2025 Cambridge IGCSE Physics (0625) 模拟试题及答案详解

The flight is divided into two phases. Phase 1 (constant acceleration): distance d1 = 0.5 * v * t1 = 0.5 * 14 m/s * 4.0 s = 28 m. Phase 2 (constant velocity): distance d2 = v * t2 = 14 m/s * 6.0 s = 84 m. The total distance travelled is d1 + d2 = 28 m + 84 m = 112 m.

C1: for calculating the distance of the acceleration phase (28 m) or showing the formula d = 0.5 * v * t1 + v * t2. A0.5: for the correct final answer of 112 m.

Using the conservation of momentum: Total initial momentum = total final momentum. (m1 * v1) + (m2 * v2) = (m1 + m2) * v. (0.50 kg * 3.0 m/s) + (1.0 kg * 0 m/s) = (0.50 kg + 1.0 kg) * v. 1.5 kg m/s = 1.5 kg * v. Solving for v gives v = 1.0 m/s.

C1: for stating the principle of conservation of momentum or showing the formula 0.5 * 3.0 = (0.5 + 1.0) * v. A0.5: for the correct final velocity of 1.0 m/s.

Cup A cools down more slowly. This is because shiny, silver-coloured surfaces are poor emitters (and good reflectors) of infrared radiation. Therefore, Cup A transfers heat to the colder surroundings at a slower rate than the dull black surface of Cup B.

B0.5: for identifying Cup A as cooling down more slowly. B1: for explaining that shiny/silver surfaces are poor emitters (or poor radiators) of infrared (or thermal) radiation.

Using Snell's law: n = sin(i) / sin(r). Rearranging to solve for the angle of refraction r: sin(r) = sin(i) / n = sin(48.0) / 1.45. sin(r) = 0.7431 / 1.45 = 0.5125. Therefore, r = arcsin(0.5125) = 30.8 degrees.

C1: for correct substitution into Snell's law to show sin(r) = sin(48.0) / 1.45. A0.5: for the correct angle of 30.8 degrees (accept 31 degrees).

According to Lenz's law, the induced current in a conductor always opposes the change that produces it. Since the approaching magnetic pole is a north pole, the top of the copper ring must also become a north pole to create a repulsive force that opposes the magnet's downward motion.

B0.5: for stating that the top of the ring becomes a north pole (or creates an upward magnetic field). B1: for explaining that Lenz's law dictates that the induced magnetic field must oppose the motion or approach of the north pole.

The activity decreases from 3200 Bq to 1600 Bq (1 half-life), then to 800 Bq (2 half-lives), and finally to 400 Bq (3 half-lives). Therefore, 3 half-lives have elapsed in 24.0 hours. The half-life is 24.0 hours / 3 = 8.0 hours.

C1: for determining that the activity has decreased by 3 half-lives (activity fraction is 1/8). A0.5: for the correct half-life of 8.0 hours.

Redshift is the increase in the observed wavelength of electromagnetic radiation emitted by a source. The fact that the light from the distant galaxy is redshifted tells astronomers that the galaxy is moving away from the Earth, which supports the theory of the expansion of the Universe.

B0.5: for defining redshift as an increase in the observed wavelength (or decrease in frequency) of light. B1: for explaining that it indicates the galaxy is moving away (receding) from Earth.

First, calculate the extension of the spring x: x = 15.0 cm - 12.0 cm = 3.0 cm. According to Hooke's law: F = k * x. Therefore, the spring constant k is k = F / x = 6.0 N / 3.0 cm = 2.0 N/cm.

C1: for calculating the extension of 3.0 cm or stating Hooke's law (F = k * x). A0.5: for the correct spring constant of 2.0 N/cm (or 200 N/m if units are converted).

