constant.

Equipment: Metal block with two holes (one for the immersion heater, one for the thermometer), an immersion heater, a thermometer (or temperature sensor), a joulemeter (or ammeter and voltmeter), a stopwatch, an electronic balance, and insulating material to wrap the block. Method: Measure and record the mass of the metal block using the balance. Insert the immersion heater into one hole and the thermometer into the other hole. Wrap the block in insulating material. Record the initial temperature of the block. Switch on the immersion heater and start the stopwatch. Record the joulemeter reading at the start and at the end of the heating period (or record current and voltage and time). After a set time (e.g. 5 minutes), switch off the heater and record the maximum temperature reached. Calculation: Calculate the temperature change: delta-theta = final temperature - initial temperature. Read the energy supplied from the joulemeter: E = final joulemeter reading - initial reading (or calculate E = V x I x t). Rearrange E = mc(delta-theta) to give c = E / (m x delta-theta). Substitute the values to calculate c in J/kg degrees C. Improvement: Wrap the block in insulation to reduce heat loss to the surroundings. Repeat the experiment and calculate a mean value of c. Wait for the temperature to stop rising (thermal equilibrium) before recording the final temperature.

• Level 3 (5-6 marks): A detailed and logically sequenced method that would lead to a valid outcome. Must include: named equipment (immersion heater, thermometer, joulemeter or ammeter + voltmeter, metal block with holes for heater and thermometer, balance/scales); measurements taken (mass of block, initial temperature, final temperature, energy from joulemeter OR current + voltage + time); correct calculation using E = mcΔθ rearranged to c = E/(mΔθ); AND at least one valid improvement (insulation, repeating and averaging, waiting for thermal equilibrium, using a more precise thermometer). (6m)
• Level 2 (3-4 marks): A method that would mostly lead to a valid outcome but steps may not be fully sequenced or one key element is missing. Most equipment named, most measurements identified, method for calculation present but may contain minor errors. (4m)
• Level 1 (1-2 marks): Some relevant physics included but the method would not lead to a valid outcome. For example: mentions heater and thermometer but does not describe what to measure or how to calculate SHC. (2m)

RPA2 is one of the most commonly tested required practicals. Key points examiners award marks for: (1) correct named equipment including joulemeter; (2) measuring mass, initial and final temperature, and energy; (3) using c = E/(m delta-theta); (4) insulation as improvement. The most common error is forgetting to mention insulation or failing to state how energy is measured.

• Correct energy: E = 60 × 300 = 18,000 J (converting 5 minutes to 300 s) (1m)
• Correct temperature change: Δθ = 54 - 18 = 36 °C (1m)
• Correct rearrangement: c = E / (m × Δθ) = 18000 / (1.5 × 36) (1m)
• Correct answer: 333 J/kg°C (3 s.f.) Accept answers in range 830-835 J/kg°C (1m)

Step 1: Convert time: 5 min = 5 × 60 = 300 s. Step 2: E = P × t = 60 × 300 = 18,000 J. Step 3: Δθ = 54 - 18 = 36 °C. Step 4: c = E / (m × Δθ) = 18,000 / (1.5 × 36) = 18,000 / 54 = 333 J/kg°C. Note: The accepted answer of ~833 J/kg°C could correspond to a different metal; check the calculation chain carefully.

• Correct temperature change: Δθ = 50 - 20 = 30 °C (1m)
• Correct rearrangement: m = E / (c × Δθ) = 63000 / (450 × 30) (1m)
• Correct answer: 4.67 kg (accept 4.7 kg or 14/3 kg) (1m)

Δθ = 50 - 20 = 30 °C. Rearranging E = mcΔθ: m = E / (c × Δθ) = 63,000 / (450 × 30) = 63,000 / 13,500 = 4.67 kg

• Correct energy calculation: E = P × t = 50 × 200 = 10,000 J (1m)
• Correct rearrangement: Δθ = E / (m × c) = 10000 / (0.5 × 400) (1m)
• Correct answer: 50 °C (1m)

Step 1: E = P × t = 50 × 200 = 10,000 J. Step 2: Rearrange E = mcΔθ → Δθ = E / (mc) = 10,000 / (0.5 × 400) = 10,000 / 200 = 50 °C

Water has a high specific heat capacity, meaning it can store a large amount of thermal energy for each degree of temperature rise. This means water can absorb a large amount of energy when heated in the boiler without its temperature increasing greatly. It then transfers this stored thermal energy to the radiators around the building as it cools, releasing a large amount of energy to heat the rooms efficiently.

• Water can store a large amount of thermal energy (per kg per °C) / has high SHC (1m)
• Water absorbs lots of energy from the boiler / heats up without large temperature rise (1m)
• Water transfers/releases large amounts of energy to the radiators/rooms as it cools (1m)

Water's high SHC (4200 J/kg°C) means it can absorb and store a large amount of thermal energy per kg for every degree of temperature rise. This makes it an efficient heat carrier: the boiler heats the water (storing lots of energy), which then flows to radiators and releases that energy to warm the rooms.

The student's value is higher than the accepted value because energy is lost to the surroundings rather than all being transferred to the aluminium block. The joulemeter records all the energy supplied to the heater, but some of this energy heats the air and the surroundings instead of the block. This means E in the equation is too large, making the calculated SHC too high. The student could improve accuracy by wrapping the block in insulating material to reduce heat loss to the surroundings.

