Question 1
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.2 — Motion (Part (a))
• Topic 1.5.1 — Effects of Forces (Part (b))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
Magnitude: 0.90 m/s; Direction: 53 ∘ to the river bank
The swimmer’s velocity relative to the bank is the vector sum of their speed across the water and the river’s flow. Using Pythagoras’ theorem: v= (0.72) 2 +(0.54) 2 =0.90 m/s. The direction relative to the bank is found via tanθ= 0.54 0.72 , giving θ=arctan(1.33)≈53 ∘ .
(a)(ii)
For the correct answer:
81 m
The total distance d travelled relative to the river bank is calculated by multiplying the resultant velocity by the total time. First, convert time to seconds: t=1.5×60=90 s. Applying the formula d=v×t, we get d=0.90 m/s×90 s=81 m.
(b)
For the correct answer:
The forward force is balanced by the drag/resistance force, resulting in zero net force.
According to Newton’s First Law, an object moves at a constant velocity when the resultant force acting on it is zero. As the swimmer moves, the water exerts a resistive drag force that increases with speed until it exactly equals the forward force produced by the swimmer. At this point, the forces are in equilibrium, acceleration is zero, and a steady speed is maintained.
Question 2
Calculate the speed of the motorcyclist at the end of the straight section.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.1 — Physical quantities and measurement techniques (Part (b)(i))
• Topic 1.2 — Motion (Parts (a)(i), (b)(ii))
• Topic 1.5.1 — Effects of Forces (Parts (a)(ii), (b)(iii))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
38 m/s
To find the final speed, use the acceleration formula a= Δt Δv . Since the motorcyclist starts from rest (u=0 m/s), the final velocity v is calculated using v=u+(a×t). Substituting the given values: v=0+(7.2 m/s 2 ×5.3 s)=38.16 m/s, which rounds to 38 m/s.
(a)(ii)
For the correct answer:
1700 N
The resultant force is determined using Newton’s second law, F=ma. Given a combined mass m=240 kg and an acceleration a=7.2 m/s 2 , the force is F=240 kg×7.2 m/s 2 =1728 N. Rounding to two significant figures as per standard convention gives 1700 N.
(b)(i)
For the correct answer:
A vector has direction (and magnitude), but a scalar only has magnitude.
Physical quantities are categorized based on their properties. A scalar quantity, such as mass or speed, is fully described by its size (magnitude) alone. In contrast, a vector quantity, such as velocity or force, requires both a magnitude and a specific direction to be completely defined.
(b)(ii)
For the correct answer:
The velocity changes because the direction of motion is changing.
Velocity is defined as speed in a given direction. Although the motorcyclist maintains a constant speed around the bend, her direction of travel is continuously shifting. Because the direction component of the vector changes, the overall velocity is considered to be changing (accelerating) throughout the turn.
(b)(iii)
For the correct answer:
A resultant force is needed to change the direction of motion (centripetal force).
According to Newton’s laws, an object will move in a straight line at a constant speed unless acted upon by a resultant force. To move in a circular path, a force perpendicular to the motion must act toward the center of the bend to change the direction; without this resultant force, the motorcyclist would continue straight.
Question 3
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 2.3.2 — Convection (Part (a))
• Topic 2.1.2 — Particle model (Parts (b)(i), (b)(ii))
▶️ Answer/Explanation
(a)
This phenomenon occurs due to convection. When the heaters warm the air, the air particles gain kinetic energy and move further apart, causing the air to expand and become less dense. The cooler, denser air sinks, while the less dense hot air rises to the top of the room. This creates a convection current that carries thermal energy rapidly to the ceiling.
(b)(i)
As the temperature increases, the helium particles gain more kinetic energy. According to the kinetic particle model, temperature is a measure of the average kinetic energy of the particles, so the particles move faster with higher speeds. The relationship can be expressed by the kinetic energy equation $E_k = \frac{1}{2}mv^2$, where an increase in $E_k$ results in a higher velocity $v$.
(b)(ii)
Pressure is defined as $p = \frac{F}{A}$, where $F$ is the force from particle collisions. While higher temperatures cause more frequent and forceful collisions, the increasing volume $V$ provides more surface area $A$ and increases the distance between particles. These effects balance out, keeping the rate of change of momentum per unit area constant, thus maintaining a constant pressure as described by the relation $pV = \text{constant}$ at constant temperature.
