2025 Cambridge IGCSE Physics (0625) 模擬試題連答案詳解

We use the relationship between weight, mass, and gravitational field strength: \(g = \frac{W}{m}\). Substituting the given values: \(g = \frac{18.5\text{ N}}{5.0\text{ kg}} = 3.7\text{ N/kg}\).

- [1.0 mark] State or use formula \(W = mg\) or \(g = W/m\). - [0.5 mark] Correct final calculation of \(3.7\text{ N/kg}\) (or \(3.7\text{ m/s}^2\)).

Impulse is defined as the change in momentum: \(I = \Delta p = m(v - u)\). If we take the initial direction as negative and final as positive, then \(u = -15\text{ m/s}\) and \(v = 25\text{ m/s}\). Hence, the change in velocity is \(\Delta v = 25 - (-15) = 40\text{ m/s}\). The impulse is \(I = 0.060\text{ kg} \times 40\text{ m/s} = 2.4\text{ N s}\).

- [1.0 mark] Correctly identify the change in velocity as \(40\text{ m/s}\) or use the impulse formula with opposite signs. - [0.5 mark] Correct calculation of impulse as \(2.4\text{ N s}\).

The matte black can cools down faster because dark, matte surfaces are much better emitters of thermal radiation (infrared radiation) than shiny, light surfaces. This allows the hot water inside the matte black can to lose thermal energy to the surroundings at a higher rate. The primary process responsible is thermal radiation.

- [1.0 mark] Identify the matte black can as cooling faster and explain that matte black is a better emitter of infrared/thermal radiation. - [0.5 mark] Correctly name radiation as the primary transfer mechanism.

First, convert time into seconds: \(t = 5.0\text{ minutes} \times 60\text{ s/minute} = 300\text{ s}\). Use the formula relating charge, current, and time: \(Q = I \times t\). Substituting the values: \(Q = 0.12\text{ A} \times 300\text{ s} = 36\text{ C}\).

- [0.5 mark] Correct conversion of time to \(300\text{ s}\). - [1.0 mark] Correct formula usage and final calculation of \(36\text{ C}\).

First, calculate the operating current using \(I = \frac{P}{V}\). Since \(P = 2.3\text{ kW} = 2300\text{ W}\), the current is \(I = \frac{2300\text{ W}}{230\text{ V}} = 10\text{ A}\). The fuse rating must be slightly higher than the operating current to prevent melting under normal use. Therefore, a \(13\text{ A}\) fuse is the most appropriate selection.

- [0.5 mark] Correct calculation of the operating current as \(10\text{ A}\). - [1.0 mark] Correct choice of the \(13\text{ A}\) fuse with explanation that the fuse rating must exceed the operating current.

The refractive index \(n\) is given by the formula \(n = \frac{c}{v}\), where \(c\) is the speed of light in a vacuum and \(v\) is the speed of light in the medium. Rearranging for \(v\): \(v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.50} = 2.0 \times 10^8\text{ m/s}\).

- [1.0 mark] Recall and rearrange the formula \(n = c/v\) to \(v = c/n\). - [0.5 mark] Correct calculation of \(2.0 \times 10^8\text{ m/s}\).

The lower number (proton number, \(Z\)) is 6, indicating there are 6 protons. The upper number (nucleon number, \(A\)) is 14. The number of neutrons is found by subtracting the proton number from the nucleon number: \(14 - 6 = 8\) neutrons.

- [0.5 mark] Correctly identify 6 protons. - [1.0 mark] Correctly calculate and identify 8 neutrons.

First, convert the orbital period from days to seconds: \(T = 365 \times 24 \times 3600\text{ s} = 3.1536 \times 10^7\text{ s}\). Next, calculate the orbital speed using \(v = \frac{2\pi r}{T}\). Substituting the given radius \(r = 1.50 \times 10^{11}\text{ m}\): \(v = \frac{2 \times \pi \times 1.50 \times 10^{11}}{3.1536 \times 10^7} \approx 2.99 \times 10^4\text{ m/s}\), which rounds to \(3.0 \times 10^4\text{ m/s}\).

