2026 Cambridge IGCSE Physics (0625) Practice Paper with Answers

(a) (i) Momentum is defined as the product of mass and velocity (or \(p = mv\), where \(m\) is mass and \(v\) is velocity).
(ii) Momentum is a vector quantity.

(b) (i) By conservation of linear momentum:
Total momentum before collision = Total momentum after collision
\(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\)
\((1.2\text{ kg} \times 4.5\text{ m/s}) + (0.80\text{ kg} \times 0) = (1.2\text{ kg} + 0.80\text{ kg}) v\)
\(5.4 = 2.0 v\)
\(v = \frac{5.4}{2.0} = 2.7\text{ m/s}\)

(ii) Resultant force is the rate of change of momentum:
\(F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}\)
For the block:
\(\Delta p = 0.80 \times (2.7 - 0) = 2.16\text{ kg m/s}\)
\(F = \frac{2.16\text{ N s}}{0.075\text{ s}} = 28.8\text{ N} \approx 29\text{ N}\)

(a) (i) mass \(\times\) velocity OR \(p = mv\) with symbols defined [B1]
(ii) vector [B1]

(b) (i) \(p = mv\) used OR \(1.2 \times 4.5\) OR \(5.4\text{ (kg m/s)}\) seen [C1]
\((1.2 + 0.80)v = 5.4\) OR \(v = \frac{5.4}{2.0}\) [C1]
\(v = 2.7\text{ m/s}\) [A1]

(ii) \(F = \frac{\Delta p}{\Delta t}\) OR \(F = ma\) AND \(a = \frac{\Delta v}{\Delta t}\) [C1]
\(\Delta p = 0.80 \times 2.7\text{ (= 2.16)}\) OR \(a = \frac{2.7}{0.075}\text{ (= 36)}\) [C1]
\(F = 28.8\text{ N}\) OR \(29\text{ N}\) [A1]

(a) Energy cannot be created or destroyed, only transferred from one store to another (or total energy in a closed system remains constant).

(b) (i) \(\Delta E_{\text{p}} = mg\Delta h\)
\(\Delta E_{\text{p}} = 65\text{ kg} \times 9.8\text{ N/kg} \times 3.2\text{ m} = 2038.4\text{ J} \approx 2040\text{ J}\) (or \(2.04\text{ kJ}\))

(ii) \(P = \frac{W}{t} = \frac{\Delta E}{t}\)
\(t = \frac{\Delta E_{\text{p}}}{P} = \frac{2038.4\text{ J}}{450\text{ W}} = 4.5298\text{ s} \approx 4.53\text{ s}\)

(iii) \(\text{Efficiency} = \frac{\text{useful power output}}{\text{total power input}} \times 100\%\)
\(0.72 = \frac{450}{\text{Power}_{\text{in}}}\)
\(\text{Power}_{\text{in}} = \frac{450}{0.72} = 625\text{ W}\)

(a) energy cannot be created or destroyed / total energy of an isolated/closed system remains constant / can only be transferred from one store to another [B1]

(b) (i) \((\Delta E_{\text{p}} =) mg\Delta h\) OR \(65 \times 9.8 \times 3.2\) [C1]
\(2040\text{ J}\) OR \(2038\text{ J}\) OR \(2.04\text{ kJ}\) [A1]

(ii) \(P = \frac{E}{t}\) OR \(t = \frac{\Delta E_{\text{p}}}{P}\) OR \(t = \frac{2038.4}{450}\) [C1]
\(t = 4.53\text{ s}\) OR \(4.5\text{ s}\) (ecf from (b)(i)) [A1]

(iii) \(\text{efficiency} = \frac{\text{useful output}}{\text{total input}} (\times 100\%)\) seen or implied [C1]
\(\text{input} = \frac{450}{0.72}\) OR \(\frac{450}{72} \times 100\) [C1]
\(625\text{ W}\) OR \(630\text{ W}\) [A1]

(a) The energy required per unit mass to increase the temperature by one unit of temperature (or \(c = \frac{\Delta E}{m\Delta\theta}\) with all terms defined).

