Question 1
A cyclist is travelling along a straight road. Fig. 1.1 shows the speed–time graph for the cyclist. The graph is divided into four sections labelled P, Q, R and S.
(a) Calculate the distance travelled by the cyclist in section P from time = 0 to time = 100 s.
(b) Describe the motion of the cyclist in each of sections Q, R and S shown in Fig. 1.1.
(c) The cyclist is moving north along the road. Determine the velocity of the cyclist at time = 300 s. Include the unit.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.2 — Motion (Parts (a), (b), (c))
▶️ Answer/Explanation
(a) Distance travelled in section P:
The area under the speed-time graph gives the distance travelled. For section P, the graph is a triangle.
Area = 0.5 × base × height = 0.5 × 100 s × 8 m/s = 400 m.
Answer: 400 m
(b) Motion in sections Q, R, and S:
Q: The cyclist is accelerating.
R: The cyclist is moving at a constant speed.
S: The cyclist is decelerating.
(c) Velocity at time = 300 s:
At time = 300 s, the speed is 12 m/s, and the direction is north.
Answer: 12 m/s north
Detailed Solution:
The area under a speed–time graph represents distance; section P forms a triangle with base 100 s and height 8 m/s, giving $0.5 \times 100 \times 8 = 400$ m. Section Q shows increasing speed (acceleration), R shows constant speed, and S shows decreasing speed (deceleration). At 300 s the speed is 12 m/s, and since velocity is a vector, the direction north must be included.
Question 2
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.8 — Pressure (Part (a))
• Topic 1.5.2 — Turning effect of forces (Parts (b), (c))
• Topic 1.5.3 — Centre of gravity (Part (d))
▶️ Answer/Explanation
(a) Pressure on the ground:
Pressure = Force / Area = 1540 N / (160 cm × 12 cm) = 1540 N / 1920 cm² = 0.8 N/cm².
Answer: 0.8 N/cm²
Detailed solution: Pressure is defined as the force acting per unit area ($p = \frac{F}{A}$). The contact area of the beam is its length multiplied by its width ($160 \text{ cm} \times 12 \text{ cm} = 1920 \text{ cm}^2$). Dividing the total weight (1540 N) by this area yields the pressure exerted on the ground.
(b) Moment of the 1030 N force:
Moment = Force × Distance = 1030 N × 120 cm = 123600 Ncm.
Answer: 123600 Ncm
Detailed solution: The moment of a force is the measure of its turning effect, calculated as the product of the force and the perpendicular distance from the pivot (moment = $F \times d$). Here, the 1030 N force acts at a perpendicular distance of 120 cm, resulting in a moment of $1030 \times 120 = 123600 \text{ Ncm}$.
(c) Using a smaller force:
The builder can increase the perpendicular distance from the pivot to reduce the force needed.
Detailed solution: For a required moment, the force is inversely proportional to the perpendicular distance ($F = \frac{\text{moment}}{d}$). By applying the effort further away from the pivot (increasing $d$), the magnitude of the force required to produce the same turning effect is reduced.
(d) Stability of the beam:
The beam in Fig. 2.3 is less stable because its center of gravity is higher, making it more prone to tipping over.
Detailed solution: The stability of an object depends on the position of its centre of gravity. In Fig. 2.1, the beam is flat with a low centre of gravity. In Fig. 2.3, the beam is upright, raising its centre of gravity. A higher centre of gravity means the line of action of the weight falls outside the base more easily, reducing stability.
Question 3
Electricity is distributed from wind turbines to homes and industry.
(a) Statements A–F describe the main stages in the transfer of energy from the Sun to electrical energy in a wind turbine generator. The statements A–F are not in the correct order.
A Air moves from regions of high pressure to regions of low pressure.
B The turbine blades turn a generator.
C Energy from the Sun heats the atmosphere unevenly.
D Uneven heating of the atmosphere produces regions of different atmospheric pressure.
E The generator produces electrical energy.
F Moving air turns the turbine blades.
