2025 Cambridge IGCSE Physics (0625) Practice Paper with Answers

First, calculate the change in momentum (impulse) of the ball. Taking the initial direction of motion as positive:
\( u = 12\text{ m/s} \)
\( v = -18\text{ m/s} \)

\( \Delta p = m(v - u) = 0.060\text{ kg} \times (-18\text{ m/s} - 12\text{ m/s}) = 0.060 \times (-30) = -1.8\text{ kg m/s} \)

The magnitude of the change in momentum is \( 1.8\text{ kg m/s} \) (or \( 1.8\text{ N s} \)).

Now, use the relationship between force, impulse, and time:
\( F = \frac{\Delta p}{\Delta t} \)
\( F = \frac{1.8\text{ N s}}{0.015\text{ s}} = 120\text{ N} \)

C1: for calculating change in momentum: \( \Delta p = 0.060 \times (18 + 12) = 1.8\text{ kg m/s} \) (or \( \text{N s} \))
C1: for \( F = \frac{\Delta p}{\Delta t} \) or \( F = \frac{1.8}{0.015} \)
A1: for \( 120\text{ N} \)

1. Calculate the useful work done in lifting the crate:
\( W = mgh = 45\text{ kg} \times 9.8\text{ m/s}^2 \times 12\text{ m} = 5292\text{ J} \)

2. Calculate the useful power output of the motor:
\( P_{\text{out}} = \frac{W}{t} = \frac{5292\text{ J}}{15\text{ s}} = 352.8\text{ W} \)

3. Calculate the efficiency of the motor:
\( \text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{352.8\text{ W}}{480\text{ W}} \times 100\% = 73.5\% \) (or \( 0.735 \))

C1: for calculating work done: \( W = 45 \times 9.8 \times 12 = 5292\text{ J} \)
C1: for calculating useful power output: \( P = \frac{5292}{15} = 352.8\text{ W} \) or \( \text{efficiency} = \frac{5292}{480 \times 15} \)
A1: for \( 73.5\% \) or \( 0.735 \)

1. Calculate the thermal energy supplied by the heater:
\( E = P \times t = 150\text{ W} \times (2.0 \times 60\text{ s}) = 150 \times 120 = 18000\text{ J} \)

2. Calculate the temperature rise:
\( \Delta\theta = 62^\circ\text{C} - 22^\circ\text{C} = 40^\circ\text{C} \)

3. Use the specific heat capacity formula:
\( E = mc\Delta\theta \implies c = \frac{E}{m\Delta\theta} \)
\( c = \frac{18000\text{ J}}{0.50\text{ kg} \times 40^\circ\text{C}} = \frac{18000}{20} = 900\text{ J}/(\text{kg }^\circ\text{C}) \)

C1: for calculating energy supplied: \( E = 150 \times 120 = 18000\text{ J} \)
C1: for calculating \( \Delta\theta = 40^\circ\text{C} \) and substituting into \( c = \frac{E}{m\Delta\theta} \)
A1: for \( 900\text{ J}/(\text{kg }^\circ\text{C}) \) (or \( \text{J}/(\text{kg K}) \))

The relationship between the refractive index \( n \) and the critical angle \( c \) is given by:
\( \sin(c) = \frac{1}{n} \)

Substitute \( n = 1.60 \):
\( \sin(c) = \frac{1}{1.60} = 0.625 \)

Calculate the critical angle:
\( c = \arcsin(0.625) \approx 38.7^\circ \)

C1: for formula \( \sin(c) = \frac{1}{n} \)
C1: for \( \sin(c) = \frac{1}{1.60} = 0.625 \)
A1: for \( 38.7^\circ \) or \( 39^\circ \)

1. Convert the operating time to seconds:
\( t = 15\text{ minutes} \times 60\text{ s/minute} = 900\text{ s} \)

2. Use the electrical energy formula:
\( E = VIt \)
\( E = 9.0\text{ V} \times 0.40\text{ A} \times 900\text{ s} = 3240\text{ J} \) (or \( 3.24\text{ kJ} \))

C1: for converting time to seconds: \( t = 15 \times 60 = 900\text{ s} \)
C1: for using \( E = VIt \) or calculating power \( P = VI = 3.6\text{ W} \) and substituting into \( E = Pt \)
A1: for \( 3240\text{ J} \) or \( 3.24\text{ kJ} \)

For a 100% efficient transformer, the primary power equals the secondary power:
\( V_p I_p = V_s I_s \)