(b) (i) \(E = Pt\)
\(t = 4.0 \times 60 = 240\text{ s}\)
\(E = 300\text{ W} \times 240\text{ s} = 72\,000\text{ J} = 72\text{ kJ}\)

(ii) \(\Delta\theta = 35.5\,^\circ\text{C} - 18.0\,^\circ\text{C} = 17.5\,^\circ\text{C}\)
\(E_{\text{Al}} = m_{\text{Al}} c_{\text{Al}} \Delta\theta\)
\(E_{\text{Al}} = 0.45\text{ kg} \times 900\text{ J}/(\text{kg}\,^\circ\text{C}) \times 17.5\,^\circ\text{C} = 7087.5\text{ J} \approx 7090\text{ J}\) (or \(7.09\text{ kJ}\))

(iii) Energy absorbed by the liquid:
\(E_{\text{liq}} = E_{\text{total}} - E_{\text{Al}} = 72\,000\text{ J} - 7087.5\text{ J} = 64\,912.5\text{ J}\)
\(c_{\text{liq}} = \frac{E_{\text{liq}}}{m_{\text{liq}} \Delta\theta} = \frac{64\,912.5}{0.60 \times 17.5} = \frac{64\,912.5}{10.5} = 6182.14\text{ J}/(\text{kg}\,^\circ\text{C}) \approx 6200\text{ J}/(\text{kg}\,^\circ\text{C})\)

(a) energy required per unit mass [B1]
per unit temperature change / per \(^\circ\text{C}\) / per \(\text{K}\) [B1]

(b) (i) \(t = 240\text{ (s)}\) OR \(E = Pt\) [C1]
\(72\,000\text{ J}\) OR \(72\text{ kJ}\) [A1]

(ii) \(\Delta\theta = 17.5\,(^\circ\text{C})\) seen or used [C1]
\(E_{\text{Al}} = 0.45 \times 900 \times 17.5 = 7090\text{ J}\) OR \(7088\text{ J}\) OR \(7.09\text{ kJ}\) [A1]

(iii) \(E_{\text{liq}} = 72\,000 - 7087.5\text{ (= 64 912.5 J)}\) [C1]
\(c_{\text{liq}} = \frac{64\,912.5}{0.60 \times 17.5} = 6200\text{ J}/(\text{kg}\,^\circ\text{C})\) OR \(6180\text{ J}/(\text{kg}\,^\circ\text{C})\) [A1]

(a) In a gas, particles are widely separated (randomly spaced far apart) compared to liquid particles which are touching / closely packed. Gas particles move freely and rapidly in random directions, whereas liquid particles slide or roll past one another.

(b)(i) As temperature increases, the internal energy of the gas increases and the gas particles gain kinetic energy (move faster). This causes the particles to collide with the walls of the rigid container more frequently (more collisions per unit time). Furthermore, each collision involves a greater change in momentum, exerting a larger force on the wall. Since pressure is force per unit area (\(p = \frac{F}{A}\)), the pressure increases.

(ii) The mass and volume are constant, so the average separation / distance between particles remains unchanged.

(iii) Absolute zero is \(-273\ ^\circ\text{C}\) (or \(0\text{ K}\)).

(a) Any two from:
- Arrangement: gas particles are far apart / liquid particles are close together / touching [B1]
- Motion: gas particles move freely / rapidly in all directions / liquid particles slide over each other [B1]

(b)(i)
- Particles have more kinetic energy / move faster [B1]
- Collide with the walls more frequently / more collisions per second [B1]
- Greater force per collision / greater change in momentum per collision [B1]

(b)(ii)
- Stays the same / no change [B1]

(b)(iii)
- \(-273\ ^\circ\text{C}\) [B1]

(a) Particles in the liquid have a distribution of kinetic energies. The particles with the highest kinetic energy / highest speeds near the surface overcome the attractive intermolecular forces and escape into the air as vapor. As the most energetic particles leave, the mean (average) kinetic energy of the particles remaining in the liquid decreases. Because temperature is a measure of the average kinetic energy of the particles, the temperature of the remaining water falls.

