Question 1
(i) the speed of the cyclist at $\text{time} = 6.0\text{ s}$
(ii) the maximum speed of the cyclist.
(ii) Describe how the motion of the cyclist between points B and C differs from the motion between points A and B. Give a reason for your answer.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.2$ — Motion (Parts $\mathrm{(a)}$, $\mathrm{(b)}$, $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
$3.0\text{ m/s}$
To determine the speed at a specific time, locate $6.0\text{ s}$ on the horizontal time axis and move vertically to the line of the graph. From that point, move horizontally to the vertical speed axis to read the value, which corresponds exactly to $3.0\text{ m/s}$.
(a)(ii)
For the correct answer:
$16\text{ m/s}$
The maximum speed is represented by the highest point reached on the speed-time graph. Looking at the peak of the curve, the highest y-value on the vertical axis is $16\text{ m/s}$, occurring just after point C.
(b)(i)
For the correct answer:
constant acceleration OR uniform acceleration
Between points A and B, the graph is a straight line with a positive gradient, indicating that the speed is increasing at a steady rate. Since acceleration is defined as the rate of change of speed, a straight upward slope on a speed-time graph signifies constant acceleration.
(b)(ii)
For the correct answer:
Greater acceleration between B and C because the gradient is steeper.
The acceleration between points B and C is greater than the acceleration between A and B. In a speed-time graph, the gradient (slope) represents the magnitude of acceleration; since the line segment BC is visibly steeper than the segment AB, the cyclist is increasing their speed more rapidly in that interval.
(c)
For the correct answer:
$25\text{ m}$
The distance travelled is equal to the area under the speed-time graph. Between points A ($0\text{ s}$, $0\text{ m/s}$) and B ($5\text{ s}$, $10\text{ m/s}$), the area forms a triangle. Using the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$, the calculation is $\frac{1}{2} \times 5\text{ s} \times 10\text{ m/s} = 25\text{ m}$.
Question 2
(a) State the principle of conservation of energy.
(b) Fig. 2.1 shows the energy flow diagram for a car powered by a petrol engine.
(i) Using the information in Fig. 2.1, calculate the percentage of energy transferred from the chemical store to the kinetic store.
(ii) Fig. 2.2 shows the energy flow diagram for an electric car. The electric car is driven by an electric motor which is powered by a battery.
Using the information in Fig. 2.1 and Fig. 2.2, state which car is more efficient. Give a reason for your answer.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.7.1 — Energy (Parts (a), (b)(i), (b)(ii))
▶️ Answer/Explanation
(a)
For the correct answer:
Energy cannot be created or destroyed, but can only be transferred from one store to another.
The law of conservation of energy dictates that the total energy in a closed system remains constant. While energy can change from one form (such as chemical) to another (such as kinetic or thermal), the initial amount of energy must equal the final amount of energy, including any wasted energy dissipated into the surroundings.
(b)(i)
For the correct answer:
$30\%$
Efficiency is calculated by comparing the useful energy output to the total energy input. In Fig. 2.1, the total input from the chemical store is represented as $100\%$. The diagram explicitly shows that $30\%$ of this total is successfully transferred to the kinetic store, which is the intended useful output for a moving vehicle.
(b)(ii)
For the correct answer:
The electric car is more efficient because a higher percentage ($70\%$) of energy is transferred to the kinetic store compared to the petrol car ($30\%$).
Efficiency is a measure of how much input energy is converted into useful work versus wasted as internal (thermal) or sound energy. By comparing the two diagrams, the electric car provides $70\text{ J}$ of useful kinetic energy for every $100\text{ J}$ input, whereas the petrol car provides only $30\text{ J}$, meaning the electric motor wastes significantly less energy.
Question 3
A platform rests on a pivot as shown in Fig. 3.1.
A diver sits at a distance of 1.8 m from the pivot. The weight of the diver is 1100 N.
(a) Using the information in Fig. 3.1, calculate the moment of the diver about the pivot.
(b)(i) Fig. 3.2 represents the platform without the diver.
The moment of the weight ( W ) of the platform is balanced by the moment of the spring. The spring exerts a downward force of 62 N.
Using the information in Fig. 3.2, calculate the weight ( W ) of the platform.
(ii) The graph of load against extension for a spring is shown in Fig. 3.3.
The unstretched length of the spring is 16 cm.
