2026 Cambridge IGCSE Physics (0625) 模擬試題連答案詳解

According to 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:
$$(0.40\text{ kg} \times 1.8\text{ m/s}) + (0.20\text{ kg} \times 0) = (0.40\text{ kg} + 0.20\text{ kg}) \times v$$
$$0.72 = 0.60 \times v$$
$$v = \frac{0.72}{0.60} = 1.2\text{ m/s}$$

The trolleys move together at \(1.2\text{ m/s}\) in the original direction of motion.

[C1] Correct momentum formula or substitution: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\) or \(0.40 \times 1.8 = 0.60 \times v\)
[A1] Calculation: \(v = 1.2\)
[B0.5] Correct unit: \(\text{m/s}\) or \(\text{m s}^{-1}\)

1. Identify the initial and final velocities taking direction into account:
\(u = +24\text{ m/s}\), \(v = -18\text{ m/s}\)
2. Calculate the change in momentum (\(\Delta p\)):
\(\Delta p = m(v - u) = 0.058\text{ kg} \times (-18 - 24)\text{ m/s} = 0.058 \times (-42) = -2.436\text{ kg m/s}\)
3. Calculate the average force using \(F = \frac{\Delta p}{\Delta t}\):
\(F = \frac{2.436\text{ N s}}{0.012\text{ s}} = 203\text{ N}\)
To 2 significant figures: \(200\text{ N}\) (or \(2.0 \times 10^2\text{ N}\)).

[C1] For correct formula for change of momentum or force: \(\Delta p = m \Delta v\) OR \(F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}\)
[C1] For correct substitution taking velocity directions into account: \(0.058 \times (18 - (-24))\) OR \(0.058 \times 42\) [= 2.436 N s]
[A0.5] For correct final magnitude with unit: \(203\text{ N}\) or \(200\text{ N}\)

1. Convert time to seconds:
\(t = 3.5\text{ min} \times 60\text{ s/min} = 210\text{ s}\)
2. Calculate the total thermal energy transferred (\(\Delta E\)):
\(\Delta E = P \times t = 450\text{ W} \times 210\text{ s} = 94\,500\text{ J}\)
3. Calculate the temperature change (\(\Delta \theta\)):
\(\Delta \theta = 73 - 21 = 52\text{ }^\circ\text{C}\)
4. Calculate the specific heat capacity \(c\):
\(c = \frac{\Delta E}{m \Delta \theta} = \frac{94\,500}{1.2 \times 52} = \frac{94\,500}{62.4} \approx 1514\text{ J/(kg }^\circ\text{C)}\)
To 2 or 3 significant figures: \(1500\text{ J/(kg }^\circ\text{C)}\) or \(1510\text{ J/(kg }^\circ\text{C)}\).

[C1] For calculating electrical energy input: \(E = P \times t = 450 \times (3.5 \times 60) = 94\,500\text{ J}\)
[C1] For correct formula and rearrangement: \(c = \frac{\Delta E}{m \Delta \theta}\) with \(\Delta \theta = 52\text{ }^\circ\text{C}\)
[A0.5] For correct numerical value and unit: \(1500\text{ J/(kg }^\circ\text{C)}\) or \(1510\text{ J/(kg }^\circ\text{C)}\) (accept \(\text{J}/(\text{kg K})\))

1. Calculate the refractive index \(n\) of the acrylic:
\(n = \frac{\sin i}{\sin r} = \frac{\sin 48^\circ}{\sin 29^\circ} = \frac{0.7431}{0.4848} \approx 1.533\)
2. Use the critical angle formula \(n = \frac{1}{\sin c}\) to find \(c\):
\(\sin c = \frac{1}{n} = \frac{0.4848}{0.7431} \approx 0.6524\)
3. Calculate \(c\):
\(c = \arcsin(0.6524) \approx 40.7^\circ\)
Rounding gives \(41^\circ\) (or \(40.7^\circ\)).

