Question 1
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.2$ — Motion (Parts $\mathrm{(a)}$, $\mathrm{(b)}$, $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
$11.5\text{ m/s}$
To determine the speed at $t = 2.0\text{ s}$, locate $2.0$ on the x-axis (time) and move vertically upwards to the plotted line. From that specific point on the graph, move horizontally to the left to read the corresponding value on the y-axis (speed). The intersection point aligns exactly halfway between the $11\text{ m/s}$ and $12\text{ m/s}$ grid lines.
(b)
For the correct answer:
$45\text{ m}$
The total distance travelled is equal to the area under the speed-time graph. The area can be split into a rectangle (from $t = 0\text{ s}$ to $t = 0.5\text{ s}$) and a triangle (from $t = 0.5\text{ s}$ to $t = 4.0\text{ s}$). Rectangle: $0.5 \times 20 = 10\text{ m}$. Triangle: $\frac{1}{2} \times 3.5 \times 20 = 35\text{ m}$. Total distance = $10 + 35 = 45\text{ m}$.
(c)
For the correct answer:
$5.7\text{ m/s}^2$
Deceleration is the magnitude of the gradient of the downward-sloping section. Using $a = \frac{\Delta v}{\Delta t}$, the change in speed is $20\text{ m/s}$ and the time interval is $3.5\text{ s}$. Calculation: $\frac{20}{3.5} \approx 5.71\text{ m/s}^2$.
Question 2
Fig. $2.3$ shows this capsule just before it lands in the sea.
The capsule travels at terminal velocity.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.6$ — Momentum (Parts $\mathrm{(a)}$, $\mathrm{(b)}$)
• Topic $1.5.1$ — Effects of forces (Part $\mathrm{(c)}$)
• Topic $1.3$ — Mass and weight (Part $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
$6.55 \times 10^6 \text{ kg m/s}$
The momentum $p$ of an object is defined as the product of its mass $m$ and its velocity $v$, given by the equation $p = mv$. Substituting the given values: $p = 840 \text{ kg} \times 7800 \text{ m/s} = 6552000 \text{ kg m/s}$. In standard form, this is $6.552 \times 10^6 \text{ kg m/s}$, which is approximately $6.55 \times 10^6 \text{ kg m/s}$.
(b)
For the correct answer:
$7790 \text{ m/s}$
Using the principle of conservation of momentum: $\text{Total momentum before} = \text{Total momentum after}$.
$6552000 = (720 \times v) + (120 \times 7850)$
$6552000 = 720v + 942000$
$5610000 = 720v$
$v = 7791.6 \text{ m/s}$. To $3$ significant figures, $v = 7790 \text{ m/s}$.
(c)
For the correct calculated value:
$12000 \text{ kg}$
At terminal velocity, the resultant force is zero, so Upward Force = Weight ($W$).
$W = 120 \text{ kN} = 120000 \text{ N}$.
Using $W = mg$ (taking $g = 10 \text{ N/kg}$), $m = \frac{W}{g} = \frac{120000}{10} = 12000 \text{ kg}$.
Question 3
(a) The archer pulls back on the bow string, doing a total of 110 J of work. Her hand moves a distance of 0.45 m. The bow is bent and stores energy.
Show that the average force applied by the archer in pulling the string back is approximately 240 N.
(i) Calculate the initial speed $v$ of the arrow.
(ii) Explain why the speed of the arrow as it hits the target is less than the value in (b)(i).
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.7.2$ — Work (Part $\mathrm{(a)}$)
• Topic $1.7.1$ — Energy (Part $\mathrm{(b)(i)}$)
• Topic $1.5.1$ — Effects of forces (Part $\mathrm{(b)(ii)}$)
▶️ Answer/Explanation
Correct Answer: 244 N
Detailed solution: The work done on the bow string is calculated using the equation $W = Fd$, where $W$ is the work done, $F$ is the applied force, and $d$ is the distance. We can rearrange this formula to solve for the average force: $F = \frac{W}{d}$. Substituting the given values, $F = \frac{110}{0.45}$, which calculates to 244.44 N. This correctly demonstrates that the average force is approximately 240 N.
