2026 Cambridge IGCSE Physics (0625) Practice Paper with Answers

(a) Acceleration is defined as the change in velocity per unit time (or the rate of change of velocity).

(b)(i)
$$W = mg$$
$$W = 180\text{ kg} \times 9.8\text{ N/kg} = 1764\text{ N} \approx 1760\text{ N}$$

(b)(ii)
First calculate the resultant downward force:
$$F_{\text{res}} = W - F_{\text{drag}}$$
$$F_{\text{res}} = 1764\text{ N} - 684\text{ N} = 1080\text{ N}$$

Now use Newton's second law, $F = ma$:
$$a = \frac{F_{\text{res}}}{m}$$
$$a = \frac{1080\text{ N}}{180\text{ kg}} = 6.0\text{ m/s}^2$$

(c)
- As the probe falls and gains speed, the upward drag force / air resistance increases.
- The air resistance continues to increase until it is equal and opposite to the downward weight of the probe (the resultant force becomes zero).
- Since the resultant force is zero, by $F = ma$, the acceleration is zero and the probe moves at a constant velocity (terminal velocity).

(a) [1 mark]
- Change in velocity per unit time / rate of change of velocity [B1]
(Accept: $\frac{\Delta v}{\Delta t}$ with terms defined; Reject: speed per unit time)

(b)(i) [1 mark]
- $1760\text{ N}$ or $1764\text{ N}$ [B1]
(Accept: $1800\text{ N}$ if $g=10\text{ N/kg}$ is used)

(b)(ii) [3 marks]
- $F_{\text{res}} = 1764 - 684 = 1080\text{ N}$ [C1]
- $a = \frac{F}{m}$ or substitution $\frac{1080}{180}$ [C1]
- $a = 6.0\text{ m/s}^2$ (or $5.98\text{ m/s}^2$) [A1]

(c) [3 marks]
- (As speed increases) air resistance / drag increases [B1]
- Air resistance becomes equal to weight / resultant force becomes zero [B1]
- Zero resultant force means zero acceleration / constant speed [B1]

(a) The total momentum of an isolated system remains constant provided no external resultant force acts on the system.

(b)(i)
By the principle of conservation of momentum:
$$\text{Total initial momentum} = \text{Total final momentum}$$
$$m_A u_A + m_B u_B = (m_A + m_B) v$$
$$(3.5\text{ kg} \times 2.4\text{ m/s}) + (2.5\text{ kg} \times 0) = (3.5\text{ kg} + 2.5\text{ kg}) \times v$$
$$8.4 = 6.0 \times v$$
$$v = \frac{8.4}{6.0} = 1.4\text{ m/s}$$

(b)(ii)
$$\Delta p_A = m_A v - m_A u_A$$
$$\Delta p_A = (3.5\text{ kg} \times 1.4\text{ m/s}) - (3.5\text{ kg} \times 2.4\text{ m/s})$$
$$\Delta p_A = 4.9 - 8.4 = -3.5\text{ kg}\cdot\text{m/s}$$
$$\text{Magnitude of impulse} = 3.5\text{ N s} \text{ (or } 3.5\text{ kg}\cdot\text{m/s})$$

(c)
$$F = \frac{\Delta p}{\Delta t}$$
$$F = \frac{3.5\text{ N s}}{0.14\text{ s}} = 25\text{ N}$$
Since trolley A decelerates, the force acting on it is directed to the left (opposite to its initial direction of motion).

(a) [1 mark]
- Total momentum remains constant (or is conserved) in a closed system / when no resultant external force acts [B1]

(b)(i) [3 marks]
- Calculation of total initial momentum: $p_i = 3.5 \times 2.4 = 8.4\text{ kg m/s}$ [C1]
- Conservation of momentum equation: $8.4 = (3.5 + 2.5)v$ or $8.4 = 6.0v$ [C1]
- $v = 1.4\text{ m/s}$ [A1]

(b)(ii) [2 marks]
- Calculation of final momentum of A ($3.5 \times 1.4 = 4.9\text{ kg m/s}$) or $\Delta p = m\Delta v = 3.5 \times (1.4 - 2.4)$ [C1]
- $3.5\text{ N s}$ or $3.5\text{ kg m/s}$ [A1]

(c) [2 marks]
- $F = \frac{\Delta p}{\Delta t} = \frac{3.5}{0.14} = 25\text{ N}$ [A1]
- Direction: to the left / opposite to initial motion of trolley A [B1]

(a) For an object to be in equilibrium:
1. The resultant force on the object must be zero (no net force in any direction).
2. The resultant moment (turning effect) about any point must be zero (total clockwise moments = total anticlockwise moments).