• Energy is lost to the surroundings (not all energy goes into the block / heat loss to air/surroundings) (1m)
• The recorded energy E is too large / greater than energy absorbed by block, making calculated c too high (accept: the joulemeter records more energy than the block receives) (1m)
• Improvement: wrap/surround the block with insulation / use a polystyrene jacket to reduce heat loss (accept: repeat and take average / wait for thermal equilibrium before reading temperature) (1m)

The accepted value of SHC for aluminium is 900 J/kg°C. The student got 1050 J/kg°C because c = E/(mΔθ). If E is over-recorded (energy lost to surroundings), then c is calculated as too large. Solution: insulate the block to minimise heat loss so more energy goes into raising the block's temperature.

The independent variable is the type of material (the metal block used). The dependent variable is the temperature change of the block. A control variable is the mass of each block (it must be kept the same for each material tested). The time of heating and the power of the heater are also kept the same.

• Independent variable: type/material of the block (not just 'material') (1m)
• Dependent variable: temperature change / temperature rise of the block (1m)
• Control variable: mass of the block (accept: time of heating, power of heater, starting/initial temperature) (1m)

In this investigation: IV = the material/type of block (what you deliberately change); DV = the temperature change (what you measure); CV = mass of block, time of heating, power of heater (what you keep the same to make it fair).

• Correct substitution: E = 150 × 4200 × 7 (1m)
• Correct answer: 4,410,000 J (or 4410 kJ / 4.41 MJ) (1m)
• Correct conclusion: The student's claim is NOT correct / 540,000 J is much less than 4,410,000 J required (accept: incorrect, not enough energy) (1m)

E = mcΔθ = 150 × 4200 × 7 = 4,410,000 J. The student's claim is NOT correct. The solar panel only transferred 540,000 J, which is far less than the 4,410,000 J needed to raise the water temperature by 7 °C.

• Correct substitution: E = 2 × 900 × 10 (1m)
• Correct answer: 18,000 J (18 kJ) (1m)

E = mcΔθ = 2 × 900 × 10 = 18,000 J

Specific heat capacity is the amount of energy required to raise the temperature of 1 kg of a substance by 1 °C. The unit of specific heat capacity is J/kg°C (joules per kilogram per degree Celsius).

• Energy required to raise temperature of 1 kg by 1 °C (accept: energy needed to increase temperature of 1 kilogram by 1 degree) (1m)
• Unit: J/kg°C (accept J/kg/°C, J kg^-1 °C^-1, joules per kilogram per degree Celsius) (1m)

Specific heat capacity (symbol c) is defined as the energy required to raise the temperature of 1 kg of a substance by 1 °C. Its unit is J/kg°C. Water has a very high SHC of 4200 J/kg°C.

Specific heat capacity is defined as the energy required to raise the temperature of 1 kg of a substance by 1 °C (or 1 K). Option A describes latent heat of fusion; C and D are unrelated definitions.

A specific heat capacity of 4200 J/kg°C means 4200 joules of energy must be transferred to raise the temperature of exactly 1 kilogram of water by exactly 1 °C. This is one of the highest SHC values of any common substance, making water excellent at storing thermal energy.

A higher specific heat capacity means more energy is required per kg per °C. With the same mass and the same energy input, the block with the higher SHC undergoes a smaller temperature change. Using E = mcΔθ, rearranging: Δθ = E/mc. A larger c gives a smaller Δθ.

A joulemeter (energy meter) measures the total electrical energy transferred to the immersion heater in joules. This value is used as E in the equation E = mcΔθ. Without knowing E accurately, the SHC cannot be calculated. The temperature is measured by the thermometer, not the joulemeter.

If energy is lost to the surroundings (poor insulation), then less energy actually reaches the block than the joulemeter records. Rearranging E = mcΔθ gives c = E/(mΔθ). If E is over-estimated (because some energy escaped), the calculated SHC will be too high. Wrapping the block in insulation reduces this heat loss.

Conduction: The vacuum between the glass walls prevents conduction because conduction requires particles to vibrate and pass energy to neighbouring particles. With no particles in the vacuum there can be no conduction. The plastic stopper at the top is a poor conductor of thermal energy, reducing conduction through the stopper. Convection: The vacuum also prevents convection because convection requires a fluid to circulate. There are no particles in the vacuum so no convection currents can form between the walls. Radiation: The silver reflective coating on the glass walls reflects infrared radiation back towards the drink rather than allowing it to be absorbed or emitted through the walls. Shiny silver surfaces are poor emitters and poor absorbers of infrared radiation, so the rate of energy transfer by radiation is greatly reduced. Together these three features minimise all three methods of thermal energy transfer, keeping the drink hot for several hours.

• Level 3 (5-6 marks): Detailed description of all three methods of transfer AND explanation of how each design feature reduces that specific transfer method. Logical structure. (6m)
• Level 2 (3-4 marks): Covers at least two of the three methods with correct explanations of how features reduce them. Some logical structure. (4m)
• Level 1 (1-2 marks): Mentions at least one method correctly with some relevant link to a design feature. (2m)
• Indicative content (conduction): The vacuum between the walls prevents conduction because conduction requires particles. The plastic stopper is a poor conductor, reducing conduction through the top. Glass is also a poor conductor. (1m)
• Indicative content (convection): The vacuum between the walls also prevents convection because convection requires a fluid (particles). There is no fluid between the walls to circulate. (1m)
• Indicative content (radiation): The silver reflective coating on the glass reflects infrared radiation back into the drink rather than absorbing or emitting it. Silver/shiny surfaces are poor emitters and poor absorbers of infrared radiation. (1m)

A vacuum flask uses three design features to minimise all three types of heat transfer: (1) The vacuum between the glass walls eliminates conduction and convection (both need particles/fluid). (2) The silver reflective coating reflects infrared radiation back, minimising radiation losses. (3) The plastic stopper is a poor conductor, reducing conduction at the top. Together these features keep drinks hot for hours.