- [0.5 mark] Correct conversion of the period to seconds. - [1.0 mark] Correct substitution into \(v = 2\pi r / T\) and final calculation yielding around \(2.99 \times 10^4\text{ m/s}\) or \(3.0 \times 10^4\text{ m/s}\).

The distance covered during acceleration is given by: \(d_1 = 0.5 \times v \times t_1 = 0.5 \times 12 \times 4.0 = 24\text{ m}\). The distance covered during the constant speed phase is: \(d_2 = v \times t_2 = 12 \times 6.0 = 72\text{ m}\). The total distance is \(d = 24 + 72 = 96\text{ m}\). The total time is 10.0 s. The average speed is: \(v = \frac{\text{total distance}}{\text{total time}} = \frac{96}{10} = 9.6\text{ m/s}\).

0.5 marks for calculating total distance (96 m) or demonstrating correct method to find area under the v-t graph. 1.0 marks for correct average speed with unit (9.6 m/s).

The extension of the spring is: \(x = 21.0\text{ cm} - 15.0\text{ cm} = 6.0\text{ cm} = 0.060\text{ m}\). Using Hooke's Law \(F = kx\), we can solve for the spring constant: \(k = \frac{F}{x} = \frac{24.0\text{ N}}{0.060\text{ m}} = 400\text{ N/m}\).

0.5 marks for calculating the extension in meters (0.060 m) or stating the formula F = kx. 1.0 marks for the correct final spring constant value with unit (400 N/m).

First, calculate the useful power output of the motor: \(P_{\text{out}} = 0.75 \times 600\text{ W} = 450\text{ W}\). The force needed to lift the mass at constant speed is equal to its weight: \(F = m \times g = 15\text{ kg} \times 9.8\text{ N/kg} = 147\text{ N}\). Using the power formula \(P = F \times v\), the vertical speed is: \(v = \frac{P_{\text{out}}}{F} = \frac{450}{147} \approx 3.06\text{ m/s}\), which rounds to 3.1 m/s.

0.5 marks for finding the useful power output (450 W) or writing the power-force-velocity relation. 1.0 marks for the correct speed value with unit (3.1 m/s or 3.06 m/s).

Conduction in copper involves two main processes: 1) Lattice vibrations: Ions at the heated end gain kinetic energy, vibrate with larger amplitude, and pass energy to neighboring ions through collisions. 2) Free electron diffusion: Delocalised free electrons gain kinetic energy and move rapidly through the metal structure, colliding with ions and transferring energy quickly.

0.5 marks for mentioning vibrating ions transferring energy to neighbors. 1.0 marks for explaining the role of free/delocalised electrons in diffusing and transferring energy through collisions.

First, convert the time to seconds: \(t = 8.0 \times 60 = 480\text{ s}\). Calculate the total charge passed: \(Q = I \times t = 0.45\text{ A} \times 480\text{ s} = 216\text{ C}\). The number of electrons is: \(N = \frac{Q}{e} = \frac{216}{1.6 \times 10^{-19}} = 1.35 \times 10^{21}\).

0.5 marks for calculating the total charge Q = 216 C or using Q = It. 1.0 marks for the correct number of electrons (1.35 x 10^21 or 1.4 x 10^21).

Common methods of demagnetisation are: 1) Heating the magnet to a high temperature, causing thermal agitation to disrupt magnetic domain alignment. 2) Placing the magnet inside a solenoid connected to an alternating current (a.c.) supply, then slowly withdrawing it along the axis of the solenoid. 3) Hammering the magnet repeatedly when oriented in an East-West direction.

0.5 marks for stating one valid method of demagnetisation. 1.0 marks for stating two valid methods.

Using the formula for critical angle: \(\sin(c) = \frac{1}{n}\), we substitute the refractive index: \(\sin(c) = \frac{1}{1.52} \approx 0.6579\). Taking the inverse sine: \(c = \sin^{-1}(0.6579) \approx 41.1^{\circ}\).

0.5 marks for stating or using the formula sin(c) = 1/n. 1.0 marks for the correct critical angle with unit (41.1 degrees or 41 degrees).

