Question 1
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.2 — Motion: deceleration as negative acceleration, distance–time graphs, speed as gradient (Parts (a), (c)(i), (c)(ii))
• Topic 1.5.1 — Effects of forces: resultant force, F = ma, reaction time (Parts (b), (d)(i), (d)(ii))
▶️ Answer/Explanation
(a)
Deceleration is a negative acceleration — it describes the rate at which the velocity of an object decreases over time. More precisely, it is the magnitude of the change in velocity per unit time when an object is slowing down. An object decelerating still has acceleration in the physics sense, but the acceleration acts in the opposite direction to the motion, causing the speed to reduce.
(b)
The car does not begin to decelerate immediately at t = 0 because of the driver’s reaction time (also called thinking time). There is a short delay between the driver seeing the obstruction and actually pressing the brake pedal — during this time the car continues at constant speed since no braking force has yet been applied. This is why the distance–time graph shows a straight line (constant speed) before the gradient begins to decrease.
(c)(i)
The property of a distance–time graph that corresponds to speed is the gradient (slope) of the graph. A steeper gradient indicates a higher speed, a horizontal line (zero gradient) indicates the object is at rest, and a straight line indicates constant speed. This is because speed = distance ÷ time, which is exactly what the gradient of a distance–time graph calculates.
(c)(ii)
The initial speed is found from the gradient of the straight-line section of the distance–time graph (before braking begins). Reading from Fig. 1.1: the car travels approximately 40 m in 2 s during the constant-speed phase, giving v = 40/2 = 20 m/s. The exact value depends on the scale of Fig. 1.1, but full marks require identifying two suitable points on the straight section and dividing the change in distance by the change in time.
(d)(i)
The deceleration is greater than F/m because, in addition to the braking force F from the brakes, there are other resistive forces also acting on the car — such as air resistance (drag) and rolling friction from the tyres on the road. The total resistive force is therefore greater than F alone, and by Newton’s Second Law (a = Ftotal/m), the actual deceleration is correspondingly greater than F/m.
(d)(ii)
The deceleration is not constant because the air resistance (drag) decreases as the car slows down. Air resistance is dependent on the speed of the car — the faster the car moves, the greater the drag force. As the car decelerates and its speed decreases, the air resistance decreases too, so the total resistive force decreases. Since the braking force F is constant but the total resistive force changes, the net deceleration (= total force / mass) also changes and is therefore not constant throughout the braking process.
Question 2
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.7.1 — Energy: gravitational potential energy stored in the reservoir water (Part (a))
• Topic 1.8 — Pressure: total pressure = atmospheric pressure + ρgΔh; force = pressure × area (Parts (b)(i), (b)(ii))
• Topic 1.7.1 — Energy: kinetic energy Ek = ½mv²; constant speed in a pipe of constant cross-section (Part (c))
▶️ Answer/Explanation
(a)
The form of energy stored in the water in the reservoir is gravitational potential energy. Because the water is held at a significant height above the turbine, it possesses energy by virtue of its position in Earth’s gravitational field. As the water descends 150 m through the pipe to the turbine, this gravitational potential energy is converted into kinetic energy, which in turn drives the turbine and generator to produce electrical energy.
(b)(i)
Total pressure = 1.6 × 10⁶ Pa
The total pressure at the turbine is the sum of atmospheric pressure and the pressure due to the column of water above it. Using Δp = ρgΔh: pressure due to water = 1000 × 10 × 150 = 1.5 × 10⁶ Pa. Adding atmospheric pressure: total pressure = 1.0 × 10⁵ + 1.5 × 10⁶ = 1.6 × 10⁶ Pa. Both contributions must be included — omitting atmospheric pressure is a common error that loses a mark.
(b)(ii)
Force = 5.6 × 10⁶ N
Using the definition of pressure as force per unit area, rearranged to F = pA: F = 1.6 × 10⁶ × 3.5 = 5.6 × 10⁶ N. This enormous force — equivalent to over 560 tonnes — illustrates why hydroelectric turbines are engineered to extremely high structural specifications, and why water pressure at depth is so powerful and practically significant in engineering design.