(b) Increasing the temperature of the liquid / surroundings, increasing the surface area of the water, increasing air movement (draught/wind) across the surface, or decreasing humidity.

(c) In solids, particles are closely packed and touching, allowing thermal vibrations to pass quickly from particle to neighbouring particle. In gases, particles are widely spaced with large distances between them, so collisions between particles occur much less frequently, resulting in a much slower rate of energy transfer.

(a)
- Particles have a range of energies / speeds and the most energetic / fastest particles escape from the surface [B1]
- Overcoming intermolecular forces of attraction [B1]
- Average kinetic energy of remaining particles decreases (so temperature decreases) [B1]

(b) Any two valid factors [1 mark each]:
- Higher temperature
- Larger surface area
- Draught / wind / air movement
- Lower humidity / drier air [B2]

(c)
- In solids, particles are closely packed / touching so vibrations / energy transfer occurs directly to neighbours [B1]
- In gases, particles are far apart so collisions between particles are infrequent [B1]

(a)(i) Using Snell's law: \(n = \frac{\sin i}{\sin r}\)
\(\sin r = \frac{\sin 42^\circ}{1.52} = \frac{0.6691}{1.52} = 0.4402\)
\(r = \arcsin(0.4402) = 26.1^\circ\)

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

(a)(iii) Total internal reflection occurs; the ray is completely reflected back into the glass and does not refract into the air.

(b)(i) Ray 1: From the top of the object, draw a ray parallel to the principal axis which refracts through the principal focus \(F\) on the other side of the lens.
Ray 2: Draw a ray from the top of the object passing directly through the optical centre of the lens, which passes straight through undeviated.

(b)(ii) Since the object distance (\(6.0\text{ cm}\)) is between \(f\) (\(4.0\text{ cm}\)) and \(2f\) (\(8.0\text{ cm}\)), the image is: 1. Real, 2. Inverted, 3. Magnified (or enlarged).

(a)(i) C1: \(n = \frac{\sin i}{\sin r}\) OR \(\sin r = \frac{\sin 42^\circ}{1.52}\)
A1: \(26.1^\circ\) or \(26^\circ\)

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

(a)(iii) B1: Total internal reflection / ray is reflected internally (do not accept 'reflection' alone without TIR or description)

(b)(i) B1: Ray parallel to the principal axis refracts through the focal point / principal focus
B1: Ray through the optical centre passes straight through / without deviation

(b)(ii) B2: Any two from: real, inverted, magnified / enlarged / larger than the object [1 mark for each correct property, max 2]

(a)(i) Wave speed formula: \(v = f \lambda\)
Rearrange for wavelength: \(\lambda = \frac{v}{f}\)
\(\lambda = \frac{1540}{3.5 \times 10^6} = 4.4 \times 10^{-4}\text{ m}\) (or \(0.44\text{ mm}\))
Unit: \(\text{m}\) or \(\text{mm}\)

(a)(ii) Distance travelled by the pulse: \(d_{\text{total}} = v \times t = 1540 \times 4.8 \times 10^{-5} = 0.07392\text{ m}\)
Since the pulse travels to the organ and back, the depth is half the total distance:
\(\text{depth} = \frac{d_{\text{total}}}{2} = \frac{0.07392}{2} = 0.03696\text{ m} \approx 0.037\text{ m}\) (or \(3.7\text{ cm}\))

(b)(i) Speed decreases; wavelength decreases; frequency remains constant / unchanged.