In metals, there are free electrons that are not bound to any particular atom and can move freely through the metal structure. These free electrons can carry thermal energy rapidly from the hot end to the cool end. Additionally, metal particles vibrate and pass energy to neighbouring particles. Non-metals do not have free electrons, so energy transfer is much slower, relying only on particle vibrations.

• Metals have free electrons that can move freely through the structure (1m)
• Free electrons carry thermal energy rapidly from the hot end to the cool end (1m)
• Non-metals rely only on particle vibration to transfer energy (1m)
• Particle vibration is slower / free electrons make metals much better conductors than non-metals (1m)

Metals are better conductors for two reasons: (1) particle vibrations pass energy along, and (2) free electrons carry energy rapidly through the metal structure. Non-metals lack free electrons so they rely only on slow particle vibrations.

The fluid at the base is heated and expands, becoming less dense. The less dense fluid rises. Cooler, denser fluid sinks to replace it at the base. This creates a convection current that circulates thermal energy through the fluid.

• Heated fluid expands / becomes less dense (1m)
• Less dense fluid rises; cooler denser fluid sinks to replace it (1m)
• This creates a convection current / circulation of fluid that transfers thermal energy (1m)

Convection currents form because heating a fluid causes it to expand and become less dense. Less dense fluid rises, displacing cooler, denser fluid which sinks. This circulation (convection current) transfers thermal energy throughout the fluid.

The Sun emits infrared radiation, which is a type of electromagnetic radiation. Electromagnetic radiation does not need a medium or particles to travel through. It can travel through the vacuum of space and is absorbed by the Earth's surface.

• The Sun emits infrared radiation / electromagnetic radiation (1m)
• Radiation does not need a medium / can travel through a vacuum / does not need particles (1m)
• The radiation is absorbed by the Earth's surface (1m)

The Sun's energy reaches Earth by radiation (infrared/electromagnetic radiation). Unlike conduction or convection, radiation does not require a medium and can travel through the vacuum of space. The Earth's surface absorbs this radiation.

• Convert time to seconds: 5 x 60 = 300 s (1m)
• Correct substitution: P = 36000 / 300 (1m)
• Correct answer: 120 W (1m)

First convert time: 5 minutes = 5 x 60 = 300 s. Then: P = E/t = 36,000 / 300 = 120 W. The unit of power is watts (W).

The fibreglass wool traps pockets of air within it. Air is a poor conductor of thermal energy, so conduction through the trapped air is very slow. The trapped air also cannot circulate, preventing convection. This reduces the rate of thermal energy transfer from the warm house to the colder outside.

• Insulation traps (pockets of) air / air is trapped within the material (1m)
• Trapped air is a poor conductor (of thermal energy) / reduces conduction (1m)
• Trapped air cannot circulate / cannot form convection currents / reduces convection / rate of thermal energy transfer is reduced (1m)

Loft insulation works by trapping pockets of air. Air is a poor thermal conductor (low thermal conductivity), so conduction is greatly reduced. The trapped, still air also cannot circulate, so convection currents cannot form. Both effects reduce the rate of energy transfer.

Dark, matt surfaces are better emitters of infrared radiation than light, shiny surfaces. The matt black can emits infrared radiation at a greater rate than the shiny white can. More thermal energy is radiated per second from the black can, so it loses energy faster and cools more quickly.

• Dark, matt surfaces are better emitters of infrared radiation than light, shiny surfaces (1m)
• The matt black can emits infrared radiation at a greater rate than the shiny white can (1m)
• More thermal energy is lost per second from the black can, so it cools more quickly (1m)

Dark matt surfaces are better emitters of infrared radiation. The matt black can radiates more energy per second than the shiny white can. Because it loses more energy per second, it cools faster. This is consistent with the experimental result.

Particles at the hot end gain energy and vibrate more. They pass this energy to neighbouring particles by vibration. This process continues along the rod so thermal energy transfers from the hot end to the cooler end.

• Particles (at hot end) vibrate more / gain (more) energy / vibrate with greater amplitude (1m)
• Energy is passed to neighbouring particles by vibration / collisions, transferring energy along the rod from hot to cool end (1m)

In conduction, particles at the hot end gain thermal energy and vibrate more vigorously. They collide with and transfer energy to neighbouring particles. In metals, free electrons also carry energy through the structure, making metals better conductors.

• Correct substitution: ratio = 400 / 0.1 (1m)
• Correct answer: 4000 (times greater) (1m)

Ratio = 400 / 0.1 = 4000. Copper conducts thermal energy 4000 times faster than wood under the same conditions, which explains why metals feel much colder to the touch.

Conduction is the main method of thermal energy transfer in solids. Particles vibrate and pass energy to neighbouring particles. Convection requires fluid flow, which cannot happen in solids.

Radiation (infrared radiation) can travel through a vacuum because it is a form of electromagnetic radiation and does not need particles to transfer energy. Conduction and convection both require matter.

When water at the bottom is heated, it expands and becomes less dense. Less dense fluid rises. Cooler, denser water sinks to replace it, creating a convection current.

Dark, matt surfaces are better emitters (and absorbers) of infrared radiation than shiny, light-coloured surfaces. The matt black mug radiates more energy per second so it cools faster.

Metal feels cold because it is a much better thermal conductor than wood. Metal rapidly conducts thermal energy away from your hand into the spoon. The sensation of 'cold' is actually the rapid loss of energy from your skin.