(c)
Because the pipe has a constant cross-sectional area, the speed of the water flowing through it must remain constant throughout — any given volume of water entering one end of the pipe must exit the other end at the same rate. Since the speed of the water does not change and the mass of water passing any point per second is also constant, the kinetic energy (Ek = ½mv²) of the water remains constant. The gravitational potential energy lost by the water as it descends is instead converted to an increase in pressure, not an increase in kinetic energy.
Question 3
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $2.2.3$ — Melting, boiling and evaporation (Part $\mathrm{(a)}$)
• Topic $2.3.1$ — Conduction [Supplement] (Part $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Faster / more energetic molecules escape from the surface of the water into the air. Because only the highest-energy molecules leave, the average kinetic energy of the remaining molecules decreases. Since temperature is directly related to average molecular kinetic energy, the temperature of the water — and therefore the cloth — falls. The slower, lower-energy molecules left behind represent a cooler liquid overall.
(b)
For the correct answer (any three points):
The metal lattice atoms/ions vibrate due to their thermal energy. These vibrating ions collide with the free (delocalised) electrons that move throughout the metal. The energised electrons travel rapidly through the metal structure, carrying energy from the warmer outer surface to the cooler inner surface in contact with the lemonade. The electrons then collide with ions on the inner wall, transferring their kinetic energy and so warming those regions — this is why metals are efficient thermal conductors compared to non-metals, which lack free electrons.
Question 4
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $2.2.2$ — Specific heat capacity / thermal properties and temperature (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$, $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
The thermocouple consists of two wires made of at least two different metals joined at two junctions — one junction (the “hot junction”) is placed inside the molten sulfur, while the other junction (the “cold junction”) is kept at a known reference temperature such as an ice-water mixture at 0 °C or simply at room temperature. A voltmeter is correctly connected across the open ends of the two wires to measure the e.m.f. produced. Both wires must be distinct metals (e.g. copper and constantan) and must form a complete loop with the voltmeter.
(a)(ii)
For the correct answer:
The e.m.f. (voltage) reading shown on the voltmeter is recorded at each point during cooling. This e.m.f. value is then converted to a temperature using a pre-prepared calibration graph or table that relates known e.m.f. values to known temperatures. As the sulfur cools, the e.m.f. decreases, and the corresponding temperature can be read off from the calibration curve at each stage.
(b)
For the correct answer (any one of):
A thermocouple can measure very high temperatures (above 400 °C) that would cause a liquid-in-glass thermometer to break or vaporise, since the metal wires do not melt at such temperatures. Alternatively: a thermocouple has a very rapid response time, a small heat capacity that minimally disturbs the system being measured, and can send its electrical output directly to a computer or data logger for continuous recording.
Question 5
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 2.2.3 — Melting, boiling and evaporation (Part (a))
• Topic 4.2.5 — Electrical energy and electrical power (Parts (b)(i), (b)(ii))
• Topic 4.4 — Electrical safety (Parts (c)(1), (c)(2))
▶️ Answer/Explanation
(a)
The boiling point is the specific temperature at which a liquid changes state into a gas (or vapour), with both the liquid and gaseous phases existing in equilibrium. At 100 °C, water molecules have sufficient kinetic energy to overcome intermolecular forces throughout the bulk of the liquid, not just at its surface, causing bubbles of vapour to form and rise through the liquid.
(b)(i)
Using the equation E = VIt: E = 230 × 13 × 60 = 1.8 × 105 J. This energy is entirely converted to thermal (heat) energy by the resistive immersion heater element. The product of voltage and current gives the power (3000 W or 3 kW), and multiplying by time gives the total energy.
(b)(ii)
The temperature rise required is ΔT = 100 − 22 = 78 °C. Using E = mcΔT, the mass of water heated per 60 s is m = 1.8 × 105 / (4200 × 78) ≈ 0.55 kg. Dividing by 60 s gives a rate of approximately 9.1 × 10−3 kg/s. This can also be found directly as rate = P / (cΔT) = 3000 / (4200 × 78).
(c)(i)
If the live wire inside the tap comes loose and touches the metal casing, the metal tap becomes live. Because the tap is earthed, a large current flows from the live tap through the earth wire to earth, creating a short circuit. This very large current causes the fuse to blow, cutting off the supply and preventing electrocution of the user.