(b)(i)
The plank is uniform, so its centre of gravity is at its geometric centre, which is at $\frac{1.60\text{ m}}{2} = 0.80\text{ m}$ from end $P$.
The pivot is located $0.50\text{ m}$ from end $P$.
$$\text{Distance from pivot to centre of gravity} = 0.80\text{ m} - 0.50\text{ m} = 0.30\text{ m}$$

(b)(ii)
Taking moments about the pivot:
- The weight of the plank acts downwards at the centre of gravity ($0.30\text{ m}$ to the right of the pivot), producing a clockwise moment:
$$\text{Moment}_{\text{plank}} = 40\text{ N} \times 0.30\text{ m} = 12.0\text{ N m}$$
- The force at end $Q$ acts downwards at a distance of $1.60\text{ m} - 0.50\text{ m} = 1.10\text{ m}$ to the right of the pivot, producing a clockwise moment:
$$\text{Moment}_Q = 15\text{ N} \times 1.10\text{ m} = 16.5\text{ N m}$$
- The load $W$ acts downwards at end $P$ ($0.50\text{ m}$ to the left of the pivot), producing an anticlockwise moment:
$$\text{Moment}_P = W \times 0.50\text{ m}$$

By the principle of moments for equilibrium:
$$\text{Total anticlockwise moment} = \text{Total clockwise moment}$$
$$W \times 0.50 = 12.0 + 16.5$$
$$W \times 0.50 = 28.5$$
$$W = \frac{28.5}{0.50} = 57\text{ N}$$

(c)
- Velocity is a vector quantity that has both magnitude (speed) and direction.
- As the car moves in a circular path, its direction of motion is continuously changing.
- Since velocity changes whenever direction changes, and acceleration is the rate of change of velocity, the car must be accelerating.

(a) [2 marks]
- Resultant / net force is zero [B1]
- Resultant / net moment is zero (or sum of clockwise moments = sum of anticlockwise moments) [B1]

(b)(i) [1 mark]
- $0.30\text{ m}$ [B1]

(b)(ii) [3 marks]
- Calculation of clockwise moments: $(40 \times 0.30) + (15 \times 1.10)$ or $12 + 16.5$ or $28.5\text{ N m}$ [C1]
- Equating clockwise and anticlockwise moments: $W \times 0.50 = 28.5$ [C1]
- $W = 57\text{ N}$ [A1]

(c) [2 marks]
- Direction (of motion) is constantly changing [B1]
- Velocity changes (since velocity is a vector / depends on direction), hence acceleration is rate of change of velocity [B1]

(a) Gas particles are in continuous, rapid, random motion and collide with the internal walls of the cylinder. Each collision exerts a force on the wall; the total force divided by the area of the wall produces pressure (\(p = \frac{F}{A}\)).

(b) Using Boyle's law at constant temperature:
\[p_1 V_1 = p_2 V_2\]
\[(1.05 \times 10^5\text{ Pa}) \times (4.8 \times 10^{-4}\text{ m}^3) = p_2 \times (1.6 \times 10^{-4}\text{ m}^3)\]
\[p_2 = \frac{1.05 \times 10^5 \times 4.8 \times 10^{-4}}{1.6 \times 10^{-4}} = 3.15 \times 10^5\text{ Pa} = 3.2 \times 10^5\text{ Pa}\]

(c) Because the volume decreases while the number and average speed (temperature) of particles remain unchanged, the particles are closer together (higher concentration/number density). Consequently, collisions with the container walls occur more frequently (more collisions per second), producing a greater total force per unit area.

(a)
- Particles collide / hit the walls [1] [B1]
- Force exerted over area (or change in momentum per second on wall creates force, \(p=F/A\)) [1] [B1]

(b)
- \(p_1 V_1 = p_2 V_2\) or \(1.05 \times 10^5 \times 4.8 \times 10^{-4} = p_2 \times 1.6 \times 10^{-4}\) [1] [C1]
- \(3.15 \times 10^5\text{ Pa}\) or \(3.2 \times 10^5\text{ Pa}\) (accept \(315\text{ kPa}\) or \(320\text{ kPa}\)) [1] [A1]

(c)
- Particles are closer together / number of particles per unit volume increases / smaller surface area for same number of particles [1] [B1]
- Collisions with the walls are more frequent / more collisions per unit time / per second (do NOT accept just 'more collisions') [1] [B1]

(a) Specific heat capacity is defined as the energy required per unit mass per unit temperature increase (\(c = \frac{\Delta E}{m \Delta\theta}\)).