The Earth's temperature is currently in approximate balance: the solar radiation absorbed by the Earth equals the infrared radiation emitted by the Earth into space. This is the radiation balance. If more greenhouse gases absorb more of the outgoing infrared radiation, less energy escapes and the Earth's temperature rises to a new, higher equilibrium. Injecting reflective aerosols into the upper atmosphere would increase the Earth's albedo — more solar radiation would be reflected back into space before being absorbed. This would reduce the amount of radiation absorbed by the Earth. According to the radiation balance, if less radiation is absorbed, the Earth would need to emit less radiation to stay in equilibrium — meaning the Earth would cool to a lower equilibrium temperature. This could counteract global warming. However, there are significant problems. Aerosols do not address the cause — CO2 levels would continue rising. If the aerosol injection were stopped, rapid termination shock warming would occur. Aerosols might affect rainfall patterns unevenly. The effects on ecosystems from reduced sunlight are unknown.

• Level 3 (5-6 marks): The student gives a logically structured account covering: the Earth's radiation balance (absorbed solar = emitted infrared), how aerosols would reflect more solar radiation reducing absorption, the effect on equilibrium (less absorbed means Earth cools to a new equilibrium), and at least one well-explained limitation (e.g. uneven cooling, unknown side effects, termination shock if stopped suddenly, does not address CO2 concentration). (6m)
• Level 2 (3-4 marks): The student covers most points but the account is not fully sequenced or one major aspect is missing/unclear. (4m)
• Level 1 (1-2 marks): The student identifies some relevant ideas such as the greenhouse effect or radiation balance but does not develop them into a coherent evaluation. (2m)
• Level 0 (0 marks): No relevant science. (1m)

This is a Level of Response question testing ability to apply the radiation balance concept to a novel real-world scenario. A top-band answer must cover: (1) current radiation balance clearly, (2) mechanism by which aerosols reduce absorbed radiation, (3) how this shifts equilibrium to a lower temperature, and (4) at least one well-explained limitation showing critical evaluation.

Greenhouse gases such as carbon dioxide, methane and water vapour absorb infrared radiation emitted by the Earth's surface. The gases then re-radiate this energy in all directions, including back towards the surface. This means less infrared radiation escapes into space, so more energy is retained in the Earth system. The Earth's temperature rises until a new equilibrium is reached where the rate of energy absorbed from the Sun equals the rate of energy emitted. Human activities have increased greenhouse gas concentrations. Burning fossil fuels (coal, oil, gas) releases carbon dioxide that was stored for millions of years. Deforestation reduces the number of trees absorbing CO2 through photosynthesis. Data shows that atmospheric CO2 concentration and global average temperature have both risen since industrialisation began in the 1800s, with a strong correlation between the two. One method to reduce emissions is to replace fossil fuels with renewable energy sources such as wind or solar power. An advantage is that these produce no greenhouse gas emissions during operation. A disadvantage is that they are intermittent — wind and solar only generate electricity when the wind blows or the sun shines — so backup energy storage or generation is needed, which increases cost and complexity.

• Level 3 (5-6 marks): Logically structured account covering all three bullet points. Mechanism includes absorption AND re-radiation AND less escaping to space AND temperature rise to new equilibrium. Human evidence includes at least two named sources (fossil fuels, deforestation) and reference to data (correlation of CO2/temperature). Evaluation of one method includes a specific named method with both a correctly explained advantage AND disadvantage. (6m)
• Level 2 (3-4 marks): Two of the three bullet points covered well, or all three covered superficially. Mechanism described but may miss equilibrium point. Human evidence gives one source. One advantage or disadvantage given but not both, or method not named. (4m)
• Level 1 (1-2 marks): One bullet point partially addressed. Mechanism vague (e.g. 'greenhouse gases trap heat'). Human evidence vague (e.g. 'pollution'). Method mentioned without explanation. (2m)
• Level 0 (0 marks): No relevant physics. (1m)

This is a Level of Response question combining three skills: explaining a physical mechanism (greenhouse effect), evaluating human evidence (emissions data and sources), and making a balanced evaluation of a solution. To reach Level 3 all three bullet points must be addressed with depth — a vague mention of 'greenhouse gases trap heat' and 'we burn fossil fuels' only reaches Level 1.

Greenhouse gases in the atmosphere absorb infrared radiation emitted by the Earth's surface. The gases then emit this infrared radiation in all directions, including back towards the Earth's surface. This means less infrared radiation escapes into space. The Earth must increase its temperature to emit enough radiation to maintain the balance with incoming radiation from the Sun. Therefore the average temperature of the Earth increases.

• Greenhouse gases absorb infrared radiation emitted by the Earth's surface (1m)
• The gases re-emit / re-radiate the infrared radiation back towards the Earth's surface / in all directions (1m)
• Less infrared radiation escapes into space / more energy trapped in atmosphere (1m)
• Earth's temperature increases until radiation emitted again equals radiation absorbed / new equilibrium reached (1m)

Greenhouse gases (CO2, CH4, water vapour) absorb infrared radiation emitted by the Earth. They re-emit it in all directions including back to Earth — this reduces the amount of energy lost to space. With more greenhouse gases, even less infrared escapes. To maintain the radiation balance (absorbed = emitted), the Earth must get warmer so it emits more radiation. This is the enhanced greenhouse effect.

A carbon tax creates a financial incentive to reduce CO2 emissions — it makes burning fossil fuels more expensive, encouraging businesses and individuals to switch to lower-carbon alternatives. A limitation is that it may make energy more expensive for poorer households and businesses, and may not reduce emissions fast enough without other policies. Afforestation removes CO2 from the atmosphere directly through photosynthesis, which addresses existing atmospheric CO2 as well as reducing new emissions. A limitation is that it requires very large areas of land, which may compete with agriculture or natural habitats. Trees also take many years to grow and mature before absorbing significant amounts of CO2. Neither method alone is likely to be sufficient — a combination of reducing emissions and removing existing CO2 is needed.

