2024 Cambridge IGCSE Physics (0625) Practice Paper with Answers

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

(b) Using the acceleration formula:
\(a = \frac{v - u}{t} = \frac{1.1\text{ m/s} - 3.5\text{ m/s}}{6.0\text{ s}} = -0.40\text{ m/s}^2\)
Thus, the deceleration is \(0.40\text{ m/s}^2\).

(c) The total distance descended is the area under the speed-time graph from \(t = 0\) to \(t = 10.0\text{ s}\):
- For the first \(6.0\text{ s}\) (trapezium area): \(s_1 = \frac{3.5 + 1.1}{2} \times 6.0 = 13.8\text{ m}\)
- For the remaining \(4.0\text{ s}\) (constant speed): \(s_2 = 1.1 \times 4.0 = 4.4\text{ m}\)
Total distance descended = \(13.8\text{ m} + 4.4\text{ m} = 18.2\text{ m}\).

(a) B1: Rate of change of velocity OR change in velocity per unit time. [1.0 mark]
B1: Mention of direction/vector nature or correct unit context if fully defined. [1.0 mark]

(b) C1: \(a = \frac{v - u}{t}\) or substitution \(\frac{1.1 - 3.5}{6.0}\). [1.0 mark]
A1: Deceleration = \(0.40\text{ m/s}^2\) (accept \(-0.40\text{ m/s}^2\) if clear context). [1.0 mark]

(c) C1: Calculation of distance in first phase: \(13.8\text{ m}\) (area of trapezium/integration). [1.0 mark]
C1: Calculation of distance in second phase: \(1.1 \times 4.0 = 4.4\text{ m}\). [1.0 mark]
A1.27: Total distance of \(18.2\text{ m}\) with correct unit. [1.27 marks]

(a) Using the transformer equation:
\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
\(\frac{690}{33\,000} = \frac{120}{N_s}\)
\(N_s = \frac{120 \times 33\,000}{690} \approx 5739\text{ turns}\)

(b) The material used for the core is soft iron. Soft iron is chosen because it is easily magnetised and demagnetised, which minimises energy losses due to alternating magnetic fields.

(c) Using the electrical power formula:
\(P = V_p \times I_p\)
\(2.4 \times 10^6\text{ W} = 690\text{ V} \times I_p\)
\(I_p = \frac{2.4 \times 10^6}{690} \approx 3478\text{ A}\) (or \(3.48\text{ kA}\)).

(a) C1: Recall of \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). [1.0 mark]
C1: Correct substitution: \(\frac{690}{33\,000} = \frac{120}{N_s}\). [1.0 mark]
A1: \(5739\) (accept range \(5730 - 5750\) due to rounding variations). [1.0 mark]

(b) B1: Soft iron. [1.0 mark]
B1: Easily magnetised/demagnetised (or reduces hysteresis/energy loss). [1.0 mark]

(c) C1: Recall of \(P = V \times I\) or substitution of \(2.4 \times 10^6\text{ W}\). [1.0 mark]
A1.27: \(3478\text{ A}\) (accept \(3500\text{ A}\) or \(3.5\text{ kA}\)) with correct unit. [1.27 marks]

(a) Using the formula for the critical angle:
\(\sin c = \frac{1}{n}\)
\(\sin c = \frac{1}{1.48} \approx 0.6757\)
\(c = \arcsin(0.6757) \approx 42.5^\circ\)

(b) Since the angle of incidence (\(40^\circ\)) is less than the critical angle (\(42.5^\circ\)), the light ray will be refracted out into the air, bending away from the normal. A faint partially reflected ray will also be observed inside the liquid.

(c) The two conditions required for total internal reflection are:
1. The light must be travelling from an optically denser medium to an optically less dense medium.
2. The angle of incidence must be greater than the critical angle.

(a) C1: Recall of \(\sin c = \frac{1}{n}\). [1.0 mark]
C1: Correct substitution: \(\sin c = \frac{1}{1.48}\). [1.0 mark]
A1: \(42.5^\circ\) (accept \(43^\circ\)). [1.0 mark]

(b) B1: Light is refracted out into the air. [1.0 mark]
B1: Bends away from the normal (since \(i < c\)). [1.0 mark]

(c) B1: Light travels from denser to less dense medium. [1.0 mark]
B1.27: Angle of incidence is greater than critical angle. [1.27 marks]

(a) This shift to a longer wavelength is called redshift. It indicates that the galaxy is moving away from the observer (Earth).

(b) This supports the Big Bang theory.

(c) CMBR is electromagnetic radiation in the microwave region received from all parts of the sky. It represents the remnant heat/radiation left over from the Big Bang. As the Universe expanded, this high-energy radiation from the early Universe was stretched to longer wavelengths (microwaves). Its highly uniform presence across space supports the Big Bang theory by showing that the Universe expanded from a single extremely hot and dense initial state.

