An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V2) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
Section Extended Theory Core & Supplement
Answer all questions. Use a calculator and take the weight of 1.0 kg to be 9.8 N.
10 Question · 80 marks
Question 1 · Calculations & Explanations
8 marks
A toy car of mass 0.50 kg is released from rest at the top of a ramp. It accelerates uniformly down the ramp at \(1.6 \text{ m/s}^{2}\) for 2.5 s. (a) (i) Calculate the velocity of the toy car at 2.5 s. (ii) Calculate the distance travelled by the toy car down the ramp during this time. (b) At the bottom of the ramp, the toy car travels onto a flat horizontal surface. A constant friction force acts on the car, bringing it to rest in a distance of 8.0 m. (i) Calculate the deceleration of the toy car as it comes to rest. (ii) State and explain the direction of the horizontal resultant force on the toy car as it decelerates.
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Worked solution
(a) (i) Using \(v = u + at\), where \(u = 0\), \(v = 1.6 \times 2.5 = 4.0 \text{ m/s}\). (ii) Using \(s = ut + \frac{1}{2}at^2\), \(s = 0 + 0.5 \times 1.6 \times (2.5)^2 = 5.0 \text{ m}\). (b) (i) Using \(v^2 = u^2 + 2as\), where \(v = 0\), \(u = 4.0 \text{ m/s}\), and \(s = 8.0 \text{ m}\): \(0 = 4.0^2 + 2 \times a \times 8.0 \Rightarrow 16a = -16 \Rightarrow a = -1.0 \text{ m/s}^2\). The deceleration is \(1.0 \text{ m/s}^2\). (ii) The resultant force acts in the opposite direction to the motion (backwards) because the only horizontal force is friction, which opposes the direction of travel.
Marking scheme
(a) (i) C1: \(v = u + at\) or correct substitution. A1: \(4.0 \text{ m/s}\). (ii) C1: \(s = \frac{1}{2}at^2\) or correct substitution. A1: \(5.0 \text{ m}\). (b) (i) C1: \(v^2 = u^2 + 2as\) or correct substitution. A1: \(1.0 \text{ m/s}^2\). (ii) B1: Direction is opposite to motion / backwards. B1: Force is due to friction / resistive force.
Question 2 · Calculations & Explanations
8 marks
A block P of mass 2.0 kg is moving to the right with a velocity of \(6.0 \text{ m/s}\). It collides head-on with a stationary block Q of mass 4.0 kg. After the collision, block P recoils to the left with a velocity of \(2.0 \text{ m/s}\). (a) (i) Define momentum in terms of mass and velocity. (ii) Calculate the velocity of block Q after the collision. (iii) Calculate the impulse exerted on block Q during the collision. (b) The collision lasts for a duration of 0.050 s. Calculate the average force exerted on block Q.
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Worked solution
(a) (i) Momentum is the product of mass and velocity (\(p = mv\)). (ii) By conservation of momentum, total initial momentum = total final momentum. \(2.0 \times 6.0 + 4.0 \times 0 = 2.0 \times (-2.0) + 4.0 \times v_Q \Rightarrow 12 = -4.0 + 4.0 v_Q \Rightarrow 16 = 4.0 v_Q \Rightarrow v_Q = 4.0 \text{ m/s}\) (to the right). (iii) Impulse = change in momentum of Q = \(m_Q v_Q - m_Q u_Q = 4.0 \times 4.0 - 0 = 16 \text{ N s}\) (or \(16 \text{ kg m/s}\)). (b) Average force \(F = \frac{\text{Impulse}}{t} = \frac{16}{0.050} = 320 \text{ N}\).
Marking scheme
(a) (i) B1: Mass \(\times\) velocity. (ii) C1: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\) or calculation of initial momentum (12). C1: substitution with correct signs (e.g. -4.0). A1: \(4.0 \text{ m/s}\). (iii) C1: \(m \Delta v\) used. A1: \(16 \text{ N s}\) or \(16 \text{ kg m/s}\). (b) C1: \(F = \Delta p / t\) or substitution. A1: \(320 \text{ N}\).