[C1] For calculating refractive index: \(n = \frac{\sin 48^\circ}{\sin 29^\circ} = 1.53\) (or \(\sin c = \frac{\sin 29^\circ}{\sin 48^\circ}\))
[C1] For correct formula linking critical angle and refractive index: \(\sin c = \frac{1}{n}\)
[A0.5] For correct final angle with unit: \(41^\circ\) or \(40.7^\circ\)

1. Calculate the equivalent resistance of the two parallel resistors (\(R_p\)):
\(\frac{1}{R_p} = \frac{1}{30} + \frac{1}{20} = \frac{2 + 3}{60} = \frac{5}{60} \implies R_p = 12\text{ }\Omega\)
2. Calculate the total circuit resistance (\(R_{\text{total}}\)):
\(R_{\text{total}} = R_p + 18 = 12 + 18 = 30\text{ }\Omega\)
3. Calculate the total current (\(I\)) using Ohm's law:
\(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{30\text{ }\Omega} = 0.40\text{ A}\)

[C1] For calculating equivalent parallel resistance: \(R_p = \frac{30 \times 20}{30 + 20} = 12\text{ }\Omega\)
[C1] For total resistance and formula: \(R_{\text{total}} = 12 + 18 = 30\text{ }\Omega\) and \(I = \frac{V}{R}\)
[A0.5] For correct current with unit: \(0.40\text{ A}\) (accept \(0.4\text{ A}\))

1. (a) For the secondary circuit:
\(P_s = I_s V_s \implies I_s = \frac{P_s}{V_s} = \frac{72\text{ W}}{24\text{ V}} = 3.0\text{ A}\)
2. (b) For a \(100\%\) efficient transformer:
\(P_p = P_s = 72\text{ W}\) (or \(I_p V_p = I_s V_s\))
\(I_p = \frac{P_p}{V_p} = \frac{72\text{ W}}{230\text{ V}} \approx 0.313\text{ A}\)
To 2 significant figures: \(0.31\text{ A}\).

[C1] For secondary current calculation: \(I_s = \frac{P}{V_s} = \frac{72}{24} = 3.0\text{ A}\) with unit
[C1] For using transformer power equality: \(I_p V_p = I_s V_s\) OR \(I_p = \frac{72}{230}\)
[A0.5] For correct primary current with unit: \(0.31\text{ A}\) (or \(0.313\text{ A}\))

1. (a) Use Hubble's equation \(v = H_0 d\):
\(d = \frac{v}{H_0} = \frac{1.65 \times 10^7\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 7.5 \times 10^{24}\text{ m}\)
2. (b) Convert distance to light-years:
\(d_{\text{ly}} = \frac{7.5 \times 10^{24}\text{ m}}{9.5 \times 10^{15}\text{ m/ly}} \approx 7.89 \times 10^8\text{ light-years}\)
To 2 significant figures: \(7.9 \times 10^8\text{ light-years}\).

[C1] For correct formula and rearrangement: \(d = \frac{v}{H_0}\)
[C1] For distance in metres: \(7.5 \times 10^{24}\text{ m}\)
[A0.5] For correct conversion to light-years: \(7.9 \times 10^8\) [light-years] (or \(7.89 \times 10^8\))

(a)
\(\Delta E_p = mgh = 85 \times 9.8 \times 14 = 11\,662\text{ J}\)
\(E_k = \frac{1}{2}mv^2 = 0.5 \times 85 \times (6.0)^2 = 1530\text{ J}\)
Total mechanical energy \(E = 11\,662 + 1530 = 13\,192\text{ J} \approx 1.3 \times 10^4\text{ J}\) (or \(1.32 \times 10^4\text{ J}\)).

(b)
\(\text{Efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%\)
\(\text{Efficiency} = \frac{13\,192}{1.6 \times 10^4} \times 100\% = 82.45\% \approx 82\%\) (or \(82.5\%\) or \(0.82\)).

(a)
- C1: Calculation of \(\Delta E_p\) (\(11\,660\text{ J}\)) OR \(E_k\) (\(1530\text{ J}\))
- A1: Addition shown with unrounded value \(13\,192\text{ J}\) leading to \(1.3 \times 10^4\text{ J}\)

(b)
- C1: (\(\text{useful output} / \text{total input}\)) formula or numerical substitution (e.g. \(13\,192 / 16\,000\) or \(1.3 \times 10^4 / 1.6 \times 10^4\))
- A0.5: Correct percentage \(82\%\) or \(82.5\%\) (or decimal \(0.82\) / \(0.825\)) [ecf from (a)]

(a)
\(\Delta \theta = 75^\circ\text{C} - 22^\circ\text{C} = 53^\circ\text{C}\)
\(\Delta E = mc\Delta \theta = 0.80 \times 2100 \times 53 = 89\,040\text{ J}\)
\(89\,040\text{ J} = 8.904 \times 10^4\text{ J} \approx 8.9 \times 10^4\text{ J}\).

(b)
\(P = \frac{\Delta E}{t} \implies t = \frac{\Delta E}{P}\)
\(t = \frac{89\,040}{450} = 197.87\text{ s} \approx 198\text{ s}\) (or \(200\text{ s}\)).