Determine the length of the spring when the load on the spring is 240 N.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.5.1 — Effects of forces (Part (b)(ii))
• Topic 1.5.2 — Turning effect of forces (Parts (a), (b)(i))
▶️ Answer/Explanation
(a)
For the correct answer:
1980 N m
The moment of a force is a measure of its turning effect and is calculated by multiplying the force by the perpendicular distance from the pivot. Using the formula $\text{moment} = F \times d$, we substitute the given weight of the diver ($1100\text{ N}$) and the distance ($1.8\text{ m}$) to get $1100 \times 1.8 = 1980\text{ N m}$. This represents the clockwise turning effect created by the diver’s weight relative to the pivot point.
(b)(i)
For the correct answer:
186 N
According to the principle of moments, for an object in equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments. Here, the spring creates a clockwise moment ($62\text{ N} \times 1.2\text{ m} = 74.4\text{ N m}$), which is balanced by the anticlockwise moment of the platform’s weight ($W \times 0.4\text{ m}$). Solving the equilibrium equation $74.4 = 0.4W$ gives a weight $W$ of $186\text{ N}$.
(b)(ii)
For the correct answer:
17 cm
To find the total length, first determine the extension from the load-extension graph in Fig. 3.3. At a load of $240\text{ N}$, the graph indicates an extension of $1.0\text{ cm}$. Since the unstretched length is $16\text{ cm}$, the new length is calculated by adding the extension to the original length ($16\text{ cm} + 1.0\text{ cm} = 17\text{ cm}$). This demonstrates how an elastic solid deforms under an applied load.
Question 4
(ii) Determine the gravitational potential energy gained by the books. Give a reason for your answer.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.3 — Mass and weight (Part (a))
• Topic 1.7.1 — Energy (Parts (b)(i), (b)(ii))
▶️ Answer/Explanation
(a)
For the correct answer:
31 N
Weight is the gravitational force acting on an object and is calculated using the formula W=m×g. Given the mass m=3.2 kg and the standard gravitational field strength g=9.8 N/kg, the weight is 3.2×9.8=31.36 N. Following significant figure rules, this value is rounded to 31 N.
(b)(i)
For the correct answer:
141 J (which is approximately 140 J)
Mechanical work done is equal to the force applied multiplied by the distance moved in the direction of the force (W=F×d). To lift the books, the student must apply a force equal to their weight (31.36 N) over a vertical height of 4.5 m. Thus, Work=31.36 N×4.5 m=141.12 J, which rounds to approximately 140 J.
(b)(ii)
For the correct answer:
141 J. Reason: The work done against gravity is stored as potential energy.
The gravitational potential energy gained is 141 J because, according to the principle of conservation of energy, the mechanical work done to lift an object is directly transferred into its gravitational potential energy store (ΔE p =W). Since 141.12 J of work was performed to move the books to the top of the stairs, that exact amount of energy is now stored as GPE.
Question 5
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.7.3$ — Energy resources (Parts $\mathrm{(a)}$, $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
1. No sulfur dioxide (emission) OR acid rain (produced).
2. No CO2 / greenhouse gases (emitted) OR no / reduces (impact on) global warming.
Tidal energy is a renewable resource, meaning it does not deplete natural stores like coal. Unlike coal-fired stations, tidal turbines do not involve combustion, so they release no carbon dioxide, a primary greenhouse gas, or sulfur dioxide, which causes acid rain. This results in a significantly lower carbon footprint and reduces the environmental impact associated with air pollution and climate change.
(b)
For the correct answer:
1. Suitable locations limited OR locations remote.
2. Marine ecosystems disrupted.
Tidal power generation is restricted to specific coastal areas with high tidal ranges, which are often geographically remote from the power grid. The installation of underwater turbines can also lead to the destruction of local marine habitats and pose risks to aquatic wildlife through physical strikes or noise pollution. Furthermore, unlike coal, tidal energy is only available during specific periods of the tidal cycle, making it less flexible for immediate demand changes.
Question 6
Explain how the vacuum prevents these two processes of thermal energy transfer.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 2.3.2 — Convection (Parts (a), (b))
▶️ Answer/Explanation
(a)
For the correct answer:
Conduction and convection.
Thermal conduction involves the transfer of energy through atomic or molecular lattice vibrations and the movement of free electrons, while convection relies on density changes within a fluid. Both processes require a medium or particles to facilitate the transfer of energy. Because a vacuum is an empty space devoid of particles, energy cannot be transferred via collisions or the bulk movement of matter, effectively blocking both conduction and convection.
(b)
For the correct answer:
Heated water expands and becomes less dense, causing it to rise.
When the water is heated, the particles gain kinetic energy and move further apart, which causes the volume to increase and the density to decrease. This less dense, warmer water rises above the cooler, denser surrounding water, creating a continuous flow known as a convection current. As the water moves through the tube, it eventually cools, becomes denser again, and sinks, maintaining the cycle illustrated by the movement of the coloured crystal.