Question 3 · Calculations & Explanations
8 marks
A research submarine has a circular viewing window of radius 0.15 m. The submarine is submerged in seawater of density \(1025 \text{ kg/m}^{3}\) at a depth of 80 m. The gravitational field strength \(g\) is \(9.8 \text{ N/kg}\). (a) (i) Calculate the pressure due to the seawater at this depth. (ii) Calculate the force exerted by the seawater on the circular window. (b) State how the total pressure on the submarine's window differs from the pressure due to the seawater alone, and explain your answer.
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Worked solution
(a) (i) Using \(P = \rho g h\), \(P = 1025 \times 9.8 \times 80 = 803,600 \text{ Pa}\) (or \(8.0 \times 10^5 \text{ Pa}\)). (ii) Area of window \(A = \pi r^2 = \pi \times (0.15)^2 \approx 0.0707 \text{ m}^2\). Force \(F = P \times A = 803,600 \times 0.0707 \approx 56,800 \text{ N}\) (or \(5.7 \times 10^4 \text{ N}\)). (b) The total pressure is higher than the seawater pressure alone because the atmosphere also exerts pressure on the surface of the water, which is transmitted through the liquid.
Marking scheme
(a) (i) C1: \(P = \rho g h\). C1: substitution: \(1025 \times 9.8 \times 80\). A1: \(8.0 \times 10^5 \text{ Pa}\) (or \(800,000 \text{ Pa}\) / \(803,600 \text{ Pa}\)). (ii) C1: \(A = \pi r^2\) or calculation of area \(0.0707 \text{ m}^2\). C1: \(F = P \times A\). A1: \(5.7 \times 10^4 \text{ N}\) (or \(56,800 \text{ N}\) to \(57,000 \text{ N}\)). (b) B1: The total pressure is greater. B1: Atmospheric pressure acts on the surface of the water.
Question 4 · Calculations & Explanations
8 marks
A cylinder contains \(400 \text{ cm}^{3}\) of air at a pressure of \(1.2 \times 10^{5} \text{ Pa}\). (a) (i) The piston is pushed in slowly so that the volume is reduced to \(150 \text{ cm}^{3}\), while the temperature remains constant. Calculate the new pressure of the gas. (ii) State and explain, in terms of particles, why the pressure increases when the volume is decreased at a constant temperature. (b) The cylinder is now heated. State the effect of this heating on: (i) the average kinetic energy of the gas particles, (ii) the average speed of the gas particles.
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Worked solution
(a) (i) Using Boyle's law \(p_1 V_1 = p_2 V_2\): \(1.2 \times 10^5 \times 400 = p_2 \times 150 \Rightarrow p_2 = \frac{4.8 \times 10^7}{150} = 3.2 \times 10^5 \text{ Pa}\). (ii) Decreasing the volume means the particles are closer together. This leads to more frequent collisions of the particles with the cylinder walls. Since pressure is the total force per unit area, the increased rate of collisions results in a higher pressure. (b) (i) The average kinetic energy increases. (ii) The average speed of the particles increases.
Marking scheme
(a) (i) C1: \(p_1 V_1 = p_2 V_2\) in any form. C1: correct substitution. A1: \(3.2 \times 10^5 \text{ Pa}\). (ii) B1: Particles are closer together / volume of container is smaller. B1: Rate of collisions / frequency of collisions with walls increases. B1: Average force per unit area increases. (b) (i) B1: Increases / greater. (ii) B1: Increases / greater.