Correct Answer: 86 m/s
Detailed solution: All of the 110 J of stored energy is completely transferred into the arrow’s kinetic energy. Using the kinetic energy equation $E_k = \frac{1}{2}mv^2$, we can rearrange it to find the initial speed: $v = \sqrt{\frac{2E_k}{m}}$. Substituting the known energy and the mass of 0.030 kg into the equation gives $v = \sqrt{\frac{2 \times 110}{0.030}}$. This results in $v = \sqrt{7333}$, yielding a final initial speed of approximately 86 m/s.
Correct Answer: Air resistance / friction causes energy loss to the surroundings.
Detailed solution: As the arrow flies toward the target, it travels through the air and experiences air resistance (drag), which opposes its motion. The arrow must do mechanical work against this friction, causing a portion of its kinetic energy to be transferred into thermal (internal) energy and dissipated into the surroundings. Consequently, the arrow has less kinetic energy left when it hits the target, resulting in a lower final speed.
Question 4
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $2.3 .1$ — Conduction (Part $\mathrm{(a)}$)
• Topic $2.2 .2$ — Specific heat capacity (Parts $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Free electrons carry thermal energy; lattice vibrations transfer energy to neighbouring ions.
Copper is a metal, meaning it contains free (delocalised) electrons. When heated by the hotplate, these free electrons gain kinetic energy and move rapidly through the metal, colliding with distant ions and transferring thermal energy. Additionally, the heated metal ions vibrate more vigorously. These lattice vibrations cause adjacent, cooler ions to vibrate, further transferring thermal energy through the solid copper base.
(b) (i)
For the correct answer:
Energy transferred per unit mass per unit temperature change.
Specific heat capacity is a physical property of matter that dictates how much thermal energy is required to change its temperature. By definition, it is the amount of energy transferred (or required) per unit mass of a substance to cause a unit change in its temperature. In calculations, this relationship is expressed as $c = \frac{\Delta E}{m\Delta\theta}$.
(b) (ii)
For the correct calculated value:
$0.51\text{ J}/(\text{g}^{\circ}\text{C})$
Assuming no heat loss to the surroundings, the thermal energy lost by the hot metal equals the thermal energy gained by the water.
$\Delta E_{\text{water}} = 50 \times 4.2 \times (31 – 22) = 1890\text{ J}$.
The energy lost by the metal is $\Delta E_{\text{metal}} = 54 \times c \times (100 – 31) = c \times 3726$.
Equating both: $c \times 3726 = 1890 \implies c \approx 0.507\text{ J}/(\text{g}^{\circ}\text{C})$.
Rounding to two significant figures gives $0.51\text{ J}/(\text{g}^{\circ}\text{C})$.
Question 5
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $3.2.2$ — Refraction of light (Parts $\mathrm{(a)}$, $\mathrm{(b)}$, $\mathrm{(c)}$, $\mathrm{(d)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
The ratio of the speed of light in air to the speed of light in water.
The refractive index is a physical quantity that indicates how much light slows down when passing from one medium into another. By definition, for light travelling from air into water, it is expressed as the ratio of the wave’s speed in the first medium (air) to its speed in the second medium (water). It can also be defined mathematically as the ratio of the sine of the angle of incidence to the sine of the angle of refraction.
(b)
For the correct answer:
Ray P continues vertically upwards. Ray Q refracts away from the normal.
Ray P strikes the boundary perfectly perpendicular to the surface, meaning its angle of incidence is $0^{\circ}$. Therefore, it passes straight through into the air without changing direction. Ray Q approaches the boundary at an angle. Because light travels faster in a less dense medium (air) than in a denser medium (water), ray Q speeds up upon crossing the boundary and bends away from the normal line.
(c)
For the correct calculated value:
$1.3$
In the diagram, ray R is shown skimming along the boundary surface, meaning the angle of refraction is exactly $90^{\circ}$. This indicates that its angle of incidence, $49^{\circ}$, is the critical angle ($c$) for water. To calculate the refractive index ($n$), we use the formula $n = \frac{1}{\sin c}$. Substituting the given value, we get $n = \frac{1}{\sin(49^{\circ})}$, which equals approximately $1.325$. Rounding to two significant figures gives a refractive index of $1.3$.