Calculate the number of half-lives that have elapsed: \(n = \frac{20.0}{4.0} = 5.0\text{ half-lives}\). After 5 half-lives, the count rate is reduced by a factor of \(2^5 = 32\). Therefore, the final count rate is: \(\text{Count rate} = \frac{1600}{32} = 50\text{ counts/second}\).

0.5 marks for identifying that 5 half-lives have elapsed. 1.0 marks for the correct final count rate with appropriate units (50 counts/second or 50 Bq).

1. Calculate the mass of the space probe using its weight on Earth: \(m = \frac{W_{\text{Earth}}}{g_{\text{Earth}}} = \frac{2450\text{ N}}{9.8\text{ N/kg}} = 250\text{ kg}\). 2. Calculate the weight of the space probe on Mars: \(W_{\text{Mars}} = m \times g_{\text{Mars}} = 250\text{ kg} \times 3.7\text{ N/kg} = 925\text{ N}\).

1.0 mark: Correct calculation of mass (\(250\text{ kg}\)) or correct formula substitution. 0.5 marks: Correct final weight with unit (\(925\text{ N}\)).

1. Calculate the total initial momentum of the system: \(p_{\text{initial}} = m_1 u_1 + m_2 u_2 = (0.80\text{ kg} \times 3.0\text{ m/s}) + (1.2\text{ kg} \times 0) = 2.4\text{ kg m/s}\). 2. Use the principle of conservation of momentum: \(p_{\text{initial}} = p_{\text{final}} = (m_1 + m_2) v\). 3. Calculate the common velocity: \(2.4\text{ kg m/s} = (0.80\text{ kg} + 1.2\text{ kg}) \times v \implies v = \frac{2.4}{2.0} = 1.2\text{ m/s}\).

1.0 mark: Correct calculation of initial momentum (\(2.4\text{ kg m/s}\)) or conservation formula. 0.5 marks: Correct final velocity with unit (\(1.2\text{ m/s}\)).

First, calculate the average speed: \(v_{\text{avg}} = \frac{u + v}{2} = \frac{0 + 24}{2} = 12\text{ m/s}\). Then, use the distance formula: \(s = v_{\text{avg}} \times t = 12 \times 6.0 = 72\text{ m}\). Alternatively, calculate the acceleration first: \(a = \frac{v - u}{t} = \frac{24}{6.0} = 4.0\text{ m/s}^2\), and then find distance: \(s = \frac{1}{2} a t^2 = 0.5 \times 4.0 \times (6.0)^2 = 72\text{ m}\).

1 Mark: Correct formula for acceleration or average velocity.
1 Mark: Correct substitution of values.
0.5 Mark: Correct final answer with appropriate unit (72 m).

By the conservation of momentum, the total momentum before the collision equals the total momentum after the collision: \(m_A u_A + m_B u_B = (m_A + m_B) v\). Substituting the given values: \((1.5 \times 4.0) + (2.5 \times 0) = (1.5 + 2.5) v \Rightarrow 6.0 = 4.0 v \Rightarrow v = 1.5\text{ m/s}\).

1 Mark: State conservation of momentum or formula \(m_A u_A = (m_A + m_B) v\).
1 Mark: Correct substitution of values.
0.5 Mark: Correct final velocity value and unit (1.5 m/s).

First, calculate the useful work done, which is equal to the gravitational potential energy gained: \(\Delta E_p = mgh = 45 \times 9.8 \times 12 = 5292\text{ J}\). Next, calculate the useful power output: \(P = \frac{\text{Work done}}{\text{time}} = \frac{5292}{15} = 352.8\text{ W}\). Rounding to standard two significant figures gives \(350\text{ W}\) (or \(353\text{ W}\) to three significant figures).

1 Mark: Correct calculation of gravitational potential energy gained (5292 J) using \(E_p = mgh\).
1 Mark: Correct use of power formula \(P = \Delta E / t\) with substitution.
0.5 Mark: Correct final answer with unit (350 W or 353 W or 352.8 W).

First, convert the time from minutes to seconds: \(t = 5.0 \times 60 = 300\text{ s}\). Then, use the formula for charge: \(Q = I \times t = 0.80 \times 300 = 240\text{ C}\).