(b)
1. Thermal energy supplied by the heater:
\[\Delta E = P \times t = 72\text{ W} \times (4.0 \times 60\text{ s}) = 72 \times 240 = 17\,280\text{ J}\]

2. Temperature change of the block:
\[\Delta\theta = 50.5 - 21.0 = 29.5\text{ }^\circ\text{C}\]

3. Specific heat capacity:
\[c = \frac{\Delta E}{m \Delta\theta} = \frac{17\,280\text{ J}}{0.65\text{ kg} \times 29.5\text{ }^\circ\text{C}} \approx 901.17\text{ J}/(\text{kg }^\circ\text{C}) \approx 900\text{ J}/(\text{kg }^\circ\text{C})\]
(to 2 or 3 sig. figs.: \(9.0 \times 10^2\text{ J}/(\text{kg }^\circ\text{C})\) or \(901\text{ J}/(\text{kg }^\circ\text{C})\)).

(c) In the liquid, particles have a distribution of kinetic energies. The particles with the highest kinetic energy / highest speeds overcome the intermolecular attractive forces and escape from the surface into the gas phase. This leaves behind particles with lower average kinetic energy, which causes the temperature of the remaining liquid to decrease.

(a)
- Energy required per unit mass per unit temperature increase (or energy per kg per °C/K) [1] [B1]

(b)
- Correct calculation of energy transferred: \(E = 72 \times 240 = 17\,280\text{ J}\) [1] [C1]
- Correct rearrangement/substitution: \(c = \frac{17\,280}{0.65 \times 29.5}\) [1] [C1]
- Final answer: \(900\) or \(901\) or \(9.0 \times 10^2\) AND unit \(\text{J}/(\text{kg }^\circ\text{C})\) or \(\text{J}/(\text{kg K})\) or \(\text{J kg}^{-1}\text{ }^\circ\text{C}^{-1}\) [1] [A1]

(c)
- Most energetic / fastest-moving particles escape / break attractive forces at the surface [1] [B1]
- Average kinetic energy of remaining particles decreases (so temperature drops) [1] [B1]

(a)(i) Using Snell's law:
\(n = \frac{\sin i}{\sin r}\)
\(1.52 = \frac{\sin 38^\circ}{\sin r}\)
\(\sin r = \frac{\sin 38^\circ}{1.52} = \frac{0.6157}{1.52} = 0.4050\)
\(r = \sin^{-1}(0.4050) = 23.9^\circ\)

(a)(ii) Using the critical angle equation:
\(n = \frac{1}{\sin c}\)
\(\sin c = \frac{1}{1.52} = 0.6579\)
\(c = \sin^{-1}(0.6579) = 41.1^\circ\)

(b) The ray hits the inner surface/boundary at an angle of incidence that is greater than the critical angle \(c\). Therefore, total internal reflection occurs and all light reflects back inside the core.

(a)(i)
[C1] \(n = \frac{\sin i}{\sin r}\) OR \(\sin r = \frac{\sin 38^\circ}{1.52}\)
[A1] \(23.9^\circ\) or \(24^\circ\)

(a)(ii)
[C1] \(\sin c = \frac{1}{n}\) OR \(\sin c = \frac{1}{1.52}\)
[A1] \(41.1^\circ\) or \(41^\circ\)

(b)
[B1] Angle of incidence (at the inner boundary) is greater than the critical angle
[B1] (So) total internal reflection occurs / no light refracts out (owtte)

(a) Ultrasound refers to sound waves with a frequency greater than \(20\,000\text{ Hz}\) (or \(20\text{ kHz}\)), above the upper limit of human hearing.

(b) In a longitudinal wave, the particles of the medium vibrate/oscillate parallel to the direction of wave propagation (energy transfer).

(c)(i) Using the wave equation:
\(v = f \lambda\)
\(\lambda = \frac{v}{f} = \frac{1500}{4.8 \times 10^4} = 0.03125\text{ m} \approx 0.031\text{ m}\) (or \(3.1\text{ cm}\))

(c)(ii) The time taken for the pulse to travel to the seabed and back is \(t = 0.64\text{ s}\).
Total distance travelled \(= v \times t = 1500 \times 0.64 = 960\text{ m}\)
\(\text{Depth} = \frac{960}{2} = 480\text{ m}\)

(a)
[B1] Sound / wave with frequency greater than \(20\,000\text{ Hz}\) / \(20\text{ kHz}\) (accept above upper limit of human hearing)

(b)
[B1] Particles vibrate / oscillate parallel to the direction of wave travel / energy propagation

(c)(i)
[C1] \(v = f \lambda\) OR \(\lambda = \frac{v}{f}\) OR \(\frac{1500}{48\,000}\)
[A1] \(0.031\text{ m}\) or \(0.0313\text{ m}\) or \(3.1\text{ cm}\)

(c)(ii)
[C1] \(d = \frac{v \times t}{2}\) OR \(s = 1500 \times 0.64\) (\(960\text{ m}\))
[A1] \(480\text{ m}\)

(a) Electromotive force is the electrical work done by a source in driving a unit charge around a complete circuit.