Complete the flow chart to describe how a wind turbine uses energy from the Sun to generate electrical energy.
(b) State two disadvantages, apart from cost, of using wind turbines to produce electrical energy for homes and industry.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.7.3 — Energy resources (Parts (a), (b))
▶️ Answer/Explanation
(a) Flow chart:
C → D → A → F → B → E
Explanation: Solar radiation unevenly heats the atmosphere, creating pressure differences that cause air movement (wind). This moving air possesses kinetic energy which is transferred to turbine blades, causing rotation that drives a generator to produce electrical energy via electromagnetic induction.
(b) Disadvantages of wind turbines:
1. Wind turbines require a large area of land.
2. The energy output is intermittent and depends on wind conditions.
Question 4
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.7.2 — Work (Parts (a)(i), (a)(ii), (a)(iii))
• Topic 1.7.4 — Power (Part (b))
▶️ Answer/Explanation
(a) (i)
For the correct answer:
50 J
The work done by a force is calculated using the equation $W = Fd$. Substituting the given values, $W = 25\text{ N} \times 2.0\text{ m} = 50\text{ J}$. This represents the mechanical work done by the motor on the load to lift it against gravity.
(a) (ii)
For the correct answer:
50 J
The useful energy output of the motor is the energy transferred to the load, which is equal to the mechanical work done on it. Since the motor does 50 J of work to lift the load, the useful energy output is also 50 J.
(a) (iii)
For the correct answer:
energy is lost / wasted as heat (or thermal energy) / sound
Electric motors are not perfectly efficient. Some of the electrical energy input is converted to unwanted forms of energy, primarily heat, due to friction in the moving parts and electrical resistance in the motor’s coils. Therefore, the useful energy output is always less than the total energy input.
(b)
For the correct answer:
16 W
Power is the rate of doing work, defined by the equation $P = \frac{W}{t}$. Using the values provided, $P = \frac{80\text{ J}}{5.0\text{ s}} = 16\text{ W}$. This means the motor transfers 16 joules of energy per second to the load.
Question 5
Most-appropriate topic code (Cambridge IGCSE Physics 0625):
• Topic 2.1.2 — Particle model (Parts (a), (b))
▶️ Answer/Explanation
(a) Pressure exerted by gas particles:
Gas particles move randomly and collide with the walls of the box. Each collision exerts a force on the wall. The pressure is the total force exerted by all collisions per unit area.
Detailed Solution (a): The particles are in constant, random motion. When they strike the walls, they rebound, resulting in a change in momentum that exerts a tiny force. The cumulative effect of countless such collisions per second on a unit area of the wall manifests as the measurable pressure of the gas.
(b) Effect of increasing the number of gas particles:
If the number of gas particles increases while the temperature remains constant, the pressure will increase. This is because there are more particles colliding with the walls, resulting in a greater force per unit area.
Detailed Solution (b): Since temperature is constant, the average speed of particles remains unchanged. However, a larger number of particles in the same volume means a higher frequency of collisions with the walls per second. This increased rate of collisions results in a larger average force, thereby increasing the overall pressure exerted by the gas.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.1 — General properties of waves (Parts (a), (b), (c))
▶️ Answer/Explanation
(a) More accurate method for determining wavelength:
The student can measure the distance between multiple peaks (e.g., 5 or 10 wavelengths) and then divide the total distance by the number of wavelengths to get a more accurate average wavelength.
(b) Wave speed:
Wave speed = Frequency × Wavelength = 6.0 Hz × 4.0 cm = 24 cm/s.
Answer: 24 cm/s
(c) Waves moving from deep to shallow water:
The waves undergo refraction. As the waves enter shallow water, their speed decreases, causing the wavelength to decrease. The frequency remains constant.
Question 7
Fig. 7.1 represents two rays of light striking a thin converging lens. The rays are both parallel to the principal axis.
\( F_2 \) and \( F_1 \) are the focal points of the lens.
(a) On Fig. 7.1, continue the path of each ray beyond the lens as far as the screen.