(a) B1: Redshift. [1.0 mark]
B1: The galaxy is moving away (from Earth). [1.0 mark]

(b) B1: Big Bang theory. [1.0 mark]

(c) B1: Definition of CMBR as microwave radiation from all directions. [1.0 mark]
B1: Remnant radiation/heat from the early hot phase of the Big Bang. [1.0 mark]
B1: Stretched wavelengths due to the expansion of space. [1.0 mark]
B1.27: Highly uniform distribution supports expansion from a single point. [1.27 marks]

(a) Total distance travelled through the pillar and back = \(2 \times 0.80\text{ m} = 1.60\text{ m}\).
Using \(t = \frac{d}{v}\):
\(t = \frac{1.60\text{ m}}{3200\text{ m/s}} = 5.0 \times 10^{-4}\text{ s}\) (or \(0.50\text{ ms}\)).

(b) Ultrasound is sound waves with a frequency higher than \(20\,000\text{ Hz}\) (or \(20\text{ kHz}\)), which is above the limit of human hearing.

(c) Since \(0.35\text{ ms}\) is less than the expected \(0.50\text{ ms}\) return time, this indicates a crack is present.
Distance to the crack and back: \(d = v \times t = 3200 \times 3.5 \times 10^{-4} = 1.12\text{ m}\).
Depth of crack = \(\frac{1.12}{2} = 0.56\text{ m}\).

(a) C1: Recall of \(d = 2 \times 0.80\). [1.0 mark]
C1: Speed-distance-time rearrangement: \(t = \frac{d}{v}\). [1.0 mark]
A1: \(5.0 \times 10^{-4}\text{ s}\) (or \(0.50\text{ ms}\)) with correct unit. [1.0 mark]

(b) B1: Frequency greater than \(20\,000\text{ Hz}\) / \(20\text{ kHz}\). [1.0 mark]
B1: Beyond/above human hearing limit. [1.0 mark]

(c) B1: Explanation that \(0.35\text{ ms} < 0.50\text{ ms}\) confirms a crack. [1.0 mark]
A1.27: Correct calculation of crack depth = \(0.56\text{ m}\) with correct unit. [1.27 marks]

(a) The work done is equal to the gain in gravitational potential energy:
\(W = m \times g \times h = 450\text{ kg} \times 9.8\text{ N/kg} \times 8.0\text{ m} = 35\,280\text{ J}\) (or \(35.3\text{ kJ}\)).

(b) Useful power output:
\(P_{\text{useful}} = \frac{\text{Work Done}}{\text{time}} = \frac{35\,280\text{ J}}{30\text{ s}} = 1176\text{ W}\) (or \(1.18\text{ kW}\)).

(c) Total power input = \(1.5\text{ kW} = 1500\text{ W}\).
\(\text{Efficiency} = \frac{P_{\text{useful}}}{P_{\text{input}}} \times 100\% = \frac{1176}{1500} \times 100\% = 78.4\%\).

(a) C1: Recall of \(W = mgh\). [1.0 mark]
C1: Correct substitution of values with \(g = 9.8\). [1.0 mark]
A1: \(35\,280\text{ J}\) (or \(35.3\text{ kJ}\)). [1.0 mark]

(b) C1: \(P = \frac{W}{t}\) or substitution \(\frac{35\,280}{30}\). [1.0 mark]
A1: \(1176\text{ W}\) (accept \(1180\text{ W}\) or \(1.18\text{ kW}\)). [1.0 mark]

(c) C1: Correct matching of units (e.g. converting input power to \(1500\text{ W}\)). [1.0 mark]
A1.27: \(78.4\%\) (accept \(78.7\%\) if using \(1180\text{ W}\)). [1.27 marks]

(a) The two conditions for equilibrium are:
1. The resultant force acting on the object is zero.
2. The resultant moment acting on the object is zero.

(b) The diver's weight acts at end B, which is \(4.0\text{ m}\) from pivot A.
\(\text{Moment} = \text{Force} \times \text{distance} = 680\text{ N} \times 4.0\text{ m} = 2720\text{ Nm}\) (clockwise).

(c) Taking moments about pivot A:
- Clockwise moments: (Weight of board \(\times 2.0\text{ m}\)) + (Weight of diver \(\times 4.0\text{ m}\))
\(\text{Clockwise moment} = (350 \times 2.0) + (680 \times 4.0) = 700 + 2720 = 3420\text{ Nm}\).
- Anticlockwise moment: (Force of pillar \(F_s \times 1.2\text{ m}\)).
For equilibrium: \(F_s \times 1.2 = 3420\)
\(F_s = \frac{3420}{1.2} = 2850\text{ N}\).

