2025 Cambridge IGCSE Physics (0625) Practice Paper with Answers

Using the formula: \( W = m \times g \). We can rearrange this formula to find the gravitational field strength: \( g = \frac{W}{m} \). Substituting the given values: \( g = \frac{14.8\text{ N}}{4.0\text{ kg}} = 3.7\text{ N/kg} \).

1 Mark: Recall of the formula \( W = mg \) or rearrangement to \( g = \frac{W}{m} \). 1 Mark: Correct calculation resulting in \( 3.7 \). 0.33 Mark: Correct unit of \( \text{N/kg} \) or \( \text{m/s}^2 \).

The kinetic energy \( E_k \) is given by the formula: \( E_k = \frac{1}{2} m v^2 \). Substituting the given values into the equation: \( E_k = \frac{1}{2} \times 250\text{ kg} \times (12\text{ m/s})^2 \). Calculating this gives: \( E_k = 125 \times 144 = 18000\text{ J} \) (or \( 18\text{ kJ} \)).

1 Mark: Recall of the formula \( E_k = \frac{1}{2}mv^2 \). 1 Mark: Correct calculation to get \( 18000 \) or \( 18 \). 0.33 Mark: Correct unit of \( \text{J} \) or \( \text{kJ} \).

According to the principle of conservation of momentum: \( \text{Total initial momentum} = \text{Total final momentum} \), which can be written as: \( m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \). Given that the second car is initially stationary (\( u_2 = 0 \)): \( (0.80\text{ kg} \times 3.0\text{ m/s}) + 0 = (0.80\text{ kg} + 1.20\text{ kg}) \times v \). This simplifies to: \( 2.4\text{ kg m/s} = 2.0\text{ kg} \times v \). Solving for \( v \) gives: \( v = \frac{2.4}{2.0} = 1.2\text{ m/s} \).

1 Mark: Correct formulation of the conservation of momentum equation, e.g., \( m_1 u_1 = (m_1 + m_2)v \). 1 Mark: Correct calculation of velocity to get \( 1.2 \). 0.33 Mark: Correct unit of \( \text{m/s} \).

First, calculate the contact area \( A \) of the base: \( A = 0.40\text{ m} \times 0.30\text{ m} = 0.12\text{ m}^2 \). Next, use the pressure formula: \( P = \frac{F}{A} \). Using the weight as the downward force \( F \): \( P = \frac{600\text{ N}}{0.12\text{ m}^2} = 5000\text{ Pa} \) (or \( 5.0\text{ kPa} \)).

1 Mark: Correct calculation of base area \( 0.12\text{ m}^2 \) and recall of pressure formula \( P = \frac{F}{A} \). 1 Mark: Correct numerical calculation of pressure to get \( 5000 \) (or \( 5.0 \)). 0.33 Mark: Correct unit of \( \text{Pa} \), \( \text{N/m}^2 \), or \( \text{kPa} \).

First, convert the time from minutes to seconds: \( t = 4.0\text{ minutes} = 4.0 \times 60 = 240\text{ s} \). Then, use the relationship between charge, current, and time: \( Q = I \times t \). Substitute the given values: \( Q = 2.5\text{ A} \times 240\text{ s} = 600\text{ C} \).

1 Mark: Time conversion to seconds (\( 240\text{ s} \)) and recall of the formula \( Q = It \). 1 Mark: Correct calculation resulting in \( 600 \). 0.33 Mark: Correct unit of charge \( \text{C} \) (or \( \text{A s} \)).

First, convert the orbital period from minutes to seconds: \( T = 96\text{ minutes} = 96 \times 60 = 5760\text{ s} \). The orbital distance is the circumference of the circle: \( d = 2 \pi r \). The orbital speed \( v \) is given by: \( v = \frac{2 \pi r}{T} \). Substitute the values: \( v = \frac{2 \times \pi \times 7200\text{ km}}{5760\text{ s}} = \frac{45238.93\text{ km}}{5760\text{ s}} \approx 7.85\text{ km/s} \). Rounding to two significant figures gives \( 7.9\text{ km/s} \).

1 Mark: Recall of the orbital speed formula \( v = \frac{2 \pi r}{T} \) and correct conversion of orbital period to seconds (\( 5760\text{ s} \)). 1 Mark: Correct calculation to get \( 7.9 \) (allow \( 7.8 \) to \( 7.9 \)). 0.33 Mark: Correct unit of \( \text{km/s} \).