During the first phase (acceleration):
Distance \( d_1 = \frac{1}{2} a t_1^2 = 0.5 \times 15\text{ m/s}^2 \times (4.0\text{ s})^2 = 120\text{ m} \).
Final velocity reached \( v = a t_1 = 15\text{ m/s}^2 \times 4.0\text{ s} = 60\text{ m/s} \).

During the second phase (constant velocity):
Distance \( d_2 = v \times t_2 = 60\text{ m/s} \times 6.0\text{ s} = 360\text{ m} \).

Total distance \( d = d_1 + d_2 = 120\text{ m} + 360\text{ m} = 480\text{ m} \).

- 0.5 marks for calculating the correct distance in the first phase: \( 120\text{ m} \) or for a correct formula.
- 0.5 marks for finding the correct constant velocity: \( 60\text{ m/s} \).
- 0.5 marks for the final correct total distance with unit: \( 480\text{ m} \).

The impulse is equal to the change in momentum: \( \Delta p = m(v - u) \).
Taking the initial direction as positive, \( u = 25\text{ m/s} \) and \( v = -35\text{ m/s} \).

\( \Delta p = 0.060\text{ kg} \times (-35\text{ m/s} - 25\text{ m/s}) = 0.060\text{ kg} \times (-60\text{ m/s}) = -3.6\text{ N s} \).
The magnitude of the impulse is \( 3.6\text{ N s} \) (or \( 3.6\text{ kg m/s} \)).

- 0.5 marks for recognizing the direction change, giving \( \Delta v = 60\text{ m/s} \).
- 0.5 marks for substituting correctly into the momentum equation: \( 0.060 \times 60 \).
- 0.5 marks for the final correct magnitude of \( 3.6\text{ N s} \) (or \( \text{kg m/s} \)) with appropriate unit.

First, calculate the total thermal energy supplied:
\( E = P \times t = 120\text{ W} \times (5.0 \times 60)\text{ s} = 120 \times 300 = 36000\text{ J} \).

Next, use the specific heat capacity formula:
\( E = m c \Delta \theta \)
\( 36000 = 2.5\text{ kg} \times c \times (32^\circ\text{C} - 20^\circ\text{C}) \)
\( 36000 = 2.5 \times c \times 12 \)
\( 36000 = 30 c \)
\( c = 1200\text{ J/(kg }^\circ\text{C)} \).

- 0.5 marks for calculating the total energy supplied: \( 36000\text{ J} \).
- 0.5 marks for rearranging \( E = m c \Delta \theta \) or substituting values correctly.
- 0.5 marks for the correct value \( 1200 \) with correct units (\( \text{J/(kg }^\circ\text{C)} \) or \( \text{J/(kg K)} \)).

Use the formula relating critical angle \( c \) and refractive index \( n \):
\( \sin(c) = \frac{1}{n} \)
\( \sin(c) = \frac{1}{1.45} \approx 0.6897 \)
\( c = \arcsin(0.6897) \approx 43.6^\circ \) (or \( 44^\circ \) to 2 significant figures).

- 0.5 marks for recalling and using the formula \( \sin(c) = 1/n \).
- 0.5 marks for correctly substituting refractive index \( 1/1.45 \).
- 0.5 marks for the correct critical angle: \( 43.6^\circ \) (or \( 44^\circ \)).

For an ideal transformer (100% efficient):
\( V_p I_p = V_s I_s \) or \( \frac{I_p}{I_s} = \frac{N_s}{N_p} \).

Using the turns ratio:
\( I_p = I_s \times \frac{N_s}{N_p} = 3.0\text{ A} \times \frac{40}{800} = 3.0 \times 0.05 = 0.15\text{ A} \).

- 0.5 marks for recalling the current-turns relationship \( I_p / I_s = N_s / N_p \) or utilizing input/output power equality.
- 0.5 marks for substituting values: \( 3.0 \times (40 / 800) \).
- 0.5 marks for the correct final current: \( 0.15\text{ A} \).