1 Mark: Correctly converting time to seconds (300 s).
1 Mark: Correct formula \(Q = I t\) and substitution.
0.5 Mark: Correct final answer with unit (240 C).

First, find the extension of the spring: \(x = 19.8\text{ cm} - 15.0\text{ cm} = 4.8\text{ cm}\). Convert this extension to meters: \(x = 0.048\text{ m}\). Using Hooke's Law \(F = kx\), solve for the spring constant: \(k = \frac{F}{x} = \frac{6.0}{0.048} = 125\text{ N/m}\).

1 Mark: Finding the extension in meters (0.048 m) or centimeters (4.8 cm).
1 Mark: Hooke's law formula \(F = kx\) correctly rearranged to \(k = F/x\) and substitution.
0.5 Mark: Correct value and unit (125 N/m).

Use the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Rearrange to find \(V_s\): \(V_s = V_p \times \frac{N_s}{N_p}\). Substitute the values: \(V_s = 240 \times \frac{60}{1200} = 240 \times \frac{1}{20} = 12\text{ V}\).

1 Mark: Correct formula \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\).
1 Mark: Correct substitution and rearrangement.
0.5 Mark: Correct final answer with unit (12 V).

Determine the number of half-lives that have elapsed by halving the activity: \(800 \rightarrow 400 \rightarrow 200 \rightarrow 100\). This represents exactly \(3\) half-lives. Therefore, \(3 \times T_{1/2} = 18\text{ hours} \Rightarrow T_{1/2} = \frac{18}{3} = 6.0\text{ hours}\).

1 Mark: Show deduction that 3 half-lives have elapsed (e.g., halving process or using \(800 / 2^n = 100\)).
1 Mark: Correct calculation of half-life (e.g., \(18 / 3\)).
0.5 Mark: Correct final answer with unit (6.0 hours or 6 hours).

First, determine the orbital radius \(r = \text{radius of Earth} + \text{altitude} = 6400\text{ km} + 400\text{ km} = 6800\text{ km}\). Next, convert the orbital period to seconds: \(T = 92 \times 60 = 5520\text{ s}\). Use the orbital speed formula: \(v = \frac{2\pi r}{T} = \frac{2 \times \pi \times 6800}{5520} \approx 7.74\text{ km/s}\). Rounding to 2 significant figures gives \(7.7\text{ km/s}\).

1 Mark: Finding orbital radius (6800 km) and period in seconds (5520 s).
1 Mark: Correct formula \(v = \frac{2\pi r}{T}\) and substitution of values.
0.5 Mark: Correct final answer to 2 significant figures with unit (7.7 km/s).

1. Calculate the useful work done (energy output) to lift the mass: \(E_{\text{out}} = mgh = 5.0\text{ kg} \times 9.8\text{ N/kg} \times 6.0\text{ m} = 294\text{ J}\). 2. Calculate the total electrical energy input to the motor: \(E_{\text{in}} = P \times t = 50\text{ W} \times 8.0\text{ s} = 400\text{ J}\). 3. Calculate the efficiency: \(\text{Efficiency} = \frac{E_{\text{out}}}{E_{\text{in}}} \times 100\% = \frac{294}{400} \times 100\% = 73.5\%\).

Formula for work done/gravitational potential energy: \(E = mgh\) [1.0 mark]; Correct calculation of useful energy output of \(294\text{ J}\) or useful power output of \(36.75\text{ W}\) [0.5 mark]; Formula for efficiency: \(\text{efficiency} = \frac{\text{useful energy}}{\text{total energy}} \times 100\) [0.5 mark]; Correct final answer: \(73.5\%\) (accept \(74\%\) or \(0.735\) or \(0.74\)) [0.5 mark]

1. Calculate the refractive index \(n\) of the plastic using Snell's Law: \(n = \frac{\sin(r)}{\sin(i)} = \frac{\sin(53^\circ)}{\sin(32^\circ)} \approx \frac{0.7986}{0.5299} \approx 1.507\). 2. Use the critical angle formula: \(\sin(c) = \frac{1}{n} = \frac{1}{1.507} \approx 0.6636\). 3. Solve for the critical angle: \(c = \sin^{-1}(0.6636) \approx 41.5^\circ\).