First, calculate the thermal energy input \( \Delta E \) from the electric heater: \( \Delta E = P \times t = 120\text{ W} \times (3.0 \times 60\text{ s}) = 21600\text{ J} \). Next, calculate the temperature increase \( \Delta \theta \): \( \Delta \theta = 52\text{ }^\circ\text{C} - 20\text{ }^\circ\text{C} = 32\text{ }^\circ\text{C} \). Now use the specific heat capacity formula: \( \Delta E = m c \Delta \theta \). Rearranging for specific heat capacity \( c \): \( c = \frac{\Delta E}{m \Delta \theta} \). Substitute the values: \( c = \frac{21600\text{ J}}{1.5\text{ kg} \times 32\text{ }^\circ\text{C}} = \frac{21600}{48} = 450\text{ J/(kg }^\circ\text{C)} \).

1 Mark: Calculation of energy input \( 21600\text{ J} \) and temperature rise \( 32\text{ }^\circ\text{C} \) with formula \( \Delta E = mc\Delta \theta \). 1 Mark: Correct calculation of specific heat capacity to get \( 450 \). 0.33 Mark: Correct unit \( \text{J/(kg }^\circ\text{C)} \) or \( \text{J/(kg K)} \).

First, determine how many half-lives have elapsed. The activity decreases by half successively: \( 3200\text{ Bq} \rightarrow 1600\text{ Bq} \rightarrow 800\text{ Bq} \rightarrow 400\text{ Bq} \rightarrow 200\text{ Bq} \). This is a total of \( 4 \) half-lives since \( \frac{200}{3200} = \frac{1}{16} = \left(\frac{1}{2}\right)^4 \). The total time elapsed is \( t = 12.0\text{ hours} \). The length of one half-life \( T_{1/2} \) is: \( T_{1/2} = \frac{\text{Total time}}{\text{Number of half-lives}} = \frac{12.0\text{ hours}}{4} = 3.0\text{ hours} \).

1 Mark: Identification that \( 4 \) half-lives have elapsed. 1 Mark: Division of the total time by the number of half-lives to find \( 3.0 \). 0.33 Mark: Correct unit of time, which is \( \text{hours} \) (or equivalent if converted).

First, recall the formula for orbital speed: \( v = \frac{2\pi r}{T} \). Substitute the given values into the equation: \( v = \frac{2 \times \pi \times 8.0 \times 10^{3}}{6.0 \times 10^{3}} \). Simplifying this gives: \( v = \frac{16000\pi}{6000} \approx 8.38\text{ km/s} \). Rounding to 2 significant figures gives \( 8.4\text{ km/s} \).

1 mark: Correct formula for orbital speed \( v = \frac{2\pi r}{T} \). 1 mark: Correct substitution of values. 0.33 mark: Correct final answer with appropriate units (accept 8.38 to 8.4 km/s).

First, calculate the work done (or increase in gravitational potential energy): \( \Delta E_p = mgh = 450 \times 9.8 \times 12 = 52920\text{ J} \). Then, calculate power: \( P = \frac{\Delta E_p}{t} = \frac{52920}{15} = 3528\text{ W} \). Rounding to 2 significant figures gives \( 3500\text{ W} \) (or \( 3.5\text{ kW} \)).

1 mark: Correct calculation of work done/potential energy \( 52920\text{ J} \). 1 mark: Correct power formula and substitution. 0.33 mark: Correct final answer with units (accept 3528 W, 3500 W or 3.5 kW).

First, convert time from minutes to seconds: \( t = 4.0 \times 60 = 240\text{ s} \). Then, use the relationship between charge, current, and time: \( Q = I \times t \). Substitute the values: \( Q = 0.35 \times 240 = 84\text{ C} \).

1 mark: Correct conversion of time to seconds (240 s). 1 mark: Correct formula \( Q = I \times t \) and substitution. 0.33 mark: Correct final answer with unit (84 C).

First, calculate the weight of the block, which is the force acting on the ground: \( F = W = m \times g = 120 \times 9.8 = 1176\text{ N} \). Next, use the pressure formula: \( p = \frac{F}{A} = \frac{1176}{0.25} = 4704\text{ Pa} \). Rounding to 2 significant figures gives \( 4700\text{ Pa} \).