(d)
For the correct answer:
The ray is travelling from a denser to a less dense medium and its angle of incidence exceeds the critical angle.
Total internal reflection occurs only when two specific conditions are met simultaneously. First, the light ray must be travelling from an optically denser medium (water) towards a less dense medium (air). Second, the ray must strike the boundary at an angle of incidence that is greater than the critical angle. From part (c), we know the critical angle is $49^{\circ}$. Ray S clearly hits the boundary at an angle greater than $49^{\circ}$, so it reflects entirely back into the water.
Question 6
(i) Describe how ultrasound is used to locate an object below the surface of the sea.
You may draw a labelled diagram as part of your answer.
(ii) State one other use of ultrasound.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.4 — Sound (Parts (a), (c))
• Topic 3.1 — General properties of waves (Part (b))
▶️ Answer/Explanation
(a)
For the correct answer:
Sound with a frequency higher than 20 kHz.
Ultrasound refers to sound waves that have a frequency above the upper limit of normal human hearing. For a typical human, this upper auditory threshold is approximately 20000 Hz or 20 kHz.
(b)
For the correct answer:
Vibrations (of the wave / particles) are parallel to the direction of propagation.
In a longitudinal wave, the particles of the transmitting medium oscillate or vibrate back and forth along the exact same axis as the wave itself travels. This means the direction of the particle vibrations is parallel to the direction of energy transfer.
(c)(i)
For the correct answer:
A pulse of ultrasound is sent into the water and reflects from the object. The time to travel to the object and back is measured, and distance is calculated using depth=speed× 2 time .
To locate a submerged object, a sonar transmitter emits a pulse of ultrasound. This sound wave travels downward until it strikes an object and reflects back. A receiver detects the returning wave, and the total time is recorded. The depth is calculated using d=v× 2 t because the sound travels to the object and back.
(c)(ii)
For the correct answer:
Non-destructive testing of materials OR medical scanning (of soft tissue).
Ultrasound is widely utilized in the medical field for creating images of internal organs or fetuses. It is also employed in industrial settings for the non-destructive testing of materials to detect internal cracks.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.3.2$ — Series and parallel circuits (Parts $\mathrm{(a)}$, $\mathrm{(c)}$, $\mathrm{(d)}$)
• Topic $4.2.4$ — Resistance (Part $\mathrm{(b)}$, $\mathrm{(d)}$)
▶️ Answer/Explanation
Correct Answer: 0.50 A
Detailed solution: In a series circuit, the current is identical at all points along the main path before branching occurs. The diagram indicates a current of 0.50 A flowing into the 6.0 Ω resistor. Because the ammeter is connected in series with the main power supply and the 6.0 Ω resistor, it will measure this same total circuit current of 0.50 A.
Correct Answer: 3.0 V
Detailed solution: The potential difference (p.d.) across a resistor is found using Ohm’s law, $V = IR$. We are given the current $I = 0.50\text{ A}$ and the resistance $R = 6.0\ \Omega$. Substituting these values into the formula yields $V = 0.50 \times 6.0$, which calculates to a potential difference of 3.0 V.
Correct Answer: $12 – 3.0 = 9.0\text{ V}$
Detailed solution: In a circuit, the total sum of the potential differences across components in series must equal the total electromotive force (e.m.f.) provided by the power supply. The total supply voltage is 12 V, and we calculated the p.d. across the 6.0 Ω resistor as 3.0 V. Subtracting this from the total gives the remaining p.d. across the parallel combination: $12 – 3.0 = 9.0\text{ V}$. Since components in parallel share the same voltage, the p.d. across the 20 Ω resistor is exactly 9.0 V.
Correct Answer: 180 Ω
Detailed solution: First, calculate the current $I_2$ through the 20 Ω resistor using $I = \frac{V}{R}$: $I_2 = \frac{9.0}{20} = 0.45\text{ A}$. The total current entering the parallel junction must equal the sum of the currents leaving it in each branch, meaning the current through $R$ is $I_1 = 0.50 – 0.45 = 0.05\text{ A}$. Finally, use the resistance equation $R = \frac{V}{I_1}$ with the known branch voltage (9.0 V) and branch current (0.05 A) to find the unknown resistance: $R = \frac{9.0}{0.05} = 180\ \Omega$.