Question 5 · Calculations & Explanations
8 marks
An electric heater rated at 150 W is used to heat a block of aluminium of mass 0.80 kg. The heater is switched on for 4.0 minutes. (a) (i) Calculate the thermal energy supplied by the heater in this time. (ii) The temperature of the aluminium block rises from \(22^{\circ}\text{C}\) to \(68^{\circ}\text{C}\). Calculate the experimental value for the specific heat capacity of aluminium. (b) (i) Suggest a reason why the experimental value is higher than the actual specific heat capacity of aluminium. (ii) State one modification to the experiment that would reduce this difference. (iii) Define melting point.
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Worked solution
(a) (i) Energy \(E = P \times t = 150 \times (4.0 \times 60) = 36,000 \text{ J}\). (ii) Temperature change \(\Delta \theta = 68 - 22 = 46^{\circ}\text{C}\). Using \(E = m c \Delta \theta\): \(36,000 = 0.80 \times c \times 46 \Rightarrow 36,000 = 36.8 c \Rightarrow c \approx 978 \text{ J/(kg }^{\circ}\text{C)}\) (or \(980 \text{ J/(kg }^{\circ}\text{C)}\) to 2 sig figs). (b) (i) Thermal energy is lost to the surroundings/air, meaning more energy was supplied than was absorbed by the block. (ii) Wrap insulation around the aluminium block. (iii) The temperature at which a substance changes state from solid to liquid.
Marking scheme
(a) (i) C1: \(E = P \times t\) with correct time in seconds. A1: \(36,000 \text{ J}\). (ii) C1: \(\Delta \theta = 46\). C1: \(c = E / (m \Delta \theta)\) or correct substitution. A1: \(980 \text{ J/(kg }^{\circ}\text{C)}\) (accept \(978\)). (b) (i) B1: Heat lost to surroundings. (ii) B1: Insulate the block / paint block silver. (iii) B1: Temperature at which solid turns to liquid (at constant temperature).
Question 6 · Calculations & Explanations
8 marks
A ray of light is incident on the flat surface of a semi-circular glass block at an angle of incidence of \(42^{\circ}\). The refractive index of the glass is 1.52. (a) (i) Calculate the angle of refraction of the light ray as it enters the glass. (ii) Calculate the speed of light in this glass block. The speed of light in air is \(3.0 \times 10^{8} \text{ m/s}\). (b) (i) Calculate the critical angle for the glass-air boundary. (ii) State what happens to a ray of light traveling inside the glass that meets the boundary with air at an angle of incidence of \(45^{\circ}\).
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Worked solution
(a) (i) Using Snell's law \(n = \frac{\sin i}{\sin r}\): \(1.52 = \frac{\sin 42^{\circ}}{\sin r} \Rightarrow \sin r = \frac{\sin 42^{\circ}}{1.52} \approx \frac{0.6691}{1.52} \approx 0.4402 \Rightarrow r \approx 26.1^{\circ}\) (or \(26^{\circ}\)). (ii) Speed of light \(v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.52} \approx 1.97 \times 10^8 \text{ m/s}\) (or \(2.0 \times 10^8 \text{ m/s}\)). (b) (i) Using \(\sin c = \frac{1}{n}\): \(\sin c = \frac{1}{1.52} \approx 0.6579 \Rightarrow c \approx 41.1^{\circ}\) (or \(41^{\circ}\)). (ii) The angle of incidence (\(45^{\circ}\)) is greater than the critical angle (\(41^{\circ}\)), so total internal reflection occurs.
Marking scheme
(a) (i) C1: \(n = \sin i / \sin r\). C1: correct substitution. A1: \(26^{\circ}\) (accept \(26.1^{\circ}\)). (ii) C1: \(v = c / n\). A1: \(2.0 \times 10^8 \text{ m/s}\) (accept \(1.97 \times 10^8 \text{ m/s}\)). (b) (i) C1: \(\sin c = 1 / n\). A1: \(41^{\circ}\) (accept \(41.1^{\circ}\)). (ii) B1: Total internal reflection.