1 mark: Correct calculation of force/weight (1176 N). 1 mark: Correct pressure formula and substitution. 0.33 mark: Correct final answer with unit (accept 4700 Pa or 4704 Pa).

First, convert the frequency from megahertz to hertz: \( f = 120\text{ MHz} = 120 \times 10^6\text{ Hz} = 1.2 \times 10^8\text{ Hz} \). Next, use the wave equation: \( v = f \lambda \), which rearranges to \( \lambda = \frac{v}{f} \). Substitute the values: \( \lambda = \frac{3.0 \times 10^8}{1.2 \times 10^8} = 2.5\text{ m} \).

1 mark: Correct conversion of frequency to Hz (1.2 * 10^8 Hz). 1 mark: Correct wave equation used and rearranged. 0.33 mark: Correct final answer with unit (2.5 m).

Use the principle of conservation of momentum: Total momentum before collision = Total momentum after collision. Before collision: \( p_{\text{before}} = m_1 u_1 + m_2 u_2 = 0.80 \times 3.0 + 1.2 \times 0 = 2.4\text{ kg m/s} \). After collision: \( p_{\text{after}} = (m_1 + m_2) v = (0.80 + 1.2) \times v = 2.0 v \). Equating the two: \( 2.0 v = 2.4 \), which gives \( v = 1.2\text{ m/s} \).

1 mark: Correct calculation of initial momentum (2.4 kg m/s). 1 mark: Application of conservation of momentum. 0.33 mark: Correct final velocity with unit (1.2 m/s).

First, calculate the thermal energy supplied by the heater: \( E = P \times t = 800\text{ W} \times (3.5 \times 60\text{ s}) = 800 \times 210 = 168000\text{ J} \). Next, use the specific heat capacity formula: \( E = m c \Delta \theta \). Rearrange to solve for temperature rise: \( \Delta \theta = \frac{E}{mc} \). Substitute the values: \( \Delta \theta = \frac{168000}{2.0 \times 4200} = \frac{168000}{8400} = 20\text{ }^\circ\text{C} \).

1 mark: Correct calculation of energy supplied (168000 J). 1 mark: Correct substitution into the specific heat capacity equation. 0.33 mark: Correct final temperature rise with unit (20 °C).

In beta-minus decay, a neutron within the unstable nucleus decays/transforms into a proton and an electron. The high-energy electron is ejected from the nucleus as a \(\beta^-\) particle. As a result, the number of protons (atomic number) increases by 1, while the total number of nucleons (mass number) stays the same.

1 mark: State that a neutron decays/changes into a proton (and an electron). 1 mark: State that the proton number increases by 1 (or mass number stays constant / electron is emitted from the nucleus).

When an object falls, its weight acts downwards, causing it to accelerate. As its speed increases, the air resistance acting upwards also increases. Eventually, the upward air resistance increases until it is equal in magnitude to the downward weight. At this point, the forces are balanced, the resultant force on the object is zero, and according to Newton's first law, it continues to fall at a constant speed, known as terminal velocity.

1 mark: Identify that air resistance increases with speed and eventually equals the weight (forces balance). 1 mark: Explain that the resultant force is zero, leading to zero acceleration (constant speed).

Gravitational force (weight) acts vertically downwards. The formula for gravitational potential energy gained (or work done against gravity) is \(W = F \cdot d = mgh\), where \(d\) is the distance moved in the direction of the force. Since weight is vertical, only the vertical displacement (height \(h\)) contributes to the work done against gravity. Any horizontal component of the motion is perpendicular to the force of gravity, so no work is done against gravity in the horizontal direction.

1 mark: Mention that the force of gravity acts vertically downwards. 1 mark: Explain that work done is force multiplied by distance in the direction of the force, so only vertical height \(h\) matters (or write formula \(W = mgh\) and define terms).

According to the kinetic theory, gas particles are in continuous, rapid, random motion. As they move, they collide with the inner walls of the container. During each collision, a particle rebounds, which means its velocity changes and it undergoes a change in momentum. This change in momentum exerts a force on the wall of the container. The average of all these forces exerted by many collisions over the surface area of the walls results in pressure (\(p = F/A\)).