(b) Visible light is a region of the electromagnetic spectrum. State one region of the electromagnetic spectrum which has waves of longer wavelength than waves of visible light.
(c) Gamma rays are another region of the electromagnetic spectrum.
(i) Describe one use of gamma rays.
(ii) Describe one harmful effect on people of excessive exposure to gamma rays.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.2.3 — Thin lenses (Part (a))
• Topic 3.3 — Electromagnetic spectrum (Parts (b), (c)(i), (c)(ii))
▶️ Answer/Explanation
(a)
For the correct answer:
The rays will converge at the focal point \( F_1 \) after passing through the lens.
Rays of light that are parallel to the principal axis of a converging lens are refracted so that they pass through the principal focus (focal point) on the opposite side of the lens.
(b)
For the correct answer:
Infrared or radio waves
The electromagnetic spectrum is arranged in order of increasing frequency and decreasing wavelength. Infrared waves and radio waves both have longer wavelengths (and lower frequencies) than visible light.
(c)(i)
For the correct answer:
Gamma rays are used in medical treatments, such as cancer therapy, to kill cancer cells.
Due to their high energy and ionising ability, gamma rays can be precisely targeted to destroy malignant cells in radiotherapy. They are also used to sterilise medical equipment by killing bacteria.
(c)(ii)
For the correct answer:
Excessive exposure to gamma rays can cause cell damage or mutations, leading to cancer or radiation sickness.
As highly ionising radiation, gamma rays can penetrate living tissue and alter or destroy cellular DNA. This can result in uncontrolled cell division (cancer), tissue damage, and acute radiation sickness.
Question 8
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.1 — Simple phenomena of magnetism (Parts (a), (b))
▶️ Answer/Explanation
(a) Labeling the poles:
The poles on each magnet should be labeled as North (N) and South (S). Since the magnets are repelling each other, like poles (N-N or S-S) must be facing each other.
Detailed Solution: The field lines curve away from the region between the magnets, indicating a repulsive force. This occurs when two like poles (both North or both South) are placed facing each other, as like poles repel.
(b) Plotting the magnetic field pattern:
To plot the magnetic field pattern, use a plotting compass or iron filings. Place the compass near the magnet and mark the direction of the needle. Repeat this process at various points around the magnet to map the field lines. Alternatively, sprinkle iron filings around the magnets and tap the surface to allow the filings to align with the magnetic field lines.
Detailed Solution: Place the magnets on a flat surface and cover with paper. Sprinkle iron filings evenly and tap gently; the filings align along the field lines. For direction, place a small compass at points along a line and mark arrowheads pointing from North to South.
Question 9
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.3.1 — Circuit diagrams and circuit components (Part (a), (c))
• Topic 4.2.5 — Electrical energy and electrical power (Part (b))
▶️ Answer/Explanation
(a) Other components in the circuit:
Two other components could be a switch and an ammeter.
Detailed Solution: The circuit contains a cell and a lamp as labelled. A switch is included to open or close the circuit, controlling the flow of current. An ammeter is connected in series to measure the electric current flowing through the circuit components.
(b) Power dissipated in the lamp:
Power = Current × Voltage = 0.40 A × 6.0 V = 2.4 W.
Answer: 2.4 W
Detailed Solution: Electrical power $P$ is the rate of energy transfer. Using $P = IV$, substitute the given current $I = 0.40\text{ A}$ and potential difference $V = 6.0\text{ V}$ to find $P = 0.40 \times 6.0 = 2.4\text{ W}$ dissipated as heat and light in the lamp.
(c) Lamp connected in parallel:
Draw another lamp symbol connected in parallel with the existing lamp in the circuit.
Detailed Solution: A parallel connection provides an alternative path for current. Draw a second identical lamp symbol with its two terminals connected by wires to the two terminals of the existing lamp, ensuring the new loop does not break the original series connections.