Question 4
(i) Show that the increase in the internal energy of the water is 6300 J.
(ii) In the experiment, the lamp is switched on for 500 s. The power supplied to the filament lamp is 13 W. The useful energy from the lamp is transferred as light. The energy that increases the temperature of the water is wasted energy. Determine the maximum possible efficiency of the filament lamp.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 2.2.1 — Specific heat capacity (Part (a)(i))
• Topic 1.7.3 — Energy resources – Efficiency (Parts (a)(ii), (b))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
6300 J
To find the increase in internal energy, use the thermal energy formula ΔE=mcΔθ. First, convert the mass to kilograms (0.6 kg) and calculate the temperature rise (Δθ=21.5 ∘ C−19.0 ∘ C=2.5 ∘ C). Substituting these into the equation gives ΔE=0.6 kg×4200 J/(kg ∘ C)×2.5 ∘ C=6300 J.
(a)(ii)
For the correct answer:
3.1%
Total energy input is calculated using E=P×t, which is 13 W×500 s=6500 J. The useful light energy is the total input minus the wasted thermal energy: 6500 J−6300 J=200 J. Finally, determine efficiency using Efficiency= total energy input useful energy output ×100%, resulting in 6500 200 ×100≈3.1%.
(b)
For the correct answer:
Energy lost to surroundings OR beaker absorbs heat
The calculated efficiency assumes all wasted energy stays in the water, but in reality, thermal energy is lost to the air and the glass beaker. Additionally, some light energy might escape the beaker without being absorbed or accounted for as “useful” if it’s blocked. These external losses mean the actual useful light energy produced is lower than the theoretical maximum.
Question 5
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $3.2.2$ — Refraction of light (Parts $\mathrm{(a)}$, $\mathrm{(b)}$, $\mathrm{(c)}$, $\mathrm{(d)(i)}$, $\mathrm{(d)(ii)}$)
▶️ Answer/Explanation
(a)
Monochromatic light is light consisting of a single frequency or a single wavelength, which visually appears as a single color. In wave terms, all the photons in a monochromatic beam share the same energy level $E = hf$.
(b)
The ray enters along the radius of the circular arc AB, meaning it strikes the surface at an angle of incidence of $0^\circ$ relative to the normal. According to Snell’s Law, $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$, if $\theta_1 = 0^\circ$, then $\theta_2$ must also be $0^\circ$, so the light travels undeviated.
(c)
The critical angle $c$ is the angle of incidence in the denser medium for which the angle of refraction is $90^\circ$. Using the formula $n = \frac{1}{\sin c}$, we rearrange to find $\sin c = \frac{1}{n} = \frac{1}{1.5} \approx 0.667$. Calculating the inverse sine gives $c = \arcsin(0.667) \approx 41.8^\circ$, which rounds to $42^\circ$.
(d)(i)
Since the new angle of incidence $\theta = 45^\circ$ is greater than the critical angle $c = 42^\circ$, the light undergoes Total Internal Reflection (TIR). No light is refracted into the air; instead, the entire ray reflects back into the glass block following the law of reflection where the angle of reflection equals $45^\circ$.
(d)(ii)
The light reflected from P will strike side CD; since CD is perpendicular to BC and the reflection angle is $45^\circ$, the new angle of incidence at CD is also $45^\circ$. Because $45^\circ > 42^\circ$, the light undergoes a second Total Internal Reflection at side CD, reflecting back towards side DA.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.3 — Electromagnetic spectrum (Parts $\mathrm{(a)}$, $\mathrm{(b)}$, $\mathrm{(c)(i)}$, $\mathrm{(c)(ii)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
$0.16 \text{ m}$ (or $0.158 \text{ m}$)
To find the wavelength, use the wave equation $v = f\lambda$, rearranged as $\lambda = \frac{v}{f}$. Substituting the given speed $v = 3.0 \times 10^8 \text{ m/s}$ and frequency $f = 1.9 \times 10^9 \text{ Hz}$, the calculation is $\lambda = \frac{3.0 \times 10^8}{1.9 \times 10^9} \approx 0.158 \text{ m}$. This value represents the physical distance between consecutive crests of the microwave in air.