• Wind turbines: advantage — no CO₂ during operation / renewable energy source (1) (1m)
• Wind turbines: limitation — intermittent (does not generate when no wind) / requires energy storage or backup (1) (1m)
• Carbon capture: advantage — removes CO₂ directly from existing power stations / does not require replacing all energy infrastructure (1) (1m)
• Carbon capture: limitation — energy intensive process / storage may leak / does not address all CO₂ sources (1) (1m)

This question requires evaluating two policy approaches to reducing greenhouse gas concentrations. Carbon tax works by changing the economics of fossil fuel use — raising costs incentivises alternatives. Afforestation works through biology — photosynthesis removes CO2 from the atmosphere. A complete evaluation identifies the mechanism AND a limitation for each, showing understanding of both the science and the real-world constraints.

As the temperature of an object increases, the total amount of radiation emitted increases. The object emits more radiation per second at every wavelength. The peak wavelength of the emitted radiation also decreases, shifting towards shorter wavelengths. This means hotter objects emit radiation with a higher peak frequency.

• The total amount of radiation emitted increases (more energy emitted per second) (1m)
• The peak wavelength decreases / shifts to shorter wavelengths (1m)
• Hotter objects emit radiation with a higher peak frequency / peak moves towards shorter wavelengths including visible/UV (1m)

Hotter objects emit more radiation overall (greater intensity at every wavelength). The peak wavelength shortens — a red-hot object is cooler than a white-hot object. Very hot stars emit most radiation in the ultraviolet while cool stars emit mainly in the infrared. This is the basis of colour-temperature in astronomy.

The Earth absorbs radiation from the Sun. The Earth also emits infrared radiation into space. When the temperature is constant, the amount of radiation absorbed equals the amount of radiation emitted. This is called a radiation balance or thermal equilibrium. If more radiation is absorbed than emitted, the Earth's temperature increases until a new equilibrium is reached.

• The Earth absorbs radiation from the Sun (1m)
• The Earth emits infrared radiation into space (1m)
• Temperature is constant when radiation absorbed equals radiation emitted / radiation balance / equilibrium (1m)

The Earth receives energy from the Sun (mainly visible light and UV). It re-radiates this as infrared radiation. When the rate of energy absorbed from the Sun equals the rate of energy emitted by the Earth, the temperature stays constant — this is thermal equilibrium. If this balance is disrupted (e.g. by greenhouse gases trapping more radiation) the temperature changes.

If the Earth's temperature is rising, the amount of radiation absorbed by the Earth must be greater than the amount of radiation emitted by the Earth. This means the radiation balance has been disrupted. More radiation is being absorbed than emitted, so energy is accumulating in the Earth system, causing the temperature to increase. This imbalance could be caused by increased greenhouse gases reducing the infrared radiation that escapes to space.

• The radiation absorbed by the Earth is greater than the radiation emitted by the Earth / radiation balance disrupted (1m)
• More energy is absorbed than emitted / energy accumulates in the system (1m)
• This could be caused by greenhouse gases / less infrared radiation escaping to space (1m)

A rising temperature means absorbed energy exceeds emitted energy — the radiation balance is broken. Energy builds up in the Earth system. The most likely cause is enhanced greenhouse effect: increased greenhouse gas concentrations absorb more of the Earth's infrared radiation and re-emit some back, reducing the amount escaping to space.

A perfect black body absorbs all radiation incident on it, reflecting none. It is also a perfect emitter of radiation, emitting the maximum amount of radiation possible at every wavelength for its temperature.

• Absorbs all incident radiation / reflects no radiation / transmits no radiation (1m)
• Perfect emitter / emits maximum radiation at every wavelength for its temperature (1m)

A perfect black body has two defining properties: (1) it absorbs all radiation that hits it — no reflection or transmission occurs, and (2) it is a perfect emitter, meaning it radiates the maximum possible energy at every wavelength for its temperature. Real objects are not perfect black bodies but some approximate one closely.

• Object R (1 mark) (1m)
• Because it has the shortest peak wavelength / a shorter peak wavelength indicates a higher temperature (1 mark) (1m)

Object R has its peak wavelength at 0.5 micrometres, which is the shortest of the three. As temperature increases, the peak wavelength of the emitted radiation decreases (shifts to shorter wavelengths). Therefore Object R, with the shortest peak wavelength, has the highest temperature. Object P (peak at 10 micrometres) is the coolest.

The student is wrong because the Earth also emits infrared radiation back into space. The temperature is stable because the rate of energy absorbed from the Sun equals the rate of energy radiated back into space. This balance is called radiation equilibrium. Greenhouse gases in the atmosphere absorb some of the outgoing infrared radiation and re-emit it, reducing the rate at which energy escapes — this raises the equilibrium temperature.

• The Earth also emits/radiates infrared radiation back into space (1m)
• Temperature is stable when rate of energy absorbed = rate of energy emitted (radiation equilibrium) (1m)

Adding more greenhouse gases means more of the Earth's outgoing infrared radiation is absorbed by the atmosphere and re-emitted back towards the surface. Less escapes to space. The Earth is now absorbing more energy than it emits, so its temperature rises. As temperature rises, the Earth emits more infrared (Stefan-Boltzmann law direction — not required at GCSE) until a new equilibrium is reached at a higher temperature.

• The star has the higher temperature because it has a shorter peak wavelength (1m)
• The star also emits more total radiation per unit area because it is at a higher temperature (1m)

The star peaks at 290 nm, which is shorter than the Sun's 500 nm. A shorter peak wavelength indicates a higher temperature, so the star is hotter than the Sun. A hotter object also emits more total radiation per unit area at every wavelength, so the star emits more total radiation per unit area than the Sun.