The change in wavelength \( \Delta \lambda = \lambda_{observed} - \lambda_{emitted} = 663\text{ nm} - 656\text{ nm} = 7\text{ nm} \).
Because the observed wavelength is longer than the emitted reference wavelength, the light is redshifted, indicating the galaxy is moving away from the Earth.

- 0.5 marks for calculating the correct change in wavelength: \( 7\text{ nm} \).
- 1.0 marks for stating 'moving away' and explaining that the wavelength has increased (redshifted).

Determine the number of half-lives that have passed by repeatedly halving the initial activity:
\( 1600 \to 800 \to 400 \to 200 \).
This is 3 half-lives.

Calculate the duration of one half-life:
\( 3 \times T_{1/2} = 18\text{ hours} \)
\( T_{1/2} = \frac{18}{3} = 6.0\text{ hours} \).

- 0.5 marks for showing that 3 half-lives have elapsed (e.g. \( 1/8 \) of activity remaining).
- 0.5 marks for dividing the total time by the number of half-lives: \( 18 / 3 \).
- 0.5 marks for the correct half-life: \( 6.0\text{ hours} \) (or \( 6\text{ hours} \)).

First, find the equivalent resistance of the two parallel resistors:
\( R_{parallel} = \frac{R \times R}{R + R} = \frac{R}{2} = 0.5 R \).

Next, add the series resistor to get the total resistance:
\( R_{total} = R_{parallel} + R = 0.5 R + R = 1.5 R \).

Set this equal to the given total resistance:
\( 1.5 R = 15\ \Omega \)
\( R = \frac{15}{1.5} = 10\ \Omega \).

- 0.5 marks for finding the correct resistance of the parallel branch: \( 0.5R \).
- 0.5 marks for expressing the total combined resistance as \( 1.5R \).
- 0.5 marks for the correct value of individual resistance with unit: \( 10\ \Omega \).

To find the number of loops on the secondary winding, we use the transformer turn ratio equation:

\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)

Where:
- \(V_p = 110\text{ V}\) (primary potential difference)
- \(V_s = 5.5\text{ V}\) (secondary potential difference)
- \(N_p = 2200\) (primary loops)
- \(N_s\) is the number of secondary loops

Rearranging the equation to solve for \(N_s\):

\(N_s = N_p \times \frac{V_s}{V_p}\)

\(N_s = 2200 \times \frac{5.5\text{ V}}{110\text{ V}}\)

\(N_s = 2200 \times 0.05 = 110\)

Thus, the secondary winding has \(110\) loops.

- **0.5 marks**: For state or use of transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) or correct substitution of values: \(\frac{110}{5.5} = \frac{2200}{N_s}\).
- **1.0 mark**: For final correct calculation of secondary turns, giving \(110\).

First, define one direction of motion as positive. Let the rebound direction be positive.
- Initial velocity, \(u = -25\text{ m/s}\)
- Final velocity, \(v = +35\text{ m/s}\)
- Mass, \(m = 0.060\text{ kg}\)
- Contact time, \(\Delta t = 0.012\text{ s}\)

The change in momentum (\(\Delta p\)) of the ball is given by:

\(\Delta p = m(v - u)\)

\(\Delta p = 0.060\text{ kg} \times (35\text{ m/s} - (-25\text{ m/s}))\)

\(\Delta p = 0.060\text{ kg} \times 60\text{ m/s} = 3.6\text{ kg m/s}\) (or \(3.6\text{ N s}\))

The average force (\(F\)) is defined as the rate of change of momentum:

\(F = \frac{\Delta p}{\Delta t}\)

\(F = \frac{3.6\text{ kg m/s}}{0.012\text{ s}} = 300\text{ N}\)

- **0.5 marks**: For calculating the change in velocity (\(60\text{ m/s}\)) or change in momentum (\(3.6\text{ kg m/s}\)), or for showing the force formula: \(F = \frac{m(v - u)}{t}\) with correct sign convention usage.
- **1.0 mark**: For correct calculation of force to give \(300\text{ N}\), with the unit 'N' or 'Newtons'.