(b)(ii) One side/part of the wavefront reaches the boundary and enters the shallow region first, slowing down before the remaining part of the wavefront. The differing speeds across the wavefront cause it to pivot and change direction (refract towards the normal).

(a)(i) C1: \(v = f \lambda\) OR \(\lambda = \frac{v}{f}\)
C1: \(\frac{1540}{3.5 \times 10^6}\)
A1: \(4.4 \times 10^{-4}\text{ m}\) OR \(0.44\text{ mm}\) (value and unit required for full credit)

(a)(ii) C1: \(d = v \times t\) (e.g. \(1540 \times 4.8 \times 10^{-5} = 0.07392\text{ m}\)) OR division by 2 in method
A1: \(0.037\text{ m}\) or \(0.0370\text{ m}\) / \(3.7\text{ cm}\)

(b)(i) B2: Speed decreases, wavelength decreases, frequency is unchanged (3 correct = 2 marks; 2 correct = 1 mark)

(b)(ii) B1: One end/side of the wavefront hits the shallow water first and slows down
B1: Different parts of the wave travel at different speeds across the boundary causing it to bend / rotate towards the normal

(a) (i) For two resistors in parallel:
\(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{48} + \frac{1}{72} = \frac{3 + 2}{144} = \frac{5}{144}\)
\(R = \frac{144}{5} = 28.8\ \Omega\)

(ii) Using Ohm's law:
\(I = \frac{V}{R} = \frac{240\text{ V}}{28.8\ \Omega} = 8.33\text{ A}\)
(Alternatively: \(I_1 = \frac{240}{48} = 5.0\text{ A}\), \(I_2 = \frac{240}{72} = 3.33\text{ A}\), \(I_{\text{total}} = 5.0 + 3.33 = 8.33\text{ A}\))

(iii) Power:
\(P = V \times I = 240\text{ V} \times 8.333\text{ A} = 2000\text{ W}\) (or \(2.0\text{ kW}\))
(Alternatively: \(P = \frac{V^2}{R} = \frac{240^2}{28.8} = 2000\text{ W}\))

(b) (i) If the fuse is placed in the live wire, when excessive current blows the fuse, the circuit is disconnected from the high potential/voltage supply. If the fuse were in the neutral wire, blowing the fuse would still leave the appliance live and dangerous to touch.

(ii) The earth wire provides a very low resistance path to ground. If the live wire touches the metal casing, a large current surges through the live and earth wires, melting the fuse and isolating the appliance.

(a)(i)
[C1] \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\) or \(R = \frac{R_1 R_2}{R_1 + R_2}\) with values substituted (e.g. \(\frac{48 \times 72}{48 + 72}\))
[A1] \(28.8\ \Omega\) (or \(29\ \Omega\))

(a)(ii)
[C1] \(I = \frac{V}{R}\) or \(I = I_1 + I_2\) seen with correct substitutions (e.g. \(\frac{240}{28.8}\) or \(5.0 + 3.33\))
[A1] \(8.33\text{ A}\) (or \(8.3\text{ A}\)) [Allow ecf from (a)(i)]

(a)(iii)
[C1] \(P = IV\) or \(P = \frac{V^2}{R}\) or \(P = I^2 R\) with correct substitution
[A1] \(2000\text{ W}\) / \(2.0\text{ kW}\) [Allow ecf from (a)(i)/(ii)]

(b)(i)
[B1] (When the fuse blows/melts) it disconnects/isolates the appliance from the high voltage / live supply.
[B0.5] Prevents the casing/appliance from remaining live / eliminates risk of electric shock when touched.

(b)(ii)
[B1] Causes a large / surge current to flow to earth, which blows/melts the fuse (cutting off power).