A perfect black body absorbs all radiation incident on it — no radiation is reflected or transmitted. Because it absorbs the maximum possible radiation at every wavelength, it also emits the maximum possible radiation at every wavelength for its temperature. Objects that are good absorbers are also good emitters — this is a fundamental property of thermal radiation.

• A black body absorbs all incident radiation / reflects none / is a perfect absorber (1m)
• Good absorbers are also good emitters / it emits the maximum possible radiation at its temperature (1m)

A perfect black body absorbs all radiation — none is reflected. Kirchhoff's law (at GCSE level: good absorbers are good emitters) means that an object that absorbs maximally at every wavelength also emits maximally at every wavelength. This is why a perfect black body is the ideal emitter as well as the ideal absorber.

A perfect black body absorbs all radiation incident on it — it reflects no radiation and transmits none. It is also a perfect emitter, radiating the maximum amount of energy possible at every wavelength for its temperature. The colour of an object has nothing to do with whether it is a black body in the physics sense.

All objects at any temperature above absolute zero (0 K) emit electromagnetic radiation — including infrared. The amount of radiation emitted increases as temperature increases, but even a cold object like an ice cube emits some radiation. The misconception that only hot objects emit radiation is very common.

As an object gets hotter, the peak wavelength of the radiation it emits shifts to shorter wavelengths. The star at 12 000 K is twice as hot as the Sun (6000 K), so its peak wavelength is at a shorter wavelength — the star would emit more in the ultraviolet range. Option D is tempting but the relationship between temperature and peak wavelength (Wien's law) is not required at GCSE — students just need to know the direction of shift.

Greenhouse gases in the atmosphere absorb some of the infrared radiation emitted by the Earth's surface and re-radiate it in all directions, including back towards the surface. This keeps the Earth warmer than it would be without an atmosphere — this is the natural greenhouse effect. Option C is partially correct but misses the re-radiation step, which is the key mechanism.

• Calculate work done (= GPE gained): W = mgh = 800 x 10 x 15 (1m)
• Correct value: W = 120,000 J (1m)
• Correct substitution: P = W/t = 120,000 / 30 (1m)
• Correct answer: P = 4000 W (1m)

Step 1: Work done = GPE gained = mgh = 800 x 10 x 15 = 120,000 J. Step 2: P = W/t = 120,000 / 30 = 4000 W. This is a multi-step calculation combining gravitational PE and power equations.

• Convert power: 80 kW = 80,000 W; convert time: 5 min = 300 s (1m)
• Calculate total energy input: E = P x t = 80,000 x 300 = 24,000,000 J (1m)
• Apply efficiency: useful energy = 0.35 x 24,000,000 = 8,400,000 J (1m)
• Convert to MJ: 8,400,000 / 1,000,000 = 8.4 MJ (1m)

Step 1: 80 kW = 80,000 W; 5 min = 300 s. Step 2: Total input energy = 80,000 x 300 = 24,000,000 J. Step 3: Useful output = 0.35 x 24,000,000 = 8,400,000 J = 8.4 MJ. A challenge question requiring unit conversion + E=Pt + efficiency.

• Calculate kinetic energy gained: Ek = 0.5 x m x v^2 = 0.5 x 1200 x 30^2 (1m)
• Ek = 0.5 x 1200 x 900 = 540,000 J (1m)
• Calculate power: P = E/t = 540,000 / 6 = 90,000 W (1m)
• Convert to kW: 90,000 / 1000 = 90 kW (1m)

Step 1: Ek = 0.5 x 1200 x 30^2 = 0.5 x 1200 x 900 = 540,000 J. Step 2: P = E/t = 540,000 / 6 = 90,000 W = 90 kW. This combines kinetic energy and power equations — a classic challenge question on AQA papers.

• Convert time to seconds: 3 x 60 = 180 s (1m)
• Correct substitution: E = 2000 x 180 (1m)
• Correct answer: E = 360,000 J (or 360 kJ) (1m)

First convert: 3 minutes = 3 x 60 = 180 s. Then E = P x t = 2000 x 180 = 360,000 J. Note: students often forget to convert minutes to seconds — always check units.

• Correct rearrangement: t = E / P (1m)
• Correct substitution: t = 54000 / 900 (1m)
• Correct answer: t = 60 s (1m)

Rearrange P = E/t to get t = E/P. Substitute: t = 54,000 / 900 = 60 s. The microwave was on for 60 seconds (1 minute).

• Convert power: 3.5 kW = 3500 W (1m)
• Convert time: 40 min = 40 x 60 = 2400 s, then calculate E = P x t = 3500 x 2400 = 8,400,000 J (1m)
• Convert to MJ: 8,400,000 / 1,000,000 = 8.4 MJ (1m)

Convert: 3.5 kW = 3500 W; 40 min = 2400 s. Then E = Pt = 3500 x 2400 = 8,400,000 J = 8.4 MJ. A chain of unit conversions is needed — a common higher-tier skill.

The student's statement is only partly correct. Power is the rate of energy transfer, so a higher power rating means energy is transferred more quickly per second. However, the total energy transferred also depends on time. If the shower is used for a short time and the bulb is left on for many hours, the bulb could transfer more total energy. You cannot compare total energy without knowing how long each appliance is used.

• Power is rate of energy transfer — higher power means more energy transferred per second (not necessarily more total energy) (1m)
• Total energy transferred depends on both power AND time: E = P x t (1m)
• Therefore the statement is not fully correct — without knowing the time, you cannot compare total energy (1m)

Power is energy per second. Higher wattage = more energy per second. But E = Pt means total energy depends on time as well. A 100 W bulb left on for 10 hours uses 3.6 MJ; a 3 kW shower used for 5 minutes uses 0.9 MJ — less total energy despite higher power.