Question 7 · Calculations & Explanations
8 marks
A battery of e.m.f. 12 V and negligible internal resistance is connected to three resistors. Resistor \(R_1 = 4.0 \text{ }\Omega\) is in series with a parallel combination of \(R_2 = 6.0 \text{ }\Omega\) and \(R_3 = 3.0 \text{ }\Omega\). (a) (i) Calculate the combined resistance of the parallel pair \(R_2\) and \(R_3\). (ii) Calculate the total resistance of the entire circuit. (iii) Calculate the total current in the circuit. (iv) Calculate the potential difference across the parallel pair. (b) A student connects a voltmeter across resistor \(R_1\). State the reading on the voltmeter.
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Worked solution
(a) (i) For parallel resistors: \(\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{3}{6.0} = \frac{1}{2.0} \Rightarrow R_p = 2.0 \text{ }\Omega\). (ii) Total resistance \(R_t = R_1 + R_p = 4.0 + 2.0 = 6.0 \text{ }\Omega\). (iii) Total current \(I = \frac{V}{R_t} = \frac{12}{6.0} = 2.0 \text{ A}\). (iv) Potential difference across parallel pair \(V_p = I \times R_p = 2.0 \times 2.0 = 4.0 \text{ V}\). (b) Reading on voltmeter across \(R_1\): \(V_1 = I \times R_1 = 2.0 \times 4.0 = 8.0 \text{ V}\) (or \(12 - 4.0 = 8.0 \text{ V}\)).
Marking scheme
(a) (i) C1: \(1/R_p = 1/R_1 + 1/R_2\) formula or correct substitution. A1: \(2.0 \text{ }\Omega\). (ii) A1: \(6.0 \text{ }\Omega\). (iii) C1: \(I = V / R\). A1: \(2.0 \text{ A}\). (iv) C1: \(V = I \times R_p\) or potential divider equation. A1: \(4.0 \text{ V}\). (b) A1: \(8.0 \text{ V}\).
Question 8 · Calculations & Explanations
8 marks
A sample containing a radioactive isotope has an initial activity due to the source alone of 800 counts/s. The background count rate in the laboratory is constant at 40 counts/s. (a) (i) Calculate the initial reading on the detector (which includes background). (ii) The half-life of the isotope is 12 hours. Calculate the activity of the source alone after 36 hours. (iii) Calculate the reading on the detector after 36 hours. (b) The isotope decays by emitting beta-minus (\(\beta^-\)) particles to form a stable nucleus. (i) State the nature of a \(\beta^-\)_particle. (ii) Explain the change that occurs in the nucleus during \(\beta^-\)_decay. (iii) State which of the following materials is most suitable to absorb the majority of \(\beta^-\)_radiation: a sheet of paper, a thin sheet of aluminium, or several centimetres of lead.
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Worked solution
(a) (i) Initial reading = source activity + background = \(800 + 40 = 840 \text{ counts/s}\). (ii) 36 hours represents \(36 / 12 = 3\) half-lives. Source activity after 3 half-lives = \(800 \times (0.5)^3 = 800 \times 0.125 = 100 \text{ counts/s}\). (iii) Detector reading = source activity + background = \(100 + 40 = 140 \text{ counts/s}\). (b) (i) A \(\beta^-\)_particle is a high-speed electron. (ii) During beta-minus decay, a neutron in the nucleus decays into a proton (which remains in the nucleus) and an electron (which is emitted). (iii) A thin sheet of aluminium is the most suitable absorber.
Marking scheme
(a) (i) A1: \(840 \text{ counts/s}\). (ii) C1: 3 half-lives identified. A1: \(100 \text{ counts/s}\). (iii) A1: \(140 \text{ counts/s}\) (accept ecf from (a)(ii)). (b) (i) B1: Electron. (ii) B1: Neutron changes into a proton. B1: Electron/beta particle is emitted. (iii) B1: Thin sheet of aluminium.