(b)(i) For two resistors in parallel: \(\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12}\), so \(R_p = 12\,\Omega\).

(b)(ii) The current through the parallel pair is equal to the total circuit current passing through \(R_1\): \(I = \frac{V_p}{R_p} = \frac{4.8}{12} = 0.40\text{ A}\).

(b)(iii) Potential difference across \(R_1\) is \(V_1 = I \times R_1 = 0.40 \times 15 = 6.0\text{ V}\). Total e.m.f. = \(V_1 + V_p = 6.0 + 4.8 = 10.8\text{ V}\).

(b)(iv) Energy transferred in \(R_3\): \(t = 2.0\text{ min} = 120\text{ s}\). \(E = \frac{V^2}{R_3} \times t = \frac{4.8^2}{30} \times 120 = \frac{23.04}{30} \times 120 = 92.16\text{ J} \approx 92\text{ J}\).

(a) Work done / energy transferred by a source [1]; per unit charge in driving charge around a complete circuit [1]
(b)(i) \(\frac{1}{R} = \frac{1}{20} + \frac{1}{30}\) [C1]; \(12\,\Omega\) [A1]
(b)(ii) \(I = \frac{V}{R_p} = \frac{4.8}{12}\) [C1]; \(0.40\text{ A}\) [A1]
(b)(iii) \(10.8\text{ V}\) [B1]
(b)(iv) \(E = Pt\) with substitution \(E = \frac{4.8^2}{30} \times 120\) OR \(I_3 = \frac{4.8}{30} = 0.16\text{ A}\) and \(E = 4.8 \times 0.16 \times 120\) [C1]; \(92\text{ J}\) or \(92.2\text{ J}\) [A1]

(a)(i) An alternating current produces a continuously changing magnetic field in the core. A changing magnetic field is required to cut the secondary coil and induce an alternating electromotive force (e.m.f.).

(a)(ii) A step-up transformer consists of a primary coil and a secondary coil wound on a soft iron core. The secondary coil has more turns of wire than the primary coil.

(b) Using the transformer turns ratio equation: \(\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 25\,000 \times \frac{18\,000}{800} = 25\,000 \times 22.5 = 562\,500\text{ V} = 5.625 \times 10^5\text{ V} = 562.5\text{ kV}\).

(c)(i) Current \(I = \frac{P}{V} = \frac{180 \times 10^6}{562.5 \times 10^3} = 320\text{ A}\).

(c)(ii) Power loss \(P = I^2 R = (320)^2 \times 6.5 = 102\,400 \times 6.5 = 665\,600\text{ W} = 6.66 \times 10^5\text{ W} \approx 6.7 \times 10^5\text{ W}\).

(a)(i) Alternating current produces a changing magnetic field (in the core) [B1]; which cuts the secondary coil to induce an e.m.f. / voltage [B1]
(a)(ii) Primary and secondary coils of wire on a (soft) iron core [B1]; more turns on the secondary coil than on the primary coil [B1]
(b) \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) or substitution \(V_s = 25\,000 \times \frac{18\,000}{800}\) [C1]; \(560\text{ kV}\) or \(562.5\text{ kV}\) or \(5.6 \times 10^5\text{ V}\) [A1]
(c)(i) \(I = \frac{180 \times 10^6}{562\,500} = 320\text{ A}\) [A1]
(c)(ii) \(P = I^2 R\) or \((320)^2 \times 6.5\) [C1]; \(6.7 \times 10^5\text{ W}\) or \(666\text{ kW}\) or \(665\,600\text{ W}\) [A1]

(a)(i) During \(\beta^-\)-decay, a neutron changes into a proton and an electron. The nucleon number remains 210, while the proton number increases by 1 from 83 to 84:
\[ {}^{210}_{83}\text{Bi} \rightarrow {}^{210}_{84}\text{Po} + {}^{0}_{-1}\beta \]

(a)(ii) A neutron in the nucleus decays/changes into a proton (and an electron is emitted as a beta particle).