1. Determine the work done against gravity: \(W = mgh = 150 \times 9.8 \times 12.0 = 17640\text{ J}\).
2. Calculate the average power output: \(P = \frac{W}{t} = \frac{17640}{20} = 882\text{ W}\).

C1 for stating or using \(W = mgh\) or \(P = \frac{W}{t}\)
C1 for correct substitution of values: \(\frac{150 \times 9.8 \times 12.0}{20}\)
A1 for final answer of \(882\text{ W}\) (accept \(0.882\text{ kW}\)) with correct unit.

1. Use the principle of conservation of momentum: \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f\).
2. Substitute the given values: \(60 \times 4.5 + 40 \times 0 = (60 + 40) \times v_f\).
3. Solve for the final velocity: \(270 = 100 \times v_f \implies v_f = 2.7\text{ m/s}\).

C1 for stating or using the conservation of momentum formula: \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v\)
C1 for correct substitution: \(60 \times 4.5 = 100 \times v\)
A1 for correct final answer: \(2.7\text{ m/s}\) (with correct unit).

1. Calculate the thermal energy supplied by the heater: \(E = P \times t = 150 \times (4.0 \times 60) = 36000\text{ J}\).
2. Use the specific heat capacity formula: \(\Delta E = m c \Delta \theta\).
3. Rearrange and solve for \(c\): \(c = \frac{\Delta E}{m \Delta \theta} = \frac{36000}{2.5 \times 16} = 900\text{ J/(kg }^\circ\text{C)}\).

C1 for calculating the energy input: \(150 \times 240 = 36000\text{ J}\)
C1 for rearranging or substituting into \(\Delta E = m c \Delta \theta\)
A1 for final answer of \(900\text{ J/(kg }^\circ\text{C)}\)
Reject: incorrect unit

1. Use Snell's Law: \(n = \frac{\sin(i)}{\sin(r)} 2. Rearrange to find \)\sin(r)\): \(\sin(r) = \frac{\sin(40.0^\circ)}{1.50}\).
3. Calculate the values: \(\sin(40.0^\circ) \approx 0.6428 \implies \sin(r) = \frac{0.6428}{1.50} \approx 0.4285\).
4. Find the angle \(r\): \(r = \arcsin(0.4285) \approx 25.4^\circ\).

C1 for stating or using \(n = \frac{\sin(i)}{\sin(r)}\)
C1 for calculating \(\sin(r) \approx 0.429\)
A1 for correct final angle: \(25.4^\circ\) (accept range \(25^\circ\) to \(25.4^\circ\)) with correct unit.

1. Calculate the current in the wire using Ohm's Law: \(I = \frac{V}{R} = \frac{12}{8.0} = 1.5\text{ A}\).
2. Use the relation between charge, current, and time: \(Q = I \times t\).
3. Convert time to seconds: \(t = 5.0 \times 60 = 300\text{ s}\).
4. Calculate the charge: \(Q = 1.5 \times 300 = 450\text{ C}\).

C1 for calculating the current: \(1.5\text{ A}\) (using \(I = \frac{V}{R}\))
C1 for converting time to seconds and substituting into \(Q = I t\)
A1 for final answer of \(450\text{ C}\) with correct unit.

1. Use the wave equation: \(v = f \lambda\).
2. Rearrange the equation to find wavelength: \(\lambda = \frac{v}{f}\).
3. Substitute the values: \(\lambda = \frac{5100}{850} = 6.0\text{ m}\).

C1 for stating or using \(v = f \lambda\)
C1 for correct substitution: \(\lambda = \frac{5100}{850}\)
A1 for final answer of \(6.0\text{ m}\) with correct unit.