Question 9 · Calculations & Explanations
8 marks
An electric heater rated at 60 W is placed inside a well-insulated copper beaker containing 0.15 kg of a liquid. The heater is switched on for 5.0 minutes, causing the temperature of the liquid to increase from 18 °C to 42 °C.
(a) Define the term specific heat capacity. [2]
(b) (i) Calculate the thermal energy supplied by the heater in 5.0 minutes. [2] (ii) Calculate the specific heat capacity of the liquid, assuming all the energy from the heater is transferred to the liquid. [2]
(c) Suggest and explain one reason why the actual specific heat capacity of the liquid might be lower than the value calculated in (b)(ii). [2]
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Worked solution
(a) Specific heat capacity is defined as the thermal energy required per unit mass of a substance to raise its temperature by 1 °C (or 1 K).
(b) (i) Use the formula E = P * t t = 5.0 minutes = 5.0 * 60 s = 300 s E = 60 W * 300 s = 18000 J (or 18 kJ)
(ii) Use the formula E = m * c * Δθ Δθ = 42 °C - 18 °C = 24 °C 18000 = 0.15 * c * 24 18000 = 3.6 * c c = 18000 / 3.6 = 5000 J/(kg °C)
(c) In reality, some thermal energy from the heater is transferred to the copper beaker or lost to the surroundings instead of only heating the liquid. Therefore, the actual energy absorbed by the liquid is less than 18000 J, which means the true specific heat capacity of the liquid is lower than the value calculated.
Marking scheme
(a) - energy required to raise the temperature of unit mass (1 kg) by 1 °C / 1 K [1] - formula c = ΔE / (m * Δθ) with all terms defined [1]
(b)(i) - use of t = 300 s [1] - E = 18000 J or 18 kJ [1]
(b)(ii) - Δθ = 24 °C [1] - c = 5000 J/(kg °C) (unit required) [1]
(c) - suggestion: thermal energy is absorbed by the beaker / lost to surroundings [1] - explanation: energy actually gained by liquid is less than 18000 J, resulting in a lower actual value for c [1]
Question 10 · Calculations & Explanations
8 marks
A small glider of mass 0.50 kg travels at a constant speed of 2.4 m/s along a horizontal, frictionless air track. It collides with a second, stationary glider of mass 0.30 kg. After the collision, the two gliders stick together and move with a common velocity v.
(a) State what is meant by a frictionless track in terms of resultant external forces. [1]
(b) Calculate: (i) the initial momentum of the 0.50 kg glider, [2] (ii) the common velocity v of the two gliders after the collision. [3]
(c) Explain, without further calculation, how the total kinetic energy of the gliders changes during this collision. [2]
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Worked solution
(a) A frictionless track means there are no resistive forces opposing motion, so there is zero resultant external force acting on the gliders along the horizontal direction of motion.
(b) (i) Momentum p = m * u p = 0.50 kg * 2.4 m/s = 1.2 kg m/s (or N s)
(ii) By conservation of momentum: Total momentum before collision = Total momentum after collision m1 * u1 + m2 * u2 = (m1 + m2) * v 1.2 + 0 = (0.50 + 0.30) * v 1.2 = 0.80 * v v = 1.2 / 0.80 = 1.5 m/s
(c) The total kinetic energy of the gliders decreases (energy is lost). This is because the collision is inelastic (the gliders stick together), and some of the initial kinetic energy is transferred/converted to thermal energy (heat) and sound during the impact.
Marking scheme
(a) - Zero resultant/net external force acting along the line of motion [1]
(b)(i) - formula p = m * v [1] - 1.2 kg m/s (or N s) [1]
(b)(ii) - statement of conservation of momentum or equating initial and final momentum [1] - 1.2 = 0.80 * v [1] - v = 1.5 m/s (unit required) [1]
(c) - kinetic energy decreases / is lost [1] - collision is inelastic / energy converted to thermal energy or sound [1]
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