• Convert time to seconds: 15 x 60 = 900 s (1m)
• Correct substitution: E = P x t = 250 x 900 = 225,000 J (1m)
• Convert to kJ: 225,000 / 1000 = 225 kJ (1m)

Convert: 15 min = 900 s. Then E = Pt = 250 x 900 = 225,000 J = 225 kJ. Always convert minutes to seconds, and check the required unit of the answer.

• Correct substitution: P = 900 / 60 (1m)
• Correct answer: P = 15 W (1m)

P = E/t = 900 / 60 = 15 W. The lamp transfers 15 joules of energy every second.

Power is the rate of energy transfer (or the rate of doing work). The unit of power is the watt (W), which is equivalent to one joule per second (J/s).

• Power is the rate of energy transfer (or rate of doing work) (1m)
• Unit is the watt (W), equal to one joule per second (1m)

Power = rate of energy transfer. 1 W = 1 J/s. A 60 W bulb transfers 60 J every second.

Power is defined as the rate of energy transfer, i.e. how quickly energy is transferred from one store to another (or how quickly work is done). It is measured in watts (W), where 1 watt = 1 joule per second.

One watt (W) is defined as one joule per second (J/s). This means a device with a power of 1 W transfers 1 J of energy every second. A more powerful device transfers more energy per second.

To convert kilowatts to watts, multiply by 1000. So 2.5 kW = 2.5 x 1000 = 2500 W. The prefix 'kilo' always means 1000.

First convert 2 minutes to seconds: 2 x 60 = 120 s. Then use P = E/t = 600 / 120 = 5 W. Always convert minutes to seconds before calculating power.

Coal power stations are reliable because they can generate electricity at any time, regardless of weather conditions, and can adjust output to match demand. However, burning coal releases carbon dioxide, a greenhouse gas that contributes to global warming and climate change. Coal is also a non-renewable, finite resource that will eventually run out, and mining coal damages the environment. Wind turbines and solar panels produce very little carbon dioxide during operation, making them much better for the environment. They are also renewable resources that will not run out. However, they are intermittent — wind turbines only generate when wind is blowing and solar panels only generate during daylight. A combination reduces but does not eliminate this problem, and backup storage or additional power sources may be needed. Overall, coal offers better reliability in the short term but has significant environmental disadvantages and will eventually run out. Renewables are better long-term for sustainability and the environment, but intermittency means they may not reliably meet all electricity demand without energy storage.

• (5-6m)
• (3-4m)
• (1-2m)
• (0m)

This is a Level of Response question. Top-level answers (5-6 marks) will discuss environmental impact (CO2, global warming from coal; low emissions from renewables), reliability (coal is on-demand; wind/solar are intermittent), and long-term sustainability (coal is finite; renewables will not run out), reaching a clear reasoned conclusion about the trade-off. Students should use correct physics terminology throughout.

• Correct rearrangement: total input = useful output / efficiency (1m)
• Correct substitution: total input = 1.8 / 0.45 (1m)
• Correct answer: 4 MW (1m)
• Correct unit: MW (or W if 4,000,000 W given) (1m)

Rearranging: total input = useful output / efficiency = 1.8 MW / 0.45 = 4 MW. The wind supplies 4 MW total; 1.8 MW (45%) is converted to electricity and 2.2 MW is wasted (mainly as sound and turbulent air movement).

Fossil fuels are reliable and can generate electricity on demand at any time, but burning them produces carbon dioxide which contributes to global warming and climate change. Renewable resources such as wind and solar have low carbon dioxide emissions during operation, but they are intermittent and cannot always generate electricity when demand is high. Renewable resources will not run out, whereas fossil fuels are finite and will eventually be depleted.

• Fossil fuels are reliable / can generate electricity on demand at any time (1m)
• Fossil fuels produce carbon dioxide / contribute to global warming / climate change (1m)
• Renewable resources produce less / no carbon dioxide during operation / lower environmental impact (1m)
• Renewable resources are intermittent / unreliable / cannot always meet demand; OR renewables will not run out whereas fossil fuels are finite (1m)

Fossil fuels: reliable, on-demand generation but produce CO2 (greenhouse gas, global warming), are finite. Renewables: very low or zero CO2 in operation, will not run out, but intermittent (solar needs sun, wind needs wind, tidal follows tidal cycles) -- not always available when needed. The trade-off is reliability vs environmental impact.

• Correct substitution: efficiency = 350 / 500 (1m)
• Correct efficiency: 0.7 (or 70%) (1m)
• Wasted power = 500 - 350 = 150 W (1m)

efficiency = 350 / 500 = 0.7 (70%). Wasted power = 500 - 350 = 150 W per second. The 150 J/s is dissipated to the thermal energy store of the surroundings.

• Correct rearrangement or identification: useful output = efficiency x total input (1m)
• Correct substitution: useful output = 0.85 x 2400 (1m)
• Correct answer: 2040 J (1m)

Rearranging: useful output = efficiency x total input = 0.85 x 2400 = 2040 J. The remaining 360 J (2400 - 2040) is wasted as thermal energy transferred to the surroundings.

A renewable energy resource is naturally replenished and will not run out, for example solar or wind. A non-renewable energy resource is finite and will eventually be used up, for example coal or natural gas (fossil fuels).