Formula for refractive index: \(n = \frac{\sin(r)}{\sin(i)}\) [1.0 mark]; Correct calculation of refractive index \(n = 1.51\) (or \(1.5\)) [0.5 mark]; Formula for critical angle: \(\sin(c) = \frac{1}{n}\) [0.5 mark]; Correct final critical angle: \(41.5^\circ\) (accept range \(41^\circ\) to \(42^\circ\)) [0.5 mark]

1. Thermal radiation (infrared radiation) does not require a medium to travel and can pass through a vacuum. 2. Shiny, silvered surfaces are excellent reflectors of radiation and very poor emitters of radiation. 3. The silvered surface facing the inside reflects radiation back into the hot liquid, while the silvered surface facing outwards minimizes the emission of radiation across the vacuum gap, thereby reducing thermal energy loss.

M1: Identify that thermal radiation (or infrared) is the mode of heat transfer that can travel across the vacuum. [1] M2: State that shiny/silvered surfaces are poor emitters of radiation (or good reflectors of radiation). [1] M3: Explain that this reflection back into the flask or poor emission across the vacuum reduces thermal energy loss. [1]

1. The light ray enters the core and strikes the boundary between the core and the cladding. Because the core has a higher refractive index than the cladding, the light is traveling in an optically denser medium towards an optically less dense medium. 2. When the angle of incidence at this boundary is greater than the critical angle, total internal reflection (TIR) occurs, and no light refracts out of the core. 3. The light ray is repeatedly reflected along the inside of the fiber, enabling it to travel long distances with minimal loss of signal.

M1: State that the light undergoes total internal reflection (TIR) along the core. [1] M2: State condition 1: the light must travel from a medium of higher refractive index to one of lower refractive index (or denser to less dense). [1] M3: State condition 2: the angle of incidence at the boundary must be greater than the critical angle. [1]

1. While the direct current is switched on, both the soft iron and steel bars become magnetized by the magnetic field of the solenoid. 2. Soft iron is a magnetically soft material, which means it is easy to magnetize but loses its magnetism easily. Thus, when the current is switched off, it loses its magnetization. 3. Steel is a magnetically hard material, meaning it is harder to magnetize but retains its magnetism. Thus, when the current is switched off, it remains magnetized as a permanent magnet.

M1: State that both bars become magnetized when the current is on. [1] M2: Explain that soft iron is a temporary/soft magnetic material and loses its magnetism when the current is switched off. [1] M3: Explain that steel is a permanent/hard magnetic material and retains its magnetism after the current is switched off. [1]

1. Initially, only gravity (weight) acts downwards on the skydiver, causing them to accelerate downwards at \( 9.8\text{ m/s}^2 \). 2. As the skydiver's speed increases, the air resistance acting upwards increases. This reduces the resultant downward force (\( F = W - R \)), which in turn decreases the acceleration according to \( F = ma \). 3. Eventually, the upward air resistance increases until it is equal in magnitude and opposite in direction to the downward weight. The resultant force becomes zero, so the acceleration becomes zero, and the skydiver falls at a constant terminal velocity.

M1: State that air resistance increases as the skydiver's speed increases. [1] M2: Explain that the increased air resistance reduces the resultant force, causing the acceleration to decrease. [1] M3: State that terminal velocity is reached when air resistance equals weight, resulting in zero resultant force and zero acceleration. [1]

1. The centripetal force required to keep the car moving in a circular path is provided by the static friction between the car's tires and the road surface. 2. On a wet road, the film of water reduces the coefficient of friction, lowering the maximum available frictional force between the tires and the road. 3. The required centripetal force is given by \( F = \frac{mv^2}{r} \). Doubling the speed increases the required centripetal force by a factor of four (\( 2^2 \)). If this required force exceeds the maximum available friction, the car cannot sustain the circular motion and skids outwards.