Since \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \), this can be rewritten in terms of turns and currents:
\( I_p N_p = I_s N_s \)

Substitute the given values:
\( I_p \times 1800 = 4.0\text{ A} \times 90 \)
\( I_p \times 1800 = 360 \)
\( I_p = \frac{360}{1800} = 0.20\text{ A} \)

C1: for formula \( \frac{I_p}{I_s} = \frac{N_s}{N_p} \) or \( I_p N_p = I_s N_s \)
C1: for substitution: \( I_p \times 1800 = 4.0 \times 90 \)
A1: for \( 0.20\text{ A} \)

1. Subtract the background count rate to find the corrected count rates:
- Initial corrected count rate = \( 1200 - 40 = 1160\text{ counts/minute} \)
- Final corrected count rate = \( 185 - 40 = 145\text{ counts/minute} \)

2. Find the fraction of the corrected count rate remaining:
\( \text{Fraction} = \frac{145}{1160} = 0.125 = \frac{1}{8} \)

3. Express the remaining fraction as powers of a half:
\( \frac{1}{8} = \left(\frac{1}{2}\right)^3 \), which means 3 half-lives have elapsed.

4. Calculate the half-life:
\( 3 \text{ half-lives} = 15 \text{ hours} \implies 1 \text{ half-life} = 5.0 \text{ hours} \)

C1: for correcting both count rates: \( 1160 \) and \( 145\text{ counts/minute} \)
C1: for determining that 3 half-lives have elapsed (since \( \frac{145}{1160} = \frac{1}{8} \))
A1: for \( 5.0\text{ hours} \) (or \( 300\text{ minutes} \))

Use Hubble's law, which states that the recessional velocity \( v \) is proportional to distance \( d \):
\( v = H_0 \times d \)

Substitute the given values:
\( v = 2.2 \times 10^{-18}\text{ s}^{-1} \times 4.2 \times 10^{24}\text{ m} \)

Calculate \( v \):
\( v = 9.24 \times 10^6\text{ m/s} \)

C1: for formula \( v = H_0 d \)
C1: for substitution: \( v = 2.2 \times 10^{-18} \times 4.2 \times 10^{24} \)
A1: for \( 9.2 \times 10^6\text{ m/s} \) or \( 9.24 \times 10^6\text{ m/s} \)

Using the principle of conservation of momentum: Total initial momentum = Total final momentum. This gives \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\). Substituting the values: \((3.0 \text{ kg} \times 2.5 \text{ m/s}) + 0 = (3.0 \text{ kg} \times 1.0 \text{ m/s}) + (1.5 \text{ kg} \times v_2)\). Simplifying the equation: \(7.5 = 3.0 + 1.5 v_2\), which leads to \(4.5 = 1.5 v_2\). Solving for \(v_2\) gives \(v_2 = 3.0 \text{ m/s}\).

- Calculating the initial momentum or stating the conservation formula: 1 mark. - Setting up the momentum conservation equation correctly: 1 mark. - Calculating the final velocity with unit: 1 mark.

Calculate the thermal energy \(Q\) required: \(Q = m c \Delta \theta = 0.50 \text{ kg} \times 4200 \text{ J/(kg }^\circ\text{C)} \times (100 - 20) ^\circ\text{C} = 168\,000 \text{ J}\). Relating this to power and time: \(E = P \times t\). Converting 2.0 kW to 2000 W, we have: \(168\,000 \text{ J} = 2000 \text{ W} \times t\). Solving for \(t\): \(t = 168\,000 / 2000 = 84 \text{ s}\).

- Correct calculation of thermal energy required (168,000 J): 1 mark. - Converting power to 2000 W and setting up the time equation: 1 mark. - Calculating the correct time with unit (84 s): 1 mark.

First convert time to seconds: \(t = 5.0 \text{ minutes} = 300 \text{ s}\). Next, calculate current: \(I = Q / t = 360 \text{ C} / 300 \text{ s} = 1.2 \text{ A}\). Finally, use Ohm's law to find resistance: \(R = V / I = 12 \text{ V} / 1.2 \text{ A} = 10 \text{ }\Omega\).

- Correct conversion of time to seconds and calculation of current (1.2 A): 1 mark. - Recall of Ohm's law formula \(R = V / I\): 1 mark. - Correct calculation of resistance with unit (10 Ω): 1 mark.