Question 8
Calculate the charge that flows.
Calculate the charge that flows.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.2.1 — Electric charge (Parts (a)(i), (a)(ii))
• Topic 4.2.3 — Electromotive force and potential difference (Part (b))
• Topic 4.2.2 — Electric current (Part (c))
▶️ Answer/Explanation
Correct Answer: A region where an electric charge experiences a force.
Detailed solution: An electric field is formally defined as a region of space where an electric charge experiences a force. It represents the influence a charge has on its surroundings, meaning any other charged object placed within this specific region will be either attracted or repelled depending on its relative polarity.
Correct Answer: Four radial straight field lines starting at the charge, with at least one arrow pointing towards the charge.
Detailed solution: For an isolated negative point charge, the electric field lines must be drawn radially inwards. You should draw exactly four straight lines distributed evenly around the charge. The arrows on these lines must point towards the central negative charge, indicating the direction of the electrostatic force that a positive test charge would experience.
Correct Answer: 1600 C
Detailed solution: Potential difference V is defined as the energy transferred W per unit charge Q, given by the equation V= Q W . Rearranging this to solve for charge gives Q= V W . First, convert the energy to Joules: W=4.5×10 5 MJ=4.5×10 11 J. Substituting the known values yields Q= 2.9×10 8 4.5×10 11 ≈1551.7 C, which rounds to 1600 C (to two significant figures).
Correct Answer: 240 C
Detailed solution: Electric current I is the rate of flow of charge Q over time t, expressed mathematically as I= t Q . Rearranging this formula to solve for the charge gives Q=I×t. By substituting the given values of current (2.0 A) and time (120 s), we calculate the total charge that flows as Q=2.0×120=240 C.
Question 9
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.5.6$ — The transformer (Parts $\mathrm{(a)}$, $\mathrm{(b)}$, $\mathrm{(c)}$)
▶️ Answer/Explanation
Part (a)(i)
When an alternating current flows through the primary coil, it generates a continuously changing magnetic field. The soft-iron core acts as a medium to transfer this changing magnetic field to the secondary coil. As the secondary coil is exposed to this varying magnetic flux, it cuts the magnetic field lines, which induces an alternating electromotive force (e.m.f.) across it. Because the original magnetic field continuously changes direction, the induced current in the secondary coil also changes direction, resulting in an alternating current.
Part (a)(ii)
For the correct answer:
$240\text{ V}$
For a transformer, the ratio of the secondary voltage to primary voltage equals the turns ratio: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$. Given a step-up ratio of $1:20$, we have $\frac{N_s}{N_p} = 20$. Substituting the primary voltage $V_p = 12\text{ V}$ yields $\frac{V_s}{12} = 20$, therefore $V_s = 240\text{ V}$.
Part (b)
For the correct answer:
$0.16\text{ A}$
The electrical power lost as heat in a transmission cable is calculated using the formula $P = I^2R$. Rearranging this to solve for the current gives $I = \sqrt{\frac{P}{R}}$. Substituting the given values: $I = \sqrt{\frac{1.25 \times 10^{-3}}{0.050}} = \sqrt{0.025} = 0.16\text{ A}$.
Part (c)
For the correct answer:
Less power/energy losses and thinner/cheaper cables can be used.
Transmitting electrical energy at high voltages allows the same amount of power to be transferred using a much lower current ($P = VI$). A lower current significantly reduces the energy lost as heat in the cables ($P = I^2R$). Furthermore, lower currents allow for the use of thinner, less expensive cables.
Question 10
(b) It takes light $490\text{ s}$ to travel from the Sun to the Earth.
Calculate the distance from the Sun to the Earth.
Fig. 10.1 is a scatter graph showing how the speed of galaxies moving away from the Earth varies with their distances from the Earth.
A scientist draws a best-fit line on the scatter graph.
Use this best-fit line to determine a value for the Hubble constant.
(d) The speed of a receding galaxy can be estimated using redshift.