M1: Identify friction between the tires and the road as the source of the centripetal force. [1] M2: Explain that water on a wet road reduces the maximum available frictional force. [1] M3: State that required centripetal force increases with the square of the speed (\( v^2 \)), so doubling the speed requires four times the force, making it more likely to exceed available friction. [1]

1. Most of the alpha particles passed straight through the gold foil with little or no deflection. This indicates that the atom consists mostly of empty space, with the nucleus occupying an extremely small volume. 2. A very small fraction of alpha particles were deflected through very large angles (some even rebounding back). This indicates that almost all the mass of the atom is concentrated in a tiny, dense region (the nucleus). 3. Since alpha particles are positively charged, the deflection (repulsion) of these particles at large angles indicates that the nucleus also carries a concentrated positive charge.

M1: State that most particles passing straight through shows the atom is mostly empty space / the nucleus is very small. [1] M2: State that large-angle deflection / rebounding of a few particles shows that the mass is concentrated in a dense nucleus. [1] M3: Explain that repulsion (deflection) of positive alpha particles indicates the nucleus has a positive charge. [1]

1. As the magnet falls towards the copper ring, the magnetic flux passing through the ring increases. This change in magnetic flux linkage induces an electromotive force (e.m.f.) and consequently an induced current in the closed copper ring. 2. According to Lenz's law, the direction of the induced current is such that its magnetic field opposes the change that produced it. 3. To oppose the approach of the magnet's downward pole (e.g., a North pole), the induced current creates a magnetic field with a matching pole (e.g., North) at the top of the ring. The resulting repulsive force acts upwards, opposing the downward motion of the falling magnet.

M1: Explain that the falling magnet causes a changing magnetic flux linkage through the copper ring, inducing an e.m.f./current. [1] M2: State Lenz's law: the induced current/magnetic field opposes the change in magnetic flux (or opposes the motion of the magnet). [1] M3: Explain that the ring develops a magnetic pole of the same polarity facing the magnet, creating an upward repulsive force. [1]

1. Inside the Sun, hydrogen nuclei (protons) collide and fuse together to form helium nuclei. 2. The total mass of the resulting helium nuclei is slightly less than the total mass of the initial hydrogen nuclei. This lost mass (mass defect) is converted into a huge amount of energy, released as electromagnetic radiation, according to \( E = \Delta m c^2 \). 3. Nuclear fusion requires extremely high temperatures (millions of degrees Celsius) and high pressures. This high thermal energy is necessary to give the positively charged hydrogen nuclei enough kinetic energy to overcome the strong electrostatic repulsion between them so they can get close enough to fuse.

M1: State that hydrogen nuclei fuse together to form helium nuclei. [1] M2: Explain that mass is lost/decreased during fusion, and this mass is converted into energy. [1] M3: State that high temperature and pressure are required to overcome the electrostatic repulsion between the positive nuclei. [1]

The car starts from rest, so at time \(t = 0\text{ s}\), the speed is \(v = 0\text{ m/s}\). This gives the starting point \((0, 0)\). Since the acceleration is constant at \(2.0\text{ m/s}^2\), the gradient of the speed-time graph is constant, meaning the line must be straight. The final speed after \(5.0\text{ s}\) is calculated using \(v = u + at = 0 + (2.0 \times 5.0) = 10\text{ m/s}\). Thus, the ending coordinates of this straight line are \((5.0, 10)\).

[1] State that the line is straight and starts at the origin (0, 0).
[1] State that the line ends at the coordinates (5.0, 10).

To locate an image using a ray diagram for a thin converging lens, two standard rays are drawn from the top of the object: 1) A ray parallel to the principal axis, which refracts through the focal point (principal focus) on the opposite side of the lens. 2) A ray passing through the optical centre of the lens, which continues in a straight line without bending or deviation.

[1] Describe a ray parallel to the principal axis refracting through the principal focus / focal point on the other side.
[1] Describe a ray passing straight through the optical centre of the lens without bending / changing direction.

When plane waves pass through a gap that is much wider than their wavelength, very little diffraction occurs. The wavefronts passing through the middle of the gap continue to travel forward as straight, parallel lines. Slight diffraction (curving) occurs only at the outer edges of the wavefronts where they pass near the edges of the barrier.

[1] State that the wavefronts remain straight or plane in the central region.
[1] State that the wavefronts are curved or bent only at the edges.