Sara Bentley
Cambridge IGCSE™ PHYSICS (0625/42)
A student cycles from home to school. Figure 1.1 shows the distance-time graph for this journey.
(a) State the distance between home and school.
[1 Mark]
distance = 1.7 km [B1]
(b) State one point, A, B, C, D or E which identifies a time when:
[2 Marks]
1. the student accelerates
Point: B[B1]
2. the student decelerates
Point: C or D [B1]
(c) Use Figure 1.1 to calculate the student’s maximum speed in km/h.
[3 Marks]
Gradient of section BC = (1.25 – 0.4) / (7.5 – 5.0) = 0.85 / 2.5 = 0.34 km/min [C1]
Convert km/min to km/h: 0.34 × 60 = 20.4 km/h [C1]
maximum speed = 20 km/h [A3]
Figure 2.1 shows a ship made of steel floating on the sea.
(a) (i) State the equation that links density, mass and volume.
[1 Mark]
density = mass / volume OR ρ = m / V [B1]
(a) (ii) The density of steel is 7900 kg/m³. The mass of a solid steel cube is 110 kg. Calculate the volume of the steel cube.
[1 Mark]
V = m / ρ = 110 / 7900 = 0.0139 m³
volume = 0.014 m³ [B1]
(a) (iii) The density of sea water is 1030 kg/m³. State why the solid steel cube sinks when placed in sea water.
[1 Mark]
The steel is denser than sea water (1030 kg/m³ < 7900 kg/m³). [B1]
(a) (iv) Explain why the steel ship floats on the sea water.
[2 Marks]
1. The ship is not solid steel because the overall volume includes air which is much less dense than sea water. [B1]
2. The average density of the ship is less than the density of sea water. [B1]
(b) (i) The mass of the ship is 3.5 × 10⁷ kg. The ship accelerates at 0.75 m/s². Calculate the force which causes this acceleration.
[2 Marks]
F = m × a = 3.5 × 10⁷ × 0.75 [C1]
force = 2.6 × 10⁷ N [A2]
(b) (ii) Explain why the force provided by the ship’s engine must be larger than the value in 2(b)(i).
[2 Marks]
1. There is a resistive force (drag / water resistance) opposing motion. [B1]
2. The calculated force in (b)(i) is the resultant force, so engine force must overcome resistive forces. [B1]
Figure 3.1 shows a polar bear and a person standing on thick ice.
| Property | Polar Bear | Person |
|---|---|---|
| Mass / kg | 590 | 84 |
| Area in contact with ice / m² | 1.12 | 0.020 |
(a) (i) Calculate the weight of the polar bear.
[1 Mark]
Weight = m × g = 590 × 9.8 = 5782 N
weight = 5800 N [B1]
(a) (ii) Calculate the pressure that the polar bear exerts on the ice.
[2 Marks]
P = F / A = 5800 / 1.12 = 5178.5 Pa [C1]
pressure = 5200 Pa [A2]
(a) (iii) Use numbers [0.125, 7.0, 8.0, 56, 392] to complete the equations:
[2 Marks]
mass of polar bear = 7.0 × mass of person
area of polar bear in contact with ice = 56 × area of person in contact with ice [B1]
pressure exerted by person = 8.0 × pressure exerted by polar bear [B1]
(b) Under the ice is sea water (density = 1030 kg/m³). Calculate the change in pressure Δp between the top of the sea water and a depth of 2.50 m.
[2 Marks]
Δp = ρ × g × Δh = 1030 × 9.8 × 2.50 [C1]
Δp = 25 000 Pa (or 2.5 × 10⁴ Pa) [A2]
An outdoor heater warms people sitting outside.
(a) State the method of thermal energy transfer that warms the people.
[1 Mark]
radiation (or infrared radiation) [B1]
(b) (i) Complete the sentences about energy transfers that warm the table.
[2 Marks]
The heater transfers energy to the thermal (or internal) store of the table. The temperature of the table increases. The rate of energy absorbed by the table is greater than the rate of energy emitted by the table. [B1][B1]
(b) (ii) One person puts a white cloth over the table. Explain why the white cloth causes the temperature of the table to decrease.
[1 Mark]
Because white is a good reflector of radiation (OR white is a poor absorber of radiation). [B1]
(c) The heater has a power of 2.5 kW for 3 hours. Calculate the energy transferred in kWh.
[2 Marks]
E = P × t = 2.5 × 3 [C1]
energy = 7.5 kWh [A2]
(a) Atmospheric pressure is 1.0 × 10⁵ Pa. A volume of 1.02 m³ of air at atmospheric pressure is pumped into a car tyre. The pressure inside is 2.6 × 10⁵ Pa. Calculate the volume inside the tyre.
[3 Marks]
P₁V₁ = P₂V₂ [C1]
V₂ = (1.02 × 1.0 × 10⁵) / (2.6 × 10⁵) [C1]
volume = 0.39 m³ [A3]
(b) (i) Describe the structure of a gas in terms of separation, arrangement, and motion of particles.
[3 Marks]
separation: large (particles far apart) [B1]
arrangement: irregular (random) [B1]
motion: rapid (in random directions) [B1]
(b) (ii) Explain, in terms of particles, why the internal energy of gas decreases at lower temperature.