(c)(ii)
The fuse contains a thin wire with a specific current rating (e.g., 13 A). If an excessively large current flows through the circuit — for example due to a fault — the fuse wire heats up rapidly and melts, permanently breaking the circuit. This prevents further current flow and protects the wiring and connected components from overheating or catching fire.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 2.3.3 — Radiation (Parts (a), (b))
• Topic 2.1.2 — Particle model (Part (c)(i))
• Topic 2.3.2 — Convection (Part (c)(ii))
▶️ Answer/Explanation
(a)
Energy from the Sun travels to Earth in the form of electromagnetic radiation — primarily infrared radiation, visible light, and ultraviolet — which requires no medium and can therefore pass through the vacuum of space. When this radiation strikes the surface of the road, it is absorbed, converting the electromagnetic energy into internal (thermal) energy and raising the temperature of the road.
(b)
Black surfaces are very good absorbers of infrared and visible radiation, so the black road absorbs a large proportion of the electromagnetic radiation incident on it. The sea surface, by comparison, is a relatively poor absorber and also reflects a significant portion of the incoming radiation. As a result, the road gains thermal energy more rapidly than the sea and reaches a higher temperature.
(c)(i)
When the air above the road is heated, the air molecules gain kinetic energy and move faster on average. As their speed increases, the molecules collide more forcefully and move further apart from one another, causing the air to expand and its density to decrease.
(c)(ii)
The warm, less dense air above the road rises, because the denser, cooler air exerts a greater buoyancy force upwards on the less dense warm air column. As the warm air rises above the land, cooler and denser air from above the sea moves horizontally inland to replace it. This horizontal movement of cooler air from sea to land is the cool sea breeze that the cyclist experiences — a convection current on a large scale.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.2.3 — Thin lenses (Parts (a), (b), (c)(i), (c)(ii), (c)(iii))
▶️ Answer/Explanation
(a)
Parallel rays travelling along the principal axis XY are refracted by the converging lens so that they all meet at a single point on the axis called the principal focus (focal point), located 3.0 cm to the right of the lens. Beyond this point, the rays diverge outward from the focal point in straight lines. This convergence to a single point is the defining characteristic of a converging lens acting on parallel light.
(b)
The two principal focuses F are located on the principal axis XY, one on each side of the lens, each exactly 3.0 cm from the optical centre of the lens. The left-hand F is 3.0 cm to the left of the lens, and the right-hand F is 3.0 cm to the right. Both points must be labelled F on the diagram.
(c)(i)
Since the object distance (2.0 cm) is less than the focal length (3.0 cm), the nail N is placed inside the focal length of the lens. To locate the image, two standard rays are drawn from the tip of N: a ray parallel to XY which, after refraction, passes through the right-hand principal focus; and a ray directed straight through the optical centre of the lens without bending. These two refracted rays diverge on the right-hand side, so they are traced back (as dashed lines) to where they appear to meet on the left-hand side of the lens, giving the position of the virtual image I.
(c)(ii)
The image I is virtual. The refracted rays on the right-hand side of the lens are diverging and do not actually meet; they only appear to come from the point I when extended backwards. Because no real light rays pass through I, the image cannot be projected onto a screen — it exists only as an apparent location from which the diverging rays seem to originate.
(c)(iii)
When a converging lens is used with an object placed closer to it than its focal length, producing an upright, magnified, virtual image on the same side as the object, the lens is being used as a magnifying glass.
Question 8
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.7.1 — Energy (Part (a))
• Topic 1.2 — Motion (Part (a))
• Topic 4.5.1 — Electromagnetic induction (Parts (b)(i), (b)(ii))
▶️ Answer/Explanation
(a)
Since the magnet falls freely with no resistive force, it accelerates uniformly under gravity (g = 10 m/s²). The velocity on reaching the bottom is v = at = 10 × 0.67 = 6.7 m/s. Applying the kinetic energy formula: KE = ½mv² = ½ × 0.012 × (6.7)² = ½ × 0.012 × 44.89 ≈ 0.27 J. This result can also be confirmed using the gravitational potential energy lost, since no energy is lost to resistive forces in the plastic tube.