1. At the hot end, particles (atoms/ions) gain kinetic energy and vibrate with greater amplitude. 2. These vibrating particles collide with adjacent particles, transferring kinetic energy along the spoon. 3. Free (delocalised) electrons also gain kinetic energy and diffuse rapidly through the metal lattice, colliding with colder ions further away and transferring energy quickly.

1. Particles in the hot region gain kinetic energy / vibrate more [1] 2. Vibrating particles collide with neighboring particles to transfer energy [1] 3. Free / delocalised electrons gain kinetic energy and move/diffuse through the metal lattice to transfer energy to distant ions [1]

Sound travels as a longitudinal wave of pressure vibrations. In a solid, the particles are tightly packed and held by strong intermolecular forces. When one particle vibrates, it immediately exerts a force on its neighbors, passing the vibration on very quickly. In a gas, particles are widely separated and must travel a distance before colliding to pass on the vibration, resulting in a much slower speed of sound.

1. Particles in a solid are much closer together than in a gas [1] 2. Intermolecular forces between particles in a solid are much stronger than in a gas [1] 3. Vibrations are passed on to adjacent particles much more quickly/rapidly [1]

1. By Lenz's law, the direction of the induced e.m.f. and current opposes the change that causes it (the downward motion of the ring towards the N-pole). 2. To oppose this motion, the induced current in the ring must create an opposing North pole at its bottom face. 3. This creates a repulsive force acting upwards, opposing the descent of the ring.

1. Force acts upwards / is a repulsive force [1] 2. Lenz's law states that induced current/magnetic field opposes the change producing it [1] 3. Induced current creates a North pole at the bottom of the ring to repel the approaching North pole of the magnet [1]

1. Redshift occurs when light waves from a receding source are stretched, increasing their wavelength. 2. Observing that almost all distant galaxies show redshift indicates they are moving away from Earth. 3. The correlation that further galaxies have larger redshifts (Hubble's Law) shows space itself is expanding. 4. Extrapolating this expansion backwards in time suggests the Universe began from a single starting point, supporting the Big Bang theory.

1. Redshift indicates galaxies are moving away / receding [1] 2. Greater redshift at greater distances shows velocity of recession is proportional to distance / space is expanding [1] 3. Extrapolating backwards implies all matter started at a single common point in the past (supporting the Big Bang) [1]

1. Ionising power depends on charge and mass. Alpha particles have charge +2e and mass 4 u, making them highly effective at ionising atoms they pass. 2. High ionisation rate means alpha particles lose their kinetic energy over a very short distance, resulting in low penetration (stopped by paper). 3. Gamma rays are photons with no mass or charge, meaning they have a very low probability of interacting with atomic electrons, leading to weak ionising power. 4. With few interactions, gamma rays can travel great distances through matter before being absorbed, giving them high penetrating power.

1. Alpha particles have large charge/mass, causing frequent collisions/high rate of ionisation [1] 2. Frequent ionisation causes alpha particles to lose energy quickly, leading to low penetration [1] 3. Gamma rays have no charge/mass, resulting in low probability of interaction (weak ionisation) and high penetration [1]

1. Thermal energy from the hot water increases the temperature of the metal lid. 2. This causes the metal atoms to vibrate more vigorously and push further apart, causing the lid to expand. 3. Metal has a higher coefficient of thermal expansion than glass, so the lid expands more than the jar neck. 4. The gap between the lid and the jar increases, reducing friction and releasing the tight fit.

1. Heating causes particles in the metal to vibrate more and move further apart / metal lid expands [1] 2. Metal expands more than glass for the same temperature rise [1] 3. The relative expansion of the lid compared to the jar loosens the tight seal [1]

1. For total internal reflection to occur, light must travel from a medium of higher refractive index (core) towards a medium of lower refractive index (cladding). 2. The light signal must strike the boundary at an angle of incidence that is greater than the critical angle for the core-cladding interface. 3. This ensures that 100% of the light energy is reflected back into the core, guiding the signal along the length of the fibre.

1. Core has a higher refractive index / is optically denser than the cladding [1] 2. Angle of incidence at the core-cladding boundary is greater than the critical angle [1] 3. Total internal reflection occurs, keeping the light inside the core [1]

1. Initially, weight is the dominant force acting downwards, so there is a large downward resultant force and the skydiver accelerates. 2. Air resistance opposes motion and is directly proportional to speed, so it increases as the skydiver speeds up. 3. As air resistance increases, the net downward resultant force decreases, reducing the acceleration. 4. Eventually, air resistance equals weight. The resultant force is zero (forces are balanced), so acceleration becomes zero, and she falls at a constant terminal velocity.