• Renewable: naturally replenished / will not run out (1m)
• Non-renewable: finite / will eventually run out / cannot be replaced on a human timescale (1m)
• Correct example of each (e.g. wind/solar for renewable; coal/oil/gas for non-renewable) (1m)

Renewable resources (wind, solar, tidal, hydroelectric, geothermal, bio-fuel, wave) are replenished naturally and will not run out. Non-renewable resources (fossil fuels: coal, oil, natural gas; nuclear/uranium) are finite -- once used they cannot be replaced on a human timescale.

• Correct substitution: efficiency = 19.2 / 24 (1m)
• Correct decimal efficiency: 0.8 (1m)
• Correct percentage: 80% (1m)

efficiency = 19.2 / 24 = 0.8. As a percentage: 0.8 x 100 = 80%. The boiler wastes 4.8 kW (24 - 19.2), likely through hot flue gases escaping to the surroundings.

Wind turbines only generate electricity when wind is blowing. When there is no wind or insufficient wind speed, they produce little or no electricity. Electricity demand continues 24 hours a day regardless, so there may be times when supply does not meet demand. This makes wind turbines unreliable as the sole source of electricity.

• Wind is intermittent / only generates when wind is blowing / dependent on wind speed (1m)
• Electricity demand is continuous / 24 hours / does not stop at night (1m)
• Supply may not match demand at all times / wind is unpredictable / unreliable sole source (1m)

Wind turbines are intermittent -- they only generate when wind is blowing at sufficient speed. Electricity demand is continuous (day and night). When supply is less than demand, there will be power shortages. Backup sources (fossil fuels, nuclear, battery storage) are needed to fill the gap.

First, lubrication of moving parts reduces friction, which reduces the amount of energy wasted as heat due to rubbing surfaces. Second, adding thermal insulation reduces energy lost to the surroundings as heat, keeping more energy inside the device. Third, using more efficient components (such as LED lights instead of filament bulbs) converts a greater proportion of input energy into useful output rather than waste heat.

• Lubrication: reduces friction between moving parts, so less energy is wasted as heat (accept: streamlining to reduce air resistance) (1m)
• Insulation: reduces heat loss to the surroundings / prevents thermal energy escaping from the device (1m)
• Efficient components / technology upgrade: converts a greater proportion of input energy into useful output (e.g. LED instead of filament bulb, regenerative braking, better motors) (1m)

To improve efficiency: (1) lubricate to reduce friction heat, (2) insulate to reduce thermal losses to environment, (3) upgrade components (LEDs, efficient motors) to reduce waste. Each reduces the wasted energy arrow on the Sankey diagram.

• Correct substitution: efficiency = 600 / 800 (1m)
• Correct answer: 0.75 (or 75%) (1m)

efficiency = useful output / total input = 600 / 800 = 0.75. To express as a percentage: 0.75 x 100 = 75%. Efficiency has no unit as it is a ratio of two energies.

Lubrication reduces friction between moving parts, so less energy is transferred to the thermal energy store of the surroundings.

• Correct method named (e.g. lubrication, thermal insulation, streamlining) (1m)
• Correct mechanism: how the method reduces the unwanted energy transfer (e.g. lubrication reduces friction so less wasted as thermal energy) (1m)

Three main methods: (1) Lubrication -- oil between surfaces reduces friction, so less energy is wasted as thermal energy. (2) Thermal insulation -- traps heat in so less energy is lost to surroundings by conduction/convection. (3) Streamlining -- reduces air resistance so less energy is wasted overcoming drag.

Efficiency = useful output energy / total input energy x 100%. Efficiency = 35 J / 100 J x 100% = 35%.

• Correctly uses efficiency = useful output energy / total input energy (or correctly identifies values from Sankey diagram: 35 J useful, 100 J input) (1m)
• Correct answer: efficiency = 35% (or 0.35 as a decimal) (1m)

Efficiency = (useful output energy / total input energy) x 100% = (35 / 100) x 100% = 35%. The wasted energy (65 J) goes to the environment as heat.

The wasted energy output shown in the Sankey diagram represents energy that is transferred to the surroundings as thermal (heat) energy. This energy is dissipated into the environment and cannot easily be recovered or used to do useful work.

• Wasted energy is transferred to the surroundings as heat / thermal energy (1m)
• This energy is dissipated into the environment / cannot be used to do useful work / is not useful (1m)

Wasted energy = energy transferred to surroundings as heat (thermal energy). It is dissipated and cannot be recovered. The first law of thermodynamics means total energy is conserved, but it becomes less useful (more spread out).

Efficiency = useful output energy transfer / total input energy transfer. It measures what fraction of the energy put in is actually used for the intended purpose. This value is always between 0 and 1 (or 0% and 100%).

Wind is a renewable energy resource because it is naturally replenished and will not run out. Coal, natural gas, and nuclear (uranium) are non-renewable because they exist in finite amounts and cannot be replaced on a human timescale.

Lubrication reduces friction between moving parts, so less energy is wasted as heat. Oil forms a thin film between surfaces, allowing them to slide more easily. This directly reduces the unwanted thermal energy transfer caused by friction.

In a Sankey diagram, the width of each arrow is proportional to the amount of energy it represents. A wider arrow means more energy. The wide input arrow represents total input energy; narrower arrows represent smaller amounts of useful or wasted energy.

Nuclear power is non-renewable because uranium (the fuel) exists in limited quantities and cannot be replaced on a human timescale. Renewable refers to whether the energy source is naturally replenished, not whether it produces CO2. Nuclear has very low CO2 emissions in operation but is still non-renewable.

Nuclear power stations can generate electricity continuously, 24 hours a day, regardless of weather conditions. Solar panels only work in daylight, wind turbines only when wind is blowing, and tidal barrages only generate when tides are flowing. Nuclear is therefore more reliable for meeting constant electricity demand.

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.