(b)(i)
As the magnet falls through the copper tube, its magnetic field lines cut through the conducting copper walls, causing a change in magnetic flux through the copper. By the principle of electromagnetic induction, this changing magnetic flux induces an electromotive force (e.m.f.) in the copper tube. Since copper is a good electrical conductor, this e.m.f. drives a current — known as an eddy current — around the copper tube.
(b)(ii)
The induced current in the copper creates its own magnetic field which, by Lenz’s law, acts in a direction that opposes the change producing it — that is, it opposes the downward motion of the magnet. This means the copper tube’s magnetic field exerts an upward force on the falling magnet, opposing gravity. The net downward force on the magnet is therefore reduced, causing it to fall significantly more slowly through the copper tube than through the plastic tube where no such opposing force existed.
Question 9
▶️ Answer/Explanation
(a)
A digital signal can only take one of two discrete values — represented as 1 (high) or 0 (low) — and switches abruptly between them, making it highly resistant to noise and degradation over long distances. An analogue signal, by contrast, is continuously variable, taking any value within a range and varying smoothly with time, just as a sound wave or temperature reading would; this makes it more susceptible to distortion and interference during transmission.
(b)
NOR (gate) 
The truth table shows that the output is 1 only when both inputs A and B are 0; in all other input combinations the output is 0. This is the behaviour of a NOR gate — it produces a high output only when neither input is high. The NOR gate symbol is drawn as a standard OR gate shape with a small circle (bubble) at the output to indicate inversion.
(c)
(i.e. NOR gate symbol with two inputs joined seen)
To construct an OR gate from two NOR gates, the inputs A and B are first fed into gate X (a NOR gate), giving the NOR output. This NOR output is then connected to both inputs of gate Y (a second NOR gate used as a NOT gate, with both inputs tied together), so that gate Y inverts the NOR result. Since NOT(NOR(A,B)) = OR(A,B), the final output Q of gate Y is logically equivalent to an OR gate. This two-gate combination correctly reproduces the OR truth table for all four input combinations.
Question 10
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 5.1.2 — The nucleus: isotopes, proton number, nucleon number (Parts (a)(i), (a)(ii))
• Topic 5.1.2 (Supplement) — Nuclear fusion, nuclide equations and mass/energy changes (Parts (b)(i), (b)(ii))
• Topic 1.7.3 (Supplement) — Energy released by nuclear fusion in the Sun (Part (b)(i))
▶️ Answer/Explanation
(a)(i)
Both hydrogen-2 and hydrogen-3 are isotopes of the same element, hydrogen, which means their nuclei contain the same number of protons. Specifically, both nuclei contain exactly one proton, giving them both a proton number (atomic number) of Z = 1. It is this identical proton number that defines them as isotopes of the same element rather than different elements.
(a)(ii)
Hydrogen-2 has a nucleon number of 2 and a proton number of 1, so its nucleus contains 1 proton and 1 neutron. Hydrogen-3 has a nucleon number of 3 and a proton number of 1, so its nucleus contains 1 proton and 2 neutrons. Therefore, the nucleus of hydrogen-3 differs from that of hydrogen-2 by having one additional neutron, giving it a greater nuclear mass.
(b)(i)
Both nuclei carry a positive charge (due to their protons) and therefore repel each other strongly via the electrostatic (Coulomb) force as they are brought close together. An extremely high temperature is required because it gives the nuclei a very high kinetic energy, enabling them to move fast enough to overcome this electrostatic repulsion and approach close enough for the short-range strong nuclear force to take over and bind them together into a new, heavier nucleus.
(b)(ii)
Conserving nucleon number: 2 + 3 = 5, and one neutron (\(\,_{0}^{1}\textrm{n}\)) accounts for 1, so element X must have nucleon number 4. Conserving proton number: 1 + 1 = 2, so element X has proton number 2. The complete equation is therefore:
\(\,_{1}^{2}\textrm{H}\;+\;\,_{1}^{3}\textrm{H}\;\longrightarrow\;\,_{2}^{4}\textrm{He}\;+\;\,_{0}^{1}\textrm{n}\)
Element X is helium-4 (\(\,_{2}^{4}\textrm{He}\)), also known as an alpha particle.