(b)
For the correct answer:
1. Microwaves can penetrate some walls.
2. They only require short aerials/antennas for transmission and reception.
Microwaves are ideal for mobile telecommunications because their relatively short wavelengths allow for very compact antennas to be integrated into portable handsets. Furthermore, their ability to pass through obstacles like building walls ensures that signals can be received reliably indoors, unlike higher-frequency waves which might be blocked or reflected.
(c)(i)
For the correct answer:
An analogue signal varies continuously in amplitude/frequency, whereas a digital signal consists of discrete pulses (typically high/low or 1/0).
Analogue signals are continuous waves that represent information by varying physical quantities like voltage over a full range of values. In contrast, digital signals are sequences of discrete steps or pulses that represent binary data (0s and 1s). While an analogue wave mimics the original sound or data directly, a digital signal translates that data into a robust code that is easier to process.
(c)(ii)
For the correct answer:
1. Increased rate of data transmission.
2. Increased range due to accurate signal regeneration (noise removal).
Digital signaling allows for much higher data speeds because multiple streams of information can be compressed and multiplexed together. Additionally, digital signals are highly resistant to interference; because they only use two states (on/off), electronic circuits can easily filter out “noise” and perfectly regenerate the original signal over long distances without the degradation common in analogue systems.
Question 7
(i) Calculate the electromotive force (e.m.f.) of the battery.
(ii) The switch is closed. Calculate the resistance of the complete circuit.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.3.2 — Series and parallel circuits (Parts (b)(i), (b)(ii), (c))
• Topic 4.3.3 — Potential dividers (Part (a))
▶️ Answer/Explanation
(a)
A potential divider is a simple circuit that splits the total voltage (e.m.f.) from a power source between two or more resistors connected in series. The output voltage across a specific component is proportional to its resistance relative to the total circuit resistance, according to V out =V in ⋅( R total R target ).
(b)(i)
For the correct answer:
15 V
With the switch open, R 1 and R 2 are in series. Since R 1 =R 2 =40Ω, the e.m.f. is shared equally between them. The voltmeter across R 2 reads 7.5 V, so the total e.m.f. E=V R1 +V R2 =7.5 V+7.5 V=15 V.
(b)(ii)
For the correct answer:
60Ω
When closed, R 2 and R 3 are in parallel: R 23 1 = 40 1 + 40 1 = 40 2 , so R 23 =20Ω. This combination is in series with R 1 , giving a total circuit resistance of R total =R 1 +R 23 =40Ω+20Ω=60Ω.
(c)
For the correct answer:
5 V
First, calculate the total current using I= R total E = 60Ω 15 V =0.25 A. The voltmeter measures the potential difference across the parallel section (R 23 ), calculated as V=I⋅R 23 =0.25 A⋅20Ω=5 V. Alternatively, using the potential divider formula: V=15 V⋅( 40Ω+20Ω 20Ω )=5 V.
Question 8
(i) There are 450 turns on the secondary coil. Calculate the number of turns on the primary coil.
(ii) The electrical power transferred to the transformer by the high-voltage cable is 77 MW. Calculate the current in the primary coil.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.5.1 — Electromagnetic induction (Parts (a), (b)(i), (b)(ii))
▶️ Answer/Explanation
(a)
A transformer operates via mutual induction: an alternating current in the primary coil generates a continuously changing magnetic field within the soft-iron core. This fluctuating magnetic flux passes through the secondary coil, inducing a varying electromotive force (e.m.f.) or voltage across its ends. The soft-iron core is essential as it easily magnetises and demagnetises, ensuring maximum magnetic linkage between the two coils.
(b)(i)
For the correct answer:
3000 turns
The relationship between voltages and number of turns is given by the transformer equation: V s V p = N s N p . Substituting the known values V p =220000 V, V s =33000 V, and N s =450, we rearrange to find N p : N p = 33000 220000 ×450=3000. Since the voltage is stepped down, the number of turns on the primary must be greater than on the secondary.
(b)(ii)
For the correct answer:
350 A
Electrical power is the product of voltage and current, expressed by the formula P=IV. To find the primary current I p , we use the input power P=77 MW=77×10 6 W and the primary voltage V p =220000 V. Rearranging the formula gives I p = V p P = 220000 77×10 6 =350 A.
Question 9
(i) Describe how a nucleus of X2 differs from a nucleus of X1.
(ii) Suggest why isotope X2 is stable whereas X1 is not stable.
(i) Define the term half-life.