(a) (i) Total resistance of circuit: \(R_{\text{total}} = 15.0 + 45.0 = 60.0\ \Omega\)
Circuit current: \(I = \frac{V}{R_{\text{total}}} = \frac{12.0}{60.0} = 0.200\text{ A}\)
Potential difference across thermistor: \(V_{\text{therm}} = I \times R_{\text{therm}} = 0.200 \times 45.0 = 9.0\text{ V}\)
(Alternatively: \(V_{\text{therm}} = 12.0 \times \frac{45.0}{15.0 + 45.0} = 9.0\text{ V}\))

(ii) As temperature increases, the resistance of the NTC thermistor decreases.
This decreases the total circuit resistance, which increases the current in the circuit (or reduces the share of voltage across the thermistor).
Therefore, the potential difference across the fixed resistor increases (since \(V = IR\)).

(b) (i) When current flows through the resistance of the cable, thermal energy is generated (\(P = I^2 R\)). When the cable is coiled tightly, heat transfer/dissipation to the surroundings is restricted. The temperature of the cable rises significantly, which can melt the plastic insulation and cause a short-circuit or fire.

(ii) Double insulation means all electrical components are insulated and the casing/outer body is made of an electrical insulator (e.g. plastic) with no exposed metal parts. Since the casing is non-conductive, even if an internal wire becomes loose, the casing cannot become live, so an earth wire is not required to prevent electric shocks.

(a)(i)
[C1] Total resistance \(R = 60.0\ \Omega\) OR potential divider ratio formula \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{therm}}}{R_1 + R_{\text{therm}}}\)
[C0.5] Substitution of values (e.g. \(12.0 \times \frac{45.0}{60.0}\) or \(I = \frac{12.0}{60.0} = 0.20\text{ A}\))
[A1] \(9.0\text{ V}\) (or \(9.00\text{ V}\))

(a)(ii)
[B1] Resistance of thermistor decreases (so total resistance decreases / current increases)
[B1] Potential difference across the fixed resistor increases

(b)(i)
[B1] Current produces heat / thermal energy in the cable (due to resistance / \(I^2 R\))
[B1] Heat cannot easily escape / dissipate when tightly coiled, leading to overheating / melting insulation / fire risk

(b)(ii)
[B1] (Double insulation means) appliance has an insulating / plastic outer casing (or two independent layers of insulation)
[B1] No exposed metal parts can become live, so user cannot get an electric shock (making an earth wire unnecessary)

(a)(i) Balancing nucleon numbers: \(2 + 2 = 3 + A \Rightarrow A = 1\).
Balancing proton numbers: \(1 + 1 = 2 + Z \Rightarrow Z = 0\).
A particle with nucleon number 1 and proton number 0 is a neutron (\(^{1}_{0}\text{n}\)).

(a)(ii) During nuclear fusion, the total mass of the products is less than the total mass of the reactants (mass decreases). This lost mass is converted into energy.

(b)(i) The increase in observed wavelength of light from receding galaxies is called redshift.

(b)(ii) Using the Hubble equation:
\[v = H_0 d\]
\[d = \frac{v}{H_0} = \frac{1.43 \times 10^{7}\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 6.50 \times 10^{24}\text{ m}\]

(b)(iii) Distance in light-years:
\[d = \frac{6.50 \times 10^{24}\text{ m}}{9.5 \times 10^{15}\text{ m/ly}} = 6.84 \times 10^{8}\text{ light-years}\] (accept \(6.8 \times 10^{8}\) or \(6.84 \times 10^{8}\)).

(a)(i)
- A = 1 and Z = 0 [B1]
- neutron [B1]

(a)(ii)
- (total) mass decreases / mass of products is less than mass of reactants [B1]
- (mass difference) is converted into energy [B1]

(b)(i)
- redshift / cosmological redshift [B1]

(b)(ii)
- d = v / H0 OR 1.43 x 10^7 / 2.2 x 10^-18 [C1]
- 6.5 x 10^24 (m) [A1]

(b)(iii)
- (candidate's (b)(ii)) / (9.5 x 10^15) [C1]
- 6.8 x 10^8 / 6.84 x 10^8 (light-years) [A1]