(a)
- C1: Correct temperature change \(\Delta \theta = 53\,^\circ\text{C}\) and formula \(\Delta E = mc\Delta \theta\)
- A1: Explicit substitution \(0.80 \times 2100 \times 53\) and unrounded value \(89\,040\text{ J}\) shown before stating \(8.9 \times 10^4\text{ J}\)

(b)
- C1: \(t = E / P\) or substitution \(89\,040 / 450\) (or \(8.9 \times 10^4 / 450\))
- A0.5: \(198\text{ s}\) (or \(200\text{ s}\) / \(197.9\text{ s}\)) with correct unit [s]

(a)
\(n = \frac{1}{\sin c} = \frac{1}{\sin(41.8^\circ)} = \frac{1}{0.6665} = 1.5003 \approx 1.5\).

(b)
\(n = \frac{v_{\text{air}}}{v_{\text{glass}}} \implies v_{\text{glass}} = \frac{v_{\text{air}}}{n}\)
\(v_{\text{glass}} = \frac{3.0 \times 10^8}{1.5003} = 2.0 \times 10^8\text{ m/s}\).

(a)
- C1: \(n = 1 / \sin c\) and substitution \(1 / \sin(41.8^\circ)\)
- A0.5: Unrounded value \(1.50\) or \(1.500\) given before stating \(1.5\)

(b)
- C1: \(v = c / n\) or substitution \(3.0 \times 10^8 / 1.5\)
- A1: \(2.0 \times 10^8\text{ m/s}\) (accept \(2.00 \times 10^8\text{ m/s}\)) with correct unit

(a)
Parallel section resistance \(R_p\):
\(\frac{1}{R_p} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} \implies R_p = 15\,\Omega\)
Total circuit resistance \(R_{\text{total}} = R_1 + R_p = 15 + 15 = 30\,\Omega\).

(b)
Method 1 (Current):
\(I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12.0}{30} = 0.40\text{ A}\)
\(V_p = I_{\text{total}} \times R_p = 0.40 \times 15 = 6.0\text{ V}\)

Method 2 (Potential divider):
\(V_p = V \times \frac{R_p}{R_1 + R_p} = 12.0 \times \frac{15}{15 + 15} = 6.0\text{ V}\).

(a)
- C1: Calculation of parallel resistance \(1/R_p = 1/30 + 1/30 \implies R_p = 15\,\Omega\)
- A0.5: Summing in series \(15 + 15 = 30\,\Omega\)

(b)
- C1: Total current calculation \(I = 12.0 / 30 = 0.40\text{ A}\) OR potential divider formula \(12.0 \times (15 / 30)\)
- A1: \(6.0\text{ V}\) (allow \(6\text{ V}\)) with correct unit [V]

(a)
\(P = IV \implies I = \frac{P}{V}\)
\(I = \frac{1.8 \times 10^6\text{ W}}{3.0 \times 10^4\text{ V}} = 60\text{ A}\).

(b)
\(P_{\text{wasted}} = I^2 R\)
\(P_{\text{wasted}} = (60)^2 \times 3.5 = 3600 \times 3.5 = 12\,600\text{ W} = 1.26 \times 10^4\text{ W} \approx 1.3 \times 10^4\text{ W}\) (or \(12.6\text{ kW}\)).

(a)
- C1: Formula \(I = P / V\) and numerical substitution \((1.8 \times 10^6) / (3.0 \times 10^4)\)
- A0.5: Stating evaluated answer \(60\text{ A}\)

(b)
- C1: Formula \(P = I^2 R\) and substitution \((60)^2 \times 3.5\) [ecf current from (a)]
- A1: \(12\,600\text{ W}\) / \(1.26 \times 10^4\text{ W}\) / \(12.6\text{ kW}\) / \(1.3 \times 10^4\text{ W}\) with correct unit

(a)
\(v = H_0 d \implies d = \frac{v}{H_0}\)
\(d = \frac{3.3 \times 10^6}{2.2 \times 10^{-18}} = 1.5 \times 10^{24}\text{ m}\).

(b)
\(\text{Distance in light-years} = \frac{d}{\text{distance of 1 light-year}} = \frac{1.5 \times 10^{24}\text{ m}}{9.5 \times 10^{15}\text{ m/light-year}} = 1.579 \times 10^8\text{ light-years} \approx 1.6 \times 10^8\text{ light-years}\) (or \(1.58 \times 10^8\text{ light-years}\)).