1. Initially weight is greater than air resistance, producing a downward resultant force and acceleration [1] 2. Air resistance increases as speed increases, reducing the resultant force [1] 3. When air resistance equals weight, the resultant force is zero, leading to constant speed/terminal velocity [1]

1. Conduction and convection require a physical medium (particles) to transfer thermal energy.
2. Since a vacuum has no particles, transfer of thermal energy by conduction and convection cannot occur across the gap, thereby reducing the rate of heat loss from the soup.
3. Radiation (specifically infrared radiation) consists of electromagnetic waves, which do not require a medium to propagate and can therefore travel through the vacuum.

- State that conduction and convection require particles / a medium to transfer energy [1]
- Explain that the vacuum has no particles / is empty space, so conduction and convection cannot occur [1]
- Identify radiation (or infrared radiation) as the method not prevented by the vacuum (since electromagnetic waves do not require a medium) [1]

1. Redshift occurs when light from a galaxy has its wavelength stretched (shifted toward the red end of the spectrum) because the galaxy is moving away from us.
2. Observations show that almost all distant galaxies are redshifted, indicating they are moving away from Earth.
3. Furthermore, more distant galaxies show a greater redshift, meaning they are receding at a higher speed. This relationship (Hubble's Law) shows that the space between galaxies is expanding, supporting the expanding Universe theory.

- Define redshift as the increase in observed wavelength / shift of light towards the red end of the spectrum [1]
- State that redshift indicates galaxies are moving away from us (receding) [1]
- Explain that more distant galaxies show larger redshifts (moving faster), which proves the fabric of space itself is expanding [1]

1. As the magnet falls through the copper pipe, the magnetic field lines of the magnet cut across the copper conductor, leading to a changing magnetic flux linkage in the pipe.
2. This changing magnetic flux induces an electromotive force (e.m.f.) and consequently a current in the closed path of the copper pipe.
3. According to Lenz's law, the direction of the induced current is such that its magnetic field opposes the change that produced it. This produces an upward magnetic force on the falling magnet, opposing gravity and causing the magnet to fall more slowly than it would in free fall.

- Explain that the falling magnet creates a changing magnetic field / flux cutting across the copper pipe [1]
- State that this changing magnetic flux induces an e.m.f. / current in the pipe [1]
- State that, by Lenz's law, the induced current creates a magnetic field that opposes the magnet's descent, exerting an upward force and slowing it down [1]

First, calculate the change in wavelength: \(\Delta \lambda = 672.7\text{ nm} - 656.3\text{ nm} = 16.4\text{ nm}\). Next, use the redshift equation: \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\). Substitute the known values: \(\frac{16.4}{656.3} = \frac{v}{3.0 \times 10^8}\). Solving for \(v\) gives: \(v = \frac{16.4}{656.3} \times 3.0 \times 10^8 \approx 7.5 \times 10^6\text{ m/s}\).

C1 for calculating change in wavelength of 16.4 nm. C1 for correct recall and substitution into the redshift formula. A1 for the final answer of 7.5 * 10^6 m/s (accept 7.49 * 10^6 m/s) with correct unit.

Calculate the initial resistance: \(R_1 = \frac{V_1}{I_1} = \frac{4.0\text{ V}}{0.25\text{ A}} = 16.0\ \Omega\). Calculate the final resistance: \(R_2 = \frac{V_2}{I_2} = \frac{8.0\text{ V}}{0.40\text{ A}} = 20.0\ \Omega\). Find the change in resistance: \(\Delta R = R_2 - R_1 = 20.0\ \Omega - 16.0\ \Omega = 4.0\ \Omega\).

C1 for calculating initial resistance as 16.0 ohms. C1 for calculating final resistance as 20.0 ohms. A1 for correct calculation of the increase as 4.0 ohms with correct unit.

The frequency of a wave remains constant as it crosses a boundary between different media. Therefore, \(f = \frac{v_1}{\lambda_1} = \frac{v_2}{\lambda_2}\). Rearranging the formula to find the speed in shallow water: \(v_2 = v_1 \times \frac{\lambda_2}{\lambda_1}\). Substitute the values: \(v_2 = 0.36\text{ m/s} \times \frac{2.0\text{ cm}}{3.0\text{ cm}} = 0.24\text{ m/s}\).

C1 for state or use of constant frequency relation or direct ratio formula. C1 for correct substitution of values. A1 for correct final speed of 0.24 m/s with unit.