1. Determine the number of half-lives that have elapsed: \(\frac{80}{640} = \frac{1}{8} = \left(\frac{1}{2}\right)^3\). This represents \(3\) half-lives.
2. Set up the relation: \(3 \times T_{1/2} = 18\text{ hours}\).
3. Calculate the half-life: \(T_{1/2} = \frac{18}{3} = 6.0\text{ hours}\).

C1 for identifying that the activity has halved 3 times (or \(1/8\) of the initial activity remains)
C1 for relating \(3\) half-lives to the time period of \(18\text{ hours}\)
A1 for final answer of \(6.0\text{ hours}\) (accept \(6\text{ h}\) or \(21600\text{ s}\)) with correct unit.

1. Determine the period in seconds: \(T = 5.0 \times 3600 = 18000\text{ s}\).
2. Use the orbital speed formula: \(v = \frac{2 \pi r}{T}\).
3. Substitute the given values: \(v = \frac{2 \times \pi \times 8.0 \times 10^6}{18000} \approx 2792.5\text{ m/s}\).
4. Round to 2 significant figures: \(2800\text{ m/s}\).

C1 for converting orbital period into seconds: \(18000\text{ s}\)
C1 for stating or using \(v = \frac{2 \pi r}{T}\)
A1 for final answer of \(2800\text{ m/s}\) (accept range \(2790\text{ m/s}\) to \(2800\text{ m/s}\)) with correct unit.

Using the principle of conservation of momentum:

\( \text{Total initial momentum} = \text{Total final momentum} \)

\( m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \)

Substitute the given values:

\( (0.45 \text{ kg} \times 1.2 \text{ m/s}) + (0.35 \text{ kg} \times 0 \text{ m/s}) = (0.45 \text{ kg} + 0.35 \text{ kg}) \times v \)

\( 0.54 \text{ kg m/s} = 0.80 \text{ kg} \times v \)

\( v = \frac{0.54}{0.80} = 0.675 \text{ m/s} \)

Rounding to 2 significant figures gives \( 0.68 \text{ m/s} \).

C1: Use of conservation of momentum formula: \( m_1 u_1 = (m_1 + m_2) v \) (or equivalent numerical expression)
C1: Correct substitution of values: \( 0.45 \times 1.2 = (0.45 + 0.35) v \)
A1: Correct calculation of velocity with appropriate unit: \( 0.68 \text{ m/s} \) or \( 0.675 \text{ m/s} \)

First, calculate the temperature rise of the water:
\( \Delta\theta = 100\,^{\circ}\text{C} - 18\,^{\circ}\text{C} = 82\,^{\circ}\text{C} \)

Next, calculate the thermal energy required:
\( E = m c \Delta\theta \)
\( E = 0.80 \text{ kg} \times 4200 \text{ J}/(\text{kg}\,^{\circ}\text{C}) \times 82\,^{\circ}\text{C} \)
\( E = 275\,520 \text{ J} \)

Then, use the relationship between energy, power, and time:
\( E = P \times t \)
Where power \( P = 2.2 \text{ kW} = 2200 \text{ W} \)

\( t = \frac{E}{P} = \frac{275\,520 \text{ J}}{2200 \text{ W}} \approx 125.24 \text{ s} \)

Rounding to 2 significant figures gives \( 130 \text{ s} \) (or \( 125 \text{ s} \) to 3 s.f.).

C1: Correct calculation of thermal energy using \( E = m c \Delta\theta \) with \( \Delta\theta = 82\,^{\circ}\text{C} \) (shows \( 275\,520 \text{ J} \) or formula)
C1: Link power and energy using \( t = E / P \) with power converted to \( 2200 \text{ W} \)
A1: Correct calculation of time with unit: \( 130 \text{ s} \) or \( 125 \text{ s} \)

1. The dull black surface is an excellent absorber of thermal/infrared radiation from the Sun, transferring heat efficiently to the copper pipes and the water inside.
2. The glass sheet allows short-wavelength solar radiation (visible light and near-infrared) to pass through but is opaque to the longer-wavelength infrared radiation emitted by the warm copper pipes, trapping this heat (greenhouse effect).
3. The glass sheet also traps a stagnant layer of air above the pipes, which minimizes heat loss by conduction and convection to the colder surrounding atmosphere.