Question 10
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.5.1 — Electromagnetic induction (Parts (a)(i), (a)(ii))
• Topic 4.5.6 — The transformer (Part (b))
▶️ Answer/Explanation
(a) (i) Pointer deflection:
The pointer deflects because the movement of the magnet induces an electromotive force (e.m.f.) in the coil, causing a current to flow through the voltmeter. The relative motion between the magnetic field and the conductor changes the magnetic flux linking the coil, which induces the e.m.f.
(a) (ii) Increasing the deflection:
1. Increase the speed of the magnet.
2. Increase the number of turns in the coil. A faster movement or more coil turns increases the rate of change of magnetic flux linkage, producing a larger induced e.m.f. and greater deflection.
(b) Secondary voltage \( V_s \):
Using the transformer equation: \( V_s / V_p = N_s / N_p \), where \( V_p = 180 \) V, \( N_p = 4800 \), and \( N_s = 200 \).
\( V_s = (200 / 4800) × 180 = 7.5 \) V. The calculation applies the turns ratio directly to find the stepped‑down output voltage.
Answer: 7.5 V
Question 11
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 5.1.2 — The nucleus (Part (a))
• Topic 5.2.4 — Half-life (Part (b))
▶️ Answer/Explanation
(a) Completing Table 11.1:
For the correct answers:
Electron: 6 particles, orbiting outside the nucleus, charge = -1.
Neutron: 8 particles, in the nucleus, charge = 0.
Proton: 6 particles, in the nucleus, charge = +1.
Figure 11.1 depicts the nuclear model of an atom, which consists of a central nucleus surrounded by orbiting electrons. By counting the particles in the diagram, there are 6 negatively charged electrons in orbit, 6 positively charged protons clustered in the nucleus, and 8 neutral neutrons also located in the nucleus. These counts correspond to the carbon-14 isotope.
(b) Age of the wood:
For the correct answer:
11400 years
The initial mass of the radioactive isotope is 8.00 mg, and it has decayed to 2.00 mg. This reduction represents two complete half-lives (8.00 mg → 4.00 mg → 2.00 mg). Since one half-life is 5700 years, the total age of the wood is calculated as 2 × 5700 years = 11400 years.
Question 12
(a) Fig. 12.1 represents the Earth and the Sun at one point in the Earth’s orbit of the Sun.
Explain the apparent daily motion of the Sun across the sky.
(b) List the four planets closest to the Sun in order of their distance from the Sun.
(c) The Sun mostly consists of two elements. State the two elements.
(d) The Sun is a star in a galaxy. State the name of the galaxy.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 6.1.1 — The Earth (Part (a))
• Topic 6.1.2 — The Solar System (Part (b))
• Topic 6.2.1 — The Sun as a Star (Parts (c) and (d))
▶️ Answer/Explanation
(a) Apparent daily motion of the Sun:
The apparent daily motion of the Sun across the sky is due to the Earth’s rotation on its axis. The Earth completes one full rotation every 24 hours, causing the Sun to appear to move from east to west.
Detailed Solution: The Earth rotates on its tilted axis from west to east once approximately every 24 hours. This rotation causes an observer on the Earth’s surface to see the Sun rise in the east, travel across the sky, and set in the west, creating the apparent daily motion.
(b) Four planets closest to the Sun:
1. Mercury
2. Venus
3. Earth
4. Mars
Detailed Solution: The order of the eight planets from the Sun is Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. The four innermost planets are rocky and relatively small compared to the outer gaseous giants.
(c) Elements in the Sun:
The Sun is mostly composed of hydrogen and helium.
Detailed Solution: The Sun is a medium-sized star whose composition is dominated by hydrogen (approximately 74%) and helium (approximately 24%) by mass, with the remaining 2% consisting of trace heavier elements.
(d) Name of the galaxy:
The Sun is part of the Milky Way galaxy.
Detailed Solution: The Milky Way is a barred spiral galaxy containing hundreds of billions of stars. Our Solar System, including the Sun and Earth, is located in one of its minor spiral arms, known as the Orion Arm.