Question 7
When a train whistle blows, steam comes out of the whistle.
The student measures the time interval between seeing the steam coming out of the whistle and hearing the whistle.
(i) Suggest a suitable device for measuring this time interval.
(ii) The time interval is 1.6 s between the steam coming out of the whistle and the student hearing the whistle. The speed of sound in air is 340 m/s. Calculate the distance d from the whistle to the student.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.4 — Sound (Parts (a), (b)(i), (b)(ii), (c))
▶️ Answer/Explanation
(a)
For the correct answer:
Particles oscillate / vibrate back and forth parallel to the direction of energy transfer.
Sound waves are longitudinal waves. This means that the air particles do not travel with the wave but instead vibrate about a fixed position. The direction of this vibration is the same as (parallel to) the direction in which the sound wave is traveling, creating regions of high pressure (compressions) and low pressure (rarefactions).
(b)(i)
For the correct answer:
Stopwatch OR digital timer
To measure the time interval between a visual cue (steam) and an auditory cue (sound), a manual or digital stopwatch is required. Since the human reaction time is a factor, a device that can record time to at least one or two decimal places, like a standard laboratory stopwatch, is the most appropriate tool for this experiment.
(b)(ii)
For the correct answer:
544 m
The distance is calculated using the wave speed formula rearranged for distance: distance=speed×time. By substituting the given values, d=340 m/s×1.6 s, which equals 544 meters. This calculation assumes the speed of light is instantaneous compared to the speed of sound.
(c)
For the correct answer:
20 Hz to 20 kHz OR 20 Hz to 20000 Hz
The standard range of hearing for a healthy human ear covers frequencies from a low pitch of 20 Hz up to a high pitch of 20000 Hz. Frequencies below this range are known as infrasound, while frequencies above 20 kHz are defined as ultrasound, which are inaudible to humans.
Question 8
(ii) State the name of the point labelled F in Fig. 8.2.
(iii) On Fig. 8.2, draw an arrow to represent the image of O.
(iv) Using a ruler, measure the focal length of the converging lens.
(v) Describe characteristics of the image in Fig. 8.2. Choose words from the list. Tick (✓) three boxes.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.1 — General properties of waves (Part (a))
• Topic 3.2.3 — Lenses (Part (b))
▶️ Answer/Explanation
(a)
For the correct answer:
Top diagram — Diffraction; Bottom diagram — Refraction
Diffraction occurs when waves spread out after passing through a narrow gap, as seen in the top diagram. Refraction is the change in direction of a wave as it crosses a boundary between two different media, such as air and glass, which is illustrated by the bending of the wavefronts in the bottom diagram.
(b)(i) & (ii)
For the correct answer:
(i) Principal axis; (ii) Principal focus
The line XY passing through the optical centre of the lens and perpendicular to its faces is the principal axis. The point F, where rays parallel to the principal axis converge after passing through the lens, is known as the principal focus (or focal point).
(b)(iii) & (iv)
For the correct answer:
(iii) Vertical arrow at the intersection; (iv) 1.9 cm
The image is formed at the point where the refracted rays intersect; an arrow drawn from the principal axis to this intersection represents the image. The focal length is the distance between the center of the lens and the principal focus F, which measures 1.9 cm in this diagram.
(b)(v)
For the correct answer:
Enlarged ✓, Inverted ✓, Real ✓
The image is “Real” because it is formed by the actual intersection of light rays and can be projected onto a screen. It is “Inverted” as it appears upside down relative to the object and “Enlarged” because the height of the image arrow is greater than the height of the original object O.
Question 9
| Date | Meter Reading (kWh) |
|---|---|
| 1st October | 3771 |
| 31st October | 3797 |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.2.5 — Electrical energy and electrical power (Parts (a)(ii), (b))
• Topic 4.4 — Electrical safety (Part (a)(i))
▶️ Answer/Explanation
(a)(i)
To prevent the risk of electric shock or electrocution from exposed live internal components. The plastic cover serves as an electrical insulator; removing it while the appliance is connected to the 230 V mains supply poses a significant hazard to the user if they come into contact with the heating element or wiring.
(a)(ii)
3.1 A (or 3.13 A)
Using the power formula P=IV, we rearrange to find current: I=P/V. From the safety label in Fig. 9.2, the power (P) is 720 W and the voltage (V) is 230 V. The calculation is I=720 W/230 V≈3.13 A, which rounds to 3.1 A.