(a) B1: Resultant/net force is zero. [1.0 mark]
B1: Resultant/net moment is zero. [1.0 mark]

(b) C1: Moment = \(F \times d\) with \(d = 4.0\text{ m}\). [1.0 mark]
A1: \(2720\text{ Nm}\). [1.0 mark]

(c) C1: Account for the uniform weight of the board acting at center \(2.0\text{ m}\). [1.0 mark]
C1: Equation for sum of clockwise moments = \(3420\text{ Nm}\). [1.0 mark]
A1.27: Correct value of \(2850\text{ N}\) upward with unit. [1.27 marks]

(a) The decay equation is:
\(^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}\text{e}\)

(b) During beta-minus decay, a neutron inside the nucleus decays into a proton and an electron. Therefore:
- The proton number increases by 1.
- The neutron number decreases by 1.
(The mass number remains unchanged.)

(c) Number of half-lives elapsed:
\(n = \frac{17\,190}{5730} = 3\text{ half-lives}\).
Mass remaining:
\(m = 8.0\text{ \mu g} \times \left(\frac{1}{2}\right)^3 = 8.0 \times \frac{1}{8} = 1.0\text{ \mu g}\).

(a) B1: Correct symbols for reactants and products (C, N). [1.0 mark]
B1: Correct symbol and numbers for beta-particle (\(^{0}_{-1}\text{e}\) or \(^{0}_{-1}\beta\)). [1.0 mark]
B1: Equation balanced on both sides for nucleon and proton numbers. [1.0 mark]

(b) B1: Proton number increases by 1. [1.0 mark]
B1: Neutron number decreases by 1. [1.0 mark]

(c) C1: Identification of 3 half-lives. [1.0 mark]
A1.27: \(1.0\text{ \mu g}\) (or \(1.0 \times 10^{-6}\text{ g}\)) with correct unit. [1.27 marks]

(a) Momentum is defined as the product of mass and velocity (p = mv). (b) Initial momentum = m_A * u_A + m_B * u_B = 0.40 kg * 2.5 m/s + 0 = 1.0 kg m/s. Since total momentum is conserved, final velocity v = (total initial momentum) / (m_A + m_B) = 1.0 kg m/s / (0.40 kg + 0.60 kg) = 1.0 m/s. (c) Average force F = change in momentum of car B / time = (m_B * v - 0) / t = (0.60 kg * 1.0 m/s) / 0.15 s = 4.0 N.

(a) Product of mass and velocity [1] (b) Use of conservation of momentum: m_A * u_A = (m_A + m_B) * v [1]; Substitution: 0.40 * 2.5 = (0.40 + 0.60) * v [1]; v = 1.0 m/s (with unit) [1] (c) Use of F = change in momentum / time [1]; Change in momentum of B = 0.60 kg m/s [1]; F = 4.0 N (with unit) [1]

(a)(i) Monochromatic light consists of light of a single frequency or wavelength. (ii) Using Snell's Law: n = sin(i) / sin(r) -> sin(r) = sin(i) / n = sin(40) / 1.52 = 0.6428 / 1.52 = 0.4229. Thus, r = arcsin(0.4229) = 25.0 degrees. (b) Critical angle c is found using: sin(c) = 1 / n = 1 / 1.52 = 0.6579. Thus, c = arcsin(0.6579) = 41.1 degrees.

(a)(i) Light of a single frequency / wavelength [1] (ii) Use of n = sin(i) / sin(r) [1]; Substitution: sin(r) = sin(40) / 1.52 [1]; r = 25 degrees (accept 25.0) [1] (b) Use of sin(c) = 1 / n [1]; Substitution: sin(c) = 1 / 1.52 [1]; c = 41 degrees (accept 41.1) [1]

(a) Carbon-14 has a proton number (Z) of 6 and a nucleon number (A) of 14. Therefore, it has 6 protons, and 14 - 6 = 8 neutrons. (b) Nitrogen-14 has nucleon number 14 and proton number 7. The emitted beta-particle is represented as an electron with nucleon number 0 and proton number -1. (c) Since beta-particles are fast-moving electrons, they carry a negative charge. In an electric field between oppositely charged plates, negative charges are attracted towards the positive plate and repelled by the negative plate, so the beam deflects towards the positive plate.

(a) 6 protons [1]; 8 neutrons [1] (b) Nitrogen nucleon number = 14 [1]; Nitrogen proton number = 7 [1]; Beta-particle symbol with nucleon number 0 and proton number -1 [1] (c) Deflected towards the positive plate [1]; Because beta-particles are negatively charged and opposite charges attract / like charges repel [1]