B1: Explains that the dull black paint is an excellent absorber of thermal/infrared radiation.
B1: Explains that the glass sheet allows short-wavelength solar radiation to enter but traps re-radiated longer-wavelength infrared radiation (greenhouse effect).
B1: Explains that the glass sheet traps air to reduce thermal energy losses by conduction or convection.

The impulse acting on the probe is equal to the change in its momentum:
\(\text{Impulse} = F \times \Delta t = 90\text{ N} \times 5.0\text{ s} = 450\text{ N s}\)

The change in momentum is:
\(\Delta p = m \times \Delta v = 450\text{ kg m/s}\)

Therefore, the change in velocity is:
\(\Delta v = \frac{450\text{ kg m/s}}{150\text{ kg}} = 3.0\text{ m/s}\)

Thus, the final velocity of the probe is:
\(v_{\text{final}} = v_{\text{initial}} + \Delta v = 25\text{ m/s} + 3.0\text{ m/s} = 28\text{ m/s}\)

C1: Calculates impulse as \(F \times t = 90 \times 5.0 = 450\text{ N s}\) (or uses \(F = \frac{m(v-u)}{t}\))
C1: Relates impulse to the change in momentum or calculates acceleration as \(0.6\text{ m/s}^2\)
A1: Obtains final velocity of \(28\text{ m/s}\) (accept correct unit \(\text{m/s}\))

1. As the copper ring falls towards the North pole of the magnet, the magnetic flux linking the ring increases.
2. According to Faraday's law of electromagnetic induction, a current is induced in the ring. According to Lenz's law, this induced current must flow in a direction that opposes the change producing it (the falling of the ring).
3. To oppose the approach of the North pole, the induced current must create an upward-facing North pole at the top of the ring. By the right-hand grip rule, this corresponds to an anticlockwise current when viewed from above.

B1: State that the falling motion causes a change in magnetic flux through the ring, inducing an e.m.f. and current (Faraday's Law).
B1: State that Lenz's law dictates the induced current opposes the downward approach of the magnet's North pole, creating an upward North pole on the ring.
B1: Concludes that the induced current flows in an anticlockwise direction as viewed from above.

1. Redshift refers to the increase in the wavelength (and decrease in frequency) of light emitted from a source moving away from the observer, shifting the spectral lines towards the red end of the electromagnetic spectrum.
2. Observations show that light from almost all distant galaxies is redshifted, meaning they are moving away from Earth.
3. Furthermore, the redshift is directly proportional to the distance of the galaxy: more distant galaxies have greater redshifts, indicating they are moving away faster. This uniform expansion of space suggests that in the past, all matter in the Universe was concentrated at a single, incredibly hot and dense point (the Big Bang).

B1: Defines redshift as an increase in the observed wavelength of light because the source is moving away from the observer.
B1: States that more distant galaxies exhibit greater redshift, meaning they are receding faster.
B1: Explains that this uniform expansion of space implies the Universe originated from a single, extremely dense point in the past (supporting the Big Bang theory).

1. For total internal reflection to occur, the light ray must be travelling in a denser medium (higher refractive index) towards a boundary with a less dense medium (lower refractive index). Therefore, \(n_{\text{core}} > n_{\text{cladding}}\).
2. The angle of incidence \(i\) at the boundary between the core and the cladding must be strictly greater than the critical angle \(c\) for that boundary.
3. Under these conditions, the light ray undergoes total internal reflection repeatedly at the boundary, guiding the light along the fiber without any light refracting out into the cladding.