Describe what is meant by redshift.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $6.2.2$ — Stars (Part $\mathrm{(a)}$)
• Topic $6.1.2$ — The Solar System (Part $\mathrm{(b)}$)
• Topic $6.2.3$ — The Universe (Parts $\mathrm{(c)}$, $\mathrm{(d)}$)
▶️ Answer/Explanation
Correct Answer: Distance travelled in the vacuum of space by light in one year.
Detailed solution: A light-year is a unit of astronomical distance, not a measure of time. It is precisely defined as the total distance that a beam of light, traveling at a constant speed of $3.0 \times 10^8\text{ m/s}$, covers as it travels through a vacuum over the duration of one Earth year.
Correct Answer: $1.5 \times 10^{11}\text{ m}$
Detailed solution: The constant speed of light in a vacuum is $c = 3.0 \times 10^8\text{ m/s}$. Using the basic distance formula $d = v \times t$, we substitute the known speed and the given time: $d = (3.0 \times 10^8\text{ m/s}) \times 490\text{ s}$. Calculating this product yields $1.47 \times 10^{11}\text{ m}$, which is generally rounded to $1.5 \times 10^{11}\text{ m}$ to reflect appropriate significant figures.
Correct Answer: $2.5 \times 10^{-18}\text{ s}^{-1}$
Detailed solution: The Hubble constant $H_0$ is defined as the ratio of recessional velocity to distance, $H_0 = \frac{v}{d}$, which equals the gradient of the provided best-fit line. By selecting a clear coordinate point on the line, such as $(100 \times 10^{20}\text{ km}, 2.5 \times 10^4\text{ km/s})$, we calculate the gradient: $\frac{2.5 \times 10^4}{100 \times 10^{20}}$. The distance units of kilometers completely cancel out, resulting in a value of $2.5 \times 10^{-18}\text{ s}^{-1}$ for the Hubble constant.
Correct Answer: The observed increase in the wavelength of electromagnetic radiation (or light) from distant galaxies compared to the wavelength measured on Earth.
Detailed solution: As a galaxy moves away from an observer, the light waves it emits get stretched out due to the Doppler effect. This stretching manifests as an observed increase in the wavelength of the light, shifting the spectral lines towards the red end of the electromagnetic spectrum. This redshift acts as key astronomical evidence that the Universe is continuously expanding.
Question 11
(b) A protactinium (Pa) nucleus decays into a uranium (U) nucleus by the emission of a beta particle ($\beta$-particle).
(i) Complete the nuclear equation for the decay.
Give two reasons why $\alpha$-particles are more strongly ionising than $\beta$-particles.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $5.2.4$ — Half-life (Part $\mathrm{(a)}$)
• Topic $5.2.3$ — Radioactive decay (Parts $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$)
• Topic $5.2.2$ — The three types of nuclear emission (Part $\mathrm{(c)}$)
▶️ Answer/Explanation
Correct Answer: Time taken for half the nuclei (in any sample) to decay.
Detailed solution: The half-life is a key statistical measure used in nuclear physics to describe the rate of radioactive decay. It specifically refers to the time interval required for exactly half of the unstable radioactive nuclei in a given sample to undergo a decay process.
Correct Answer: $_{92}^{234}\text{U} + _{-1}^{0}\beta$
Detailed solution: In a $\beta$-decay process, a $\beta$-particle is emitted ($_{-1}^{0}\beta$). To conserve the nucleon number ($234$) and the proton number ($91$), the resulting nucleus must have $234$ nucleons and $92$ protons ($92 – 1 = 91$), which is Uranium ($\text{U}$).
Correct Answer: A neutron changes to a proton (plus an electron).
Detailed solution: During $\beta$-emission, a neutron in the nucleus transforms into a proton and an electron. The proton remains in the nucleus (increasing the atomic number), while the electron is emitted as the beta particle.
Correct Answer: $\alpha$-particles have greater kinetic energy and greater charge than $\beta$-particles.
Detailed solution: $\alpha$-particles have a $+2$ charge and a much larger mass, leading to higher kinetic energy and more frequent collisions with atoms, making them more strongly ionising than the smaller, $-1$ charged $\beta$-particles.