[2 Marks]
The average kinetic energy of particles decreases as temperature decreases. [C1][A2]
Figure 6.1 shows a full-scale drawing of object O, thin converging lens L and image I.
(a) (i) Identify line Z.
[1 Mark]
Z = principal axis [B1]
(a) (ii) Determine the focal length of the lens L.
[2 Marks]
Ray parallel to principal axis drawn from O/I to lens, and passing through focus to I/O. [B1]
focal length = 1.8 cm [B1]
(a) (iii) State two other characteristics of the image in Figure 6.1.
[2 Marks]
1. real[B1]
2. same size [B1]
(b) (i) Draw an X on Figure 6.2 for a position of object producing a magnified image.
[1 Mark]
X placed between F and the lens. [B1]
(b) (ii) Tick two statements describing the position of the magnified image.
[1 Mark]
[✓] on the same side of the lens as the object[B1]
[✓] further away from the lens than the object
Two identical cells in series with thermistor and fixed resistor (480 Ω). Voltmeter reading = 9.0 V, Ammeter = 4.6 mA.
(a) Determine the e.m.f. of one cell. Give the unit.
[2 Marks]
e.m.f. = 4.5 | unit = V [B1][B1]
(b) Complete sentence: Electric current is the amount of ______ passing a point per unit ______.
[1 Mark]
amount of charge passing a point per unit time. [B1]
(c) Resistance of fixed resistor is 480 Ω. Calculate thermistor resistance.
[3 Marks]
Total R = V / I = 9.0 / (4.6 × 10⁻³) = 1956.5 Ω [C1]
R_thermistor = 1956.5 – 480 [C1]
resistance = 1500 Ω [A3]
(d) Thermistor placed in ice (temperature decreases). State meter changes.
[1 Mark]
voltmeter: unchanged AND ammeter: decreases [B1]
Step-down transformer: Primary turns Nₚ = 6200, Secondary turns Nₛ = 43. Output voltage Vₛ = 230 V.
(a) (i) Calculate the input voltage Vₚ.
[2 Marks]
Vₚ = (Nₚ × Vₛ) / Nₛ = (6200 × 230) / 43 [C1]
input voltage = 33 000 V (or 3.3 × 10⁴ V) [A2]
(a) (ii) A 2.2 kW kettle is connected to secondary coil. Calculate primary current Iₚ.
[3 Marks]
2.2 kW = 2200 W [C1]
Iₚ = P / Vₚ = 2200 / 33 000 [C1]
current = 0.067 A (or 0.066 A) [A3]
(b) (i) State the equation for power loss in transmission.
[1 Mark]
P = I²R [B1]
(b) (ii) Explain how step-up transformers reduce power losses during transmission.
[2 Marks]
1. High voltage results in a low current for the same power. [B1]
2. Lower current produces less heating effect in transmission cables. [B1]
Beam of α-particles and γ-radiation entering electric field between plates P (+) and Q (-).
(a) Draw paths of α-particles and γ-radiation in electric field.
[3 Marks]
1. Line labeled γ passes straight through without deviation. [B1]
2. Line labeled α curves downwards towards negative plate Q. [A2]
(b) Sheet of paper placed in path before electric field. State effect.
[2 Marks]
α-particles: stopped by paper [B1]
γ-radiation: unchanged [B1]
(c) Beam replaced by β-particles. State and explain path.
[3 Marks]
statement: deflected upwards towards positive plate P[B1]
explanation: β-particles are negatively charged[B1] and experience an electrostatic force in an electric field. [B1]
(a) (i) Complete sentence: Stars expand when running out of hydrogen. A ______ forms a planetary nebula.
[1 Mark]
A red giant forms a planetary nebula. [B1]
(a) (ii) State what is at the centre of a planetary nebula.
[1 Mark]
white dwarf (star) [B1]
(b) (i) Explain why light with wavelength 5.0 × 10⁻⁷ m is blue light (compared to 6.6 × 10⁻⁷ m).
[1 Mark]
Blue light has a shorter wavelength than red light. [B1]
(b) (ii) Speed of light = 3.0 × 10⁸ m/s. Calculate frequency of blue light.
[2 Marks]
f = c / λ = (3.0 × 10⁸) / (5.0 × 10⁻⁷) [C1]
frequency = 6.0 × 10¹⁴ Hz [A2]
Average orbital speed v = 2πr / T
(a) Define terms r and T in this equation.
[2 Marks]
r is radius of orbit[B1]
T is orbital period (time taken to complete one orbit) [B1]
(b) (i) Explain why orbital speed is an average value.
[2 Marks]
1. The orbit is not circular (it is slightly elliptical), so distance r from Sun varies. [B1]
2. Gravitational pull from Sun varies along the orbit, changing speed. [B1]
(b) (ii) Saturn is 1.4 × 10¹² m from Sun, speed = 9600 m/s. Show 1 Saturn year ≈ 30 Earth years.
[4 Marks]
T_Earth = 365 × 24 × 60 × 60 = 3.15 × 10⁷ s [B1]
T_Saturn = (2π × r) / v = (2π × 1.4 × 10¹²) / 9600 = 9.16 × 10⁸ s [C1][A2]
T_Saturn in Earth years = (9.16 × 10⁸) / (3.15 × 10⁷) = 29.1 Earth years ≈ 30 Earth years [B1]
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