(b)
468 cents (or $4.68)
First, determine the total energy consumed by subtracting the initial reading from the final reading: 3797 kWh−3771 kWh=26 kWh. Next, multiply the energy used by the unit cost: 26 kWh×18 cents/kWh=468 cents. This represents the total monthly operating cost for the heater.
Question 10
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.1 — Simple phenomena of magnetism (Parts (a)(i), (a)(ii), (a)(iii))
• Topic 4.5 — Electromagnetic effects (Part (b))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
Soft iron
Soft iron is magnetically “soft,” meaning it is easily magnetized when placed in a magnetic field but loses its magnetism rapidly once the field is removed. This characteristic makes it ideal for temporary magnets, such as core materials in electromagnets, where magnetism needs to be switched on and off effectively.
(a)(ii)
For the correct answer:
Steel
Steel is a magnetically “hard” material because it requires a strong external field to become magnetized but retains that magnetism for a very long duration afterward. Due to its high retentivity and resistance to demagnetization, it is the standard choice for manufacturing permanent magnets used in compasses and motors.
(a)(iii)
For the correct answer:
The magnet does not attract the material.
A simple test for magnetism involves bringing a known magnet close to the substance; non-magnetic materials like plastic, wood, or copper will show no interaction or force of attraction. While magnetic materials are drawn toward either pole of a magnet, non-magnetic materials lack the internal structures to be influenced by external magnetic fields.
(b)
For the correct answer:
The circuit is completed and broken repeatedly.
When the switch closes, current flows through the coil, turning the iron nail into an electromagnet that attracts the springy iron strip, causing the hammer to strike the gong. This movement simultaneously pulls the strip away from the contact screw, breaking the electrical circuit and demagnetizing the nail. The spring then pulls the strip back to close the circuit again, creating a continuous cycle of motion.
Question 11
(_{……}^{……}\textrm{Be})
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $5.1.1$ — Nuclear Atom (Parts $\mathrm{(a)}$, $\mathrm{(b)}$)
• Topic $5.2$ — Radioactivity (Part $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
({4}^{9}\textrm{Be})
Nuclide notation consists of the mass number (top) and atomic number (bottom). In Fig. 11.1, the nucleus contains 4 protons (atomic number) and 5 neutrons; adding these gives a mass number of 9. Therefore, the notation is written as ({4}^{9}\textrm{Be}).
(a)(ii)
For the correct answer:
neutrons
Atoms consist of three subatomic particles: protons, neutrons, and electrons. The key already identifies electrons (orbiting the nucleus) and protons (positively charged particles in the nucleus). The remaining particles in the nucleus, which are uncharged, are the neutrons.
(b)(i) & (ii)
For the correct answer:
(i) A, B, and D; (ii) A
Isotopes of beryllium must have exactly 4 protons. Diagrams A, B, and D all contain 4 protons, making them beryllium, while C has 5. A positive ion occurs when there are fewer electrons than protons; Diagram A has 4 protons but only 3 electrons, resulting in a net positive charge.
(c)
For the correct answer:
2 hours
To find the half-life, determine how many times the count rate halved: $400 \rightarrow 200 \rightarrow 100 \rightarrow 50$. This shows that 3 half-lives occurred over a duration of 6 hours. Dividing the total time by the number of half-lives ($6 \div 3$) gives a half-life of 2 hours.
Question 12
Jupiter is …… and ……
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 6.1.1 & 6.1.2 — The Earth and the Solar System (Parts (a), (b), (c))
▶️ Answer/Explanation
(a)
For the correct answer:
1. Mercury, 2. Venus, 4. Mars, 6. Saturn, 7. Uranus, 8. Neptune
The eight planets in our Solar System follow a specific sequence based on their orbital distance from the Sun. Starting from the closest, the order is Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and finally Neptune. Memorizing this sequence is essential for understanding the basic structure of the planetary system.
(b)
For the correct answer:
Asteroids and Comets (OR Dwarf planets/Minor planets)
Beyond the eight major planets, various other natural bodies orbit the Sun. Asteroids are primarily rocky bodies found in the asteroid belt between Mars and Jupiter, while comets are icy bodies that often have highly elliptical orbits. Other examples include dwarf planets like Pluto and various minor planets or natural satellites.
(c)
For the correct answer:
Mercury is rocky and small.
Jupiter is gaseous and large.
Planets are generally categorized into two groups: the four inner planets (Mercury, Venus, Earth, Mars) which are small and rocky (terrestrial), and the four outer planets (Jupiter, Saturn, Uranus, Neptune) which are large and gaseous (gas giants). This distinction arises from the different conditions and materials available during the formation of the Solar System.