1 mark: Mention that particles collide with the walls of the container. 1 mark: Explain that these collisions involve a change in momentum, producing a force on the walls (and hence pressure is force per unit area).

When the temperature of the metallic filament increases, the positive metal ions in the lattice vibrate more rapidly and with greater amplitude. This increased vibration increases the frequency of collisions between the free (delocalised) electrons, which constitute the current, and the metal ions. Consequently, the flow of charge is resisted more, resulting in an increased electrical resistance.

1 mark: Explain that higher temperature causes metal ions/lattice to vibrate more (with larger amplitude). 1 mark: State that this increases the rate/frequency of collisions with the flowing electrons (impeding current flow).

At the equator, the Sun is high in the sky, meaning its rays strike the Earth's surface at a nearly perpendicular angle (close to \(90^\circ\)). This concentrates the solar radiation over a small, highly concentrated surface area. In contrast, at the poles, the Earth's curvature causes the Sun's rays to strike at a very shallow, oblique angle. This spreads the same quantity of solar energy over a much larger surface area, and the rays must also pass through a thicker layer of the atmosphere, leading to more absorption and scattering. Consequently, the energy received per unit area is much lower at the poles than at the equator.

1 mark: Explain that at the equator, Sun's rays are more concentrated / strike perpendicularly (or at a high angle). 1 mark: Explain that at the poles, the rays strike at an angle / are spread over a larger surface area.

When a ray of light travelling in an optically denser medium (water) meets the boundary with an optically less dense medium (air) at an angle of incidence that exceeds the critical angle, refraction cannot occur. Instead, all of the light is reflected back into the denser medium (water) according to the law of reflection (angle of incidence equals angle of reflection). This phenomenon is known as total internal reflection.

1 mark: State that the light is completely reflected back into the water (no light escapes/refracts into air). 1 mark: Identify the process as total internal reflection.

The fundamental difference lies in the direction of particle oscillation relative to energy transfer:
1. In a transverse wave (such as light or water waves), the particles of the medium oscillate (vibrate) in a direction perpendicular (at \(90^\circ\)) to the direction in which the wave is propagating (travelling).
2. In a longitudinal wave (such as sound waves), the particles of the medium oscillate in a direction parallel (along the same line) to the direction of wave propagation.

1 mark: State that in transverse waves, oscillations are perpendicular to the direction of wave travel. 1 mark: State that in longitudinal waves, oscillations are parallel to the direction of wave travel.

Metals contain a large number of free (delocalized) electrons. When a metal is heated, these free electrons gain kinetic energy and move rapidly through the metal lattice. They collide with other metal ions and pass on their kinetic energy, transferring thermal energy much more rapidly than the lattice vibrations alone that occur in non-metals.

1 mark: Identify that metals contain free or delocalized electrons (which non-metals do not have). 1 mark: Explain that these free electrons gain kinetic energy, move rapidly, and transfer energy via collisions with other ions. 0.08 mark: Contrast this with non-metals where energy transfer is restricted to slower lattice vibrations.

Pressure is defined as force per unit area, given by the formula \(P = \frac{F}{A}\). The downward force exerted by the cylinder on the ground is its weight, which remains constant. When the cylinder is placed on its narrower end, the contact area \(A\) is smaller. Since pressure is inversely proportional to area for a constant force, the smaller area results in a greater pressure exerted on the soft ground, causing it to sink deeper.

1 mark: Reference the relationship between pressure, force, and area (e.g., \(P = \frac{F}{A}\)) and state that the force (weight) is constant. 1 mark: State that the narrower end has a smaller contact area. 0.08 mark: Conclude that a smaller area increases the pressure exerted on the ground, causing it to sink deeper.

Alpha particles have a charge of \(+2e\) and a much larger mass compared to beta particles, which have a charge of \(-1e\) and negligible mass. Due to their greater charge and slower speed at the same energy, alpha particles interact strongly with air molecules, causing intense ionization. This frequent ionization causes them to lose their kinetic energy very rapidly, resulting in a short range of only a few centimeters in air.

1 mark: State that alpha particles have a greater charge (or larger mass/size) compared to beta particles. 1 mark: Explain that this greater charge or size causes a much higher rate of ionization with air molecules. 0.08 mark: Conclude that this rapid ionization causes alpha particles to lose their energy quickly, leading to a much shorter range.