(b)
Step 1: Subtract background radiation to obtain corrected count rates.
Initial corrected count rate at \(t = 0\):
\[ R_0 = 344 - 24 = 320\text{ counts/minute} \]
Corrected count rate at \(t = 18\text{ hours}\):
\[ R = 64 - 24 = 40\text{ counts/minute} \]

Step 2: Determine the number of half-lives that have passed:
\[ 320 \xrightarrow{1\text{ half-life}} 160 \xrightarrow{2\text{ half-lives}} 80 \xrightarrow{3\text{ half-lives}} 40 \]
Number of half-lives \(n = 3\).

Step 3: Calculate the half-life \(T_{1/2}\):
\[ T_{1/2} = \frac{18\text{ hours}}{3} = 6.0\text{ hours} \]

(c) Precaution: Use tongs / forceps / tweezers (or store in a lead-lined container).
Explanation: Increasing the distance from the radioactive source significantly decreases the radiation intensity reaching the body (or lead shielding absorbs/blocks the beta particles), reducing the radiation dose.

(a)(i)
- \(^{210}_{84}\text{Po}\) correct (nucleon number 210 and proton number 84) [B1]
- \(^{0}_{-1}\beta\) or \(^{0}_{-1}\text{e}\) correct [B1]

(a)(ii)
- A neutron changes / decays into a proton (and an electron) [B1]

(b)
- Subtracts background count rate to get 320 and 40 (counts/min) [C1]
- Identifies that 3 half-lives have elapsed (e.g. \(320 \rightarrow 160 \rightarrow 80 \rightarrow 40\) or \(2^n = 8\)) [C1]
- \(6.0\text{ hours}\) / \(6\text{ h}\) (or \(360\text{ minutes}\) with correct unit) [A1]

(c)
- Valid precaution: use forceps / tongs OR keep at a large distance OR use lead shielding [B1]
- Valid matching explanation: increasing distance reduces intensity / dose OR shielding absorbs / stops the radiation [B0.5]

(a)
Convert orbital period \(T\) from days to seconds:
\[ T = 200 \times 24 \times 3600\text{ s} = 1.728 \times 10^{7}\text{ s} \]
Orbital radius \(r = 1.44 \times 10^{8}\text{ km} = 1.44 \times 10^{11}\text{ m} \] Orbital speed: \[ v = \frac{2\pi r}{T} = \frac{2 \times \pi \times 1.44 \times 10^{8}\text{ km}}{1.728 \times 10^{7}\text{ s}} = 52.3598...\text{ km/s} \approx 52.4\text{ km/s} \] (b)(i) Redshift is the observed increase in the wavelength (or decrease in frequency) of light/electromagnetic radiation emitted by distant celestial objects moving away from an observer. (b)(ii) Step 1: Use Hubble's law \)v = H_0 d\) to find the distance \(d\) in metres:
\[ d = \frac{v}{H_0} = \frac{1.32 \times 10^{7}\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 6.00 \times 10^{24}\text{ m} \]

Step 2: Convert the distance to light-years:
\[ \text{Distance in light-years} = \frac{6.00 \times 10^{24}\text{ m}}{9.5 \times 10^{15}\text{ m/light-year}} = 6.3157... \times 10^{8}\text{ light-years} \approx 6.3 \times 10^{8}\text{ light-years} \]

(c) Cosmic Microwave Background Radiation (CMBR).

(a)
- Correct formula for orbital speed: \(v = \frac{2\pi r}{T}\) [C1]
- Correct conversion of time into seconds: \(T = 200 \times 24 \times 3600 = 1.728 \times 10^{7}\text{ s}\) [C1]
- Correct value for speed: \(52.4\text{ km/s}\) (accept \(52\text{ km/s}\) or \(5.24 \times 10^4\text{ m/s}\)) [A1]

(b)(i)
- (Observed) increase in wavelength (or shift towards red end of spectrum) of light from receding celestial bodies / galaxies [B1]

(b)(ii)
- Uses \(d = \frac{v}{H_0}\) or substitutes: \(\frac{1.32 \times 10^7}{2.2 \times 10^{-18}}\) [C1]
- Calculates distance in metres: \(6.0 \times 10^{24}\text{ m}\) [C0.5]
- Converts to light-years correctly: \(6.3 \times 10^{8}\text{ light-years}\) (accept \(6.32 \times 10^8\)) [A1]

(c)
- Cosmic Microwave Background Radiation / CMBR (accept CMB) [B1]