(a)
- C1: Formula \(d = v / H_0\) and substitution \((3.3 \times 10^6) / (2.2 \times 10^{-18})\)
- A0.5: Value \(1.5 \times 10^{24}\text{ m}\) stated clearly

(b)
- C1: Division of distance in metres by \(9.5 \times 10^{15}\) (substitution \((1.5 \times 10^{24}) / (9.5 \times 10^{15})\))
- A1: \(1.6 \times 10^8\) or \(1.58 \times 10^8\) [light-years]

1. Gas particles move randomly at high speeds and frequently collide with the interior walls of the container.
2. During each collision, the particles undergo a change in momentum and exert a force on the wall. The total force exerted per unit area across the walls constitutes gas pressure (\(p = \frac{F}{A}\)).

[B1] Particles move randomly / at high speed AND collide with / hit the walls (of the container)
[B1] Collisions exert a force (due to change in momentum) AND pressure is force per unit area / \(p = \frac{F}{A}\)

When water at the base of the beaker absorbs thermal energy, its particles gain kinetic energy and move further apart, causing the heated water to expand. As a result, its density decreases and the warmer water rises. Cooler, denser water near the top sinks to the bottom to take its place, creating a continuous convection current that distributes thermal energy.

[B1] Heated water expands / particles move further apart AND becomes less dense and rises
[B1] Cooler / denser water sinks (to replace it / forming a convection current)

A vibrating source causes neighbouring air particles to oscillate back and forth parallel to the direction of energy propagation. This oscillation produces compressions, where particles are pushed close together creating regions of higher pressure, and rarefactions, where particles are spread further apart creating regions of lower pressure.

[B1] Particles vibrate / oscillate parallel to the direction of wave / energy travel
[B1] (Produces alternating) compressions where particles are close together / high pressure AND rarefactions where particles are spread apart / low pressure

For a fixed electrical power being transmitted, increasing the transmission voltage reduces the electric current because \(P = IV\). Since the rate of thermal energy loss in the cables is proportional to the square of the current (\(P_{\text{loss}} = I^2 R\)), a lower current dramatically reduces energy wasted as heat in the cables.

[B1] High voltage means lower current (for the same power / \(P = IV\))
[B1] Lower current causes less thermal energy / power loss in cables because \(P = I^2 R\)

An alpha particle carries a double positive charge (\(+2\)) and has a relatively large mass compared to other emissions. This large charge and mass mean it exerts strong electrostatic forces on orbital electrons during collisions, readily stripping them from atoms. In contrast, gamma radiation is uncharged and massless, interacting far less frequently with atomic electrons.

[B1] Alpha particle has a \(+2\) charge / large charge / large mass AND gamma has no charge / is an EM wave
[B1] (Large charge/mass causes) strong electrostatic attraction/repulsion / greater force on electrons, removing/knocking them out of atoms more easily

When the parachute opens, the increased surface area causes air resistance (drag) to become much larger than the downward weight. This produces an upward resultant force, causing the skydiver to decelerate. As speed decreases, the upward air resistance also decreases until it becomes equal and opposite to weight, resulting in zero acceleration and a new, lower terminal velocity.

[B1] Upward air resistance exceeds downward weight / resultant force is upward, causing deceleration / slowing down
[B1] As speed decreases, air resistance decreases until air resistance equals weight (resultant force is zero / constant speed reached)

Redshift means that the spectral absorption lines in light emitted by distant galaxies are observed at longer wavelengths (shifted towards the red end of the visible spectrum) than when produced in a laboratory. This Doppler effect indicates that distant galaxies are receding from us, providing key evidence that the Universe is expanding.

[B1] (Observed) increase in wavelength / decrease in frequency of light (or spectral lines shifted towards red end)
[B1] Galaxies are moving away (from Earth / each other) OR the Universe is expanding

(a) The object is placed at \(u = 3.0\text{ cm}\), which is less than the focal length \(f = 5.0\text{ cm}\) (i.e. \(u < f\)). Therefore, the refracted rays diverge and form a virtual, magnified, and upright image on the same side of the lens as the object.

(b) Standard ray constructions for a converging lens:
1. A ray passing through the optical centre of the lens continues in a straight line without refraction.
2. A ray directed parallel to the principal axis refracts through the principal focus (focal point \(F\)) on the opposite side of the lens.
3. Because the two rays diverge on the opposite side of the lens, they must be projected backwards as virtual (dashed) lines; their point of intersection determines the position of the top of the virtual image.

(a) Virtual [B1] AND reason: object is closer to the lens than the principal focus / \(u < f\) / inside focal length [B1 - sub-point included in 1 mark total allocation].

(b)
- Ray through optical centre passes straight through / undeviated [B0.5]