Venus is surrounded by a very thick and dense atmosphere consisting mostly of carbon dioxide, which is a greenhouse gas. This atmosphere traps the infrared radiation (thermal energy) emitted by Venus's surface, causing a runaway greenhouse effect. In contrast, Mercury has almost no atmosphere, allowing thermal energy from its surface to readily radiate back into space, resulting in a lower average surface temperature despite being closer to the Sun.

1 mark: Identify that Venus has a thick atmosphere rich in greenhouse gases (carbon dioxide) whereas Mercury has virtually no atmosphere. 1 mark: Explain that greenhouse gases on Venus trap infrared (thermal) radiation. 0.08 mark: Explain that this greenhouse effect keeps heat in, whereas Mercury easily radiates its heat back into space.

Measure the thickness at several different positions along the wire, and calculate the average of these readings. This accounts for any non-uniformity in the wire's diameter.

1 mark: Take measurements at different positions along the wire. 1 mark: Calculate the average of these measurements or check for and subtract any zero error.

The limit of proportionality is the point up to which extension is directly proportional to load. On the graph, this is identified as the point where the linear (straight-line) relationship ceases and the line starts to curve.

1 mark: Point beyond which extension is no longer directly proportional to load (or Hooke's law is no longer obeyed). 1 mark: Identified on the graph as the point where the straight line begins to curve.

For every 100 J of input energy, 55 J is wasted. This wasted energy is transferred as thermal energy (heat) and sound energy, which are dissipated to the surroundings.

1 mark: Identifies the wasted energy form as thermal energy / heat (or sound). 1 mark: Identifies that this energy is transferred or dissipated to the surroundings.

Heating the water at the bottom causes it to expand and become less dense. This heated, less dense water rises. The cooler, denser water at the top sinks to the bottom to take its place, creating a continuous convection current.

1 mark: Heated water expands and becomes less dense, so it rises. 1 mark: Cooler, denser water sinks to the bottom.

When plane wavefronts pass through a gap comparable in size to the wavelength, they undergo significant diffraction, causing the waves to spread out and the wavefronts to become circular.

1 mark: Identifies that the waves spread out / diffract. 1 mark: Identifies that the wavefronts become circular / curved.

As the potential difference increases, the temperature of the filament rises. This causes the metal ions in the lattice of the filament to vibrate more rapidly and with larger amplitude, increasing the frequency of collisions with conduction electrons and thus increasing the resistance.

1 mark: State that resistance increases. 1 mark: Explain that the lattice ions/atoms vibrate more (due to increased temperature), causing more collisions with conduction electrons.

Alpha particles (positive charge) and beta particles (negative charge) experience forces in opposite directions, causing them to deflect in opposite directions. Since beta particles have a much smaller mass than alpha particles, they undergo a much greater deflection (with a smaller radius of curvature). Gamma rays are uncharged and unaffected.

1 mark: Deflected in the opposite direction. 1 mark: Deflected much more / more sharply / with a smaller radius of curvature.

A planet's orbit around the Sun is elliptical. The orbital speed varies such that the planet travels fastest when it is closest to the Sun (perihelion) and slowest when it is furthest away (aphelion).

1 mark: Identifies the orbit shape as elliptical / an ellipse. 1 mark: States that orbital speed increases when closer to the Sun / decreases when further away.

When water waves travel from deep to shallow water, their speed decreases. Since the wave frequency remains constant, the wavelength must decrease, meaning the spacing between wavefronts decreases. Because the wavefronts meet the boundary at an angle, the portion of each wavefront entering the shallow water first slows down first, causing the direction of travel to bend towards the normal.

1 mark: Spacing decreases (or wavefronts get closer together). 1 mark: Direction of travel bends towards the normal (or wavefronts turn to become closer to parallel with the boundary).

The magnetic field around a straight current-carrying wire forms concentric circular field lines centered on the wire. According to the right-hand grip rule, if the thumb points upwards in the direction of the current, the fingers curl in an anticlockwise direction when viewed from above.

1 mark: Shape is concentric circles (centered on the wire). 1 mark: Direction is anticlockwise / counter-clockwise (when viewed from above).