B1: Explains that total internal reflection can only occur when light travels from a higher refractive index medium (core) to a lower refractive index medium (cladding).
B1: States that the angle of incidence at the core-cladding interface must be greater than the critical angle.
B1: Explains that this causes all light to be reflected back into the core, preventing escape (guiding the light).

1. An alpha particle consists of 2 protons and 2 neutrons, which is represented as \(^{4}_{2}\text{He}\) or \(^{4}_{2}\alpha\).
2. Due to the conservation of nucleon number, the nucleon number of the daughter nucleus is \(226 - 4 = 222\).
3. Due to the conservation of proton number (charge), the proton number of the daughter nucleus is \(88 - 2 = 86\). This corresponds to Radon (\(^{222}_{86}\text{Rn}\)).

The decay equation is:
\(^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + ^{4}_{2}\alpha\)

B1: Identifies that an alpha particle removes 4 nucleons and 2 protons from the parent nucleus.
B1: States the daughter nucleus has a nucleon number of 222 and proton number of 86.
B1: Presents the correct balanced nuclear equation: \(^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + ^{4}_{2}\alpha\) (or \(^{4}_{2}\text{He}\)).

1. When the temperature decreases, the resistance of the thermistor increases.
2. This increases the total resistance of the series circuit, which causes the electric current in the circuit to decrease.
3. Since the voltmeter measures the potential difference across the fixed resistor, and the current has decreased, the potential difference \(V = IR\) across the fixed resistor decreases.

B1: Identifies that a decrease in temperature causes the resistance of the thermistor to increase.
B1: Explains that this increases the total resistance, thereby decreasing the current in the circuit.
B1: Concludes that the potential difference across the fixed resistor decreases (since \(V = IR\) and \(R\) is constant).

1. Immediately after jumping, the only force acting on the skydiver is their weight (acting downwards). Because there is a large resultant downward force, the skydiver accelerates downwards at \(g = 9.8\text{ m/s}^2\).
2. As the skydiver's velocity increases, the upward air resistance force acting on them also increases. This reduces the resultant downward force, so the downward acceleration decreases (though speed is still increasing).
3. Eventually, the upward air resistance increases to become equal in magnitude to the downward weight of the skydiver. At this point, the resultant force on the skydiver is zero, the acceleration becomes zero, and they continue to fall at a constant maximum velocity called terminal velocity.

B1: Explains that initially, the only significant force is weight, resulting in maximum downward acceleration.
B1: States that as speed increases, the upward air resistance increases, which decreases the net resultant force (and acceleration).
B1: Explains that terminal velocity is reached when upward air resistance equals downward weight, resulting in zero acceleration.

1. The cooling of the liquid phase is represented by a downward sloping line or curve, showing that temperature is decreasing over time as thermal energy is lost to the surroundings.
2. The solidification process is represented by a horizontal (flat) line at a constant temperature (the freezing point), showing that the temperature remains constant while the substance changes phase and releases latent heat.

1 mark: Downward sloping line/curve to represent the liquid cooling (temperature decreasing with time).
1 mark: Horizontal / flat line at constant temperature to represent the phase change during solidification.

1. At the critical angle, the angle of refraction is exactly \(90^\circ\). Therefore, the refracted ray must be drawn traveling directly along the glass–air boundary.
2. There is also partial reflection at the boundary, so a weaker reflected ray must be drawn inside the glass, reflecting back into the medium at an angle of reflection equal to the angle of incidence.

1 mark: Refracted ray drawn along the boundary surface (at \(90^\circ\) to the normal).
1 mark: Weak reflected ray drawn back into the glass obeying the law of reflection (angle of reflection equal to angle of incidence).

1. To represent a uniform field, the magnetic field lines must be drawn as straight, parallel lines that are equally spaced from one another, indicating constant field strength and direction.
2. To show the correct direction of the field, arrows must be drawn on the lines pointing from the North pole to the South pole.

1 mark: Lines drawn straight, parallel, and with equal spacing.
1 mark: Direction of the field lines shown pointing from North to South.