(ii) Suggest one reason why isotopes with very short half-lives are especially hazardous.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 5.1.2 — The nucleus (Part (a))
• Topic 5.2.2 — The three types of nuclear emission (Part (b)(i))
• Topic 5.2.3 — Radioactive decay (Part (b)(ii))
• Topic 5.2.4 — Half-life (Parts (c)(i), (c)(ii))
▶️ Answer/Explanation
(a)
For the correct answer:
7
In a neutral atom, the number of protons equals the number of electrons orbiting the nucleus. By counting the electrons in the diagram, we find 5 protons, and counting the total particles in the nucleus gives the nucleon number A=12. The number of neutrons N is calculated using the equation N=A−Z, which results in 12−5=7 neutrons.
(b)(i)
For the correct answer:
X2 has one more proton and one fewer neutron than X1.
During beta (β − ) decay, a neutron in the nucleus decays into a proton and an electron (the beta particle). Consequently, the daughter nucleus X2 gains a proton while losing a neutron, increasing the atomic number by one while the nucleon number remains constant. This transformation is expressed as 5 12 X1→ 6 12 X2+ −1 0 β.
(b)(ii)
For the correct answer:
X2 has a more stable proton-to-neutron ratio.
Nuclear stability depends on the balance between the strong nuclear force and the electrostatic repulsion of protons. Isotope X1 has an excess of neutrons (7 neutrons to 5 protons), making it unstable. Isotope X2 is stable because its 1:1 ratio (6 protons and 6 neutrons) provides a lower energy state and a more balanced configuration of nuclear particles.
(c)(i)
For the correct answer:
The time taken for half the nuclei of a sample of a radioactive isotope to decay.
Half-life is a statistical measure of the rate of radioactive decay, representing the duration required for the initial number of undecayed nuclei (N 0 ) to reduce to N 0 /2. Alternatively, it can be defined as the time taken for the activity (count rate) of a radioactive source to decrease to half of its original value.
(c)(ii)
For the correct answer:
They release a large amount of radiation in a very short period.
A very short half-life indicates a high activity level, meaning a vast number of decay events occur per second. This results in a high dose rate of ionising radiation being delivered to surroundings or living tissue almost instantly, which can cause severe cellular damage or mutations compared to isotopes that release radiation more slowly over time.
Question 10
The orbital speed of Pluto varies as it orbits the Sun.
(i) Describe how the speed of Pluto varies as it moves from X to Y and then back to X.
(ii) Explain, in terms of energy transfers, why the speed of Pluto varies in this way.
(i) Convert this temperature to a value in degrees Celsius (°C).
(ii) Pluto has a white surface, as shown in Fig. 10.2. As Pluto rotates, the white surface alternately faces towards and away from the Sun.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 6.1.1 & 6.1.2 — The Earth and the Solar System (Parts (a)(i), (a)(ii))
• Topic 2.3.4 — Consequences of energy transfer (Parts (b)(i), (b)(ii))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
The speed decreases from X to Y and increases from Y to X.
As Pluto moves from its closest point (perihelion) at X to its furthest point (aphelion) at Y, it is moving against the Sun’s gravitational pull, causing it to slow down. Conversely, as it returns from Y to X, gravity pulls it closer, causing its velocity to increase.
(a)(ii)
For the correct answer:
Interchange of G.P.E. and K.E. while total energy remains constant.
In an elliptical orbit, energy is conserved such that E total =E k +E p . Moving from X to Y increases gravitational potential energy (E p ) at the expense of kinetic energy (E k ), causing the speed to decrease. From Y to X, E p is converted back into E k , resulting in an increase in speed.
(b)(i)
For the correct answer:
−230 ∘ C
Temperature in Celsius is calculated by subtracting 273 from the Kelvin value: T( ∘ C)=T(K)−273. Substituting the given value, 43−273=−230 ∘ C. This relationship is fundamental to the thermodynamic scale where absolute zero is 0 K or −273 ∘ C.
(b)(ii)
For the correct answer:
White surfaces are poor absorbers and poor emitters of radiation.
A white surface is a poor absorber of infrared radiation; thus, when facing the Sun, it reflects most energy, leading to a smaller temperature rise. When facing away, white is a poor emitter, meaning it loses heat slowly to space. Consequently, the rotation causes less extreme temperature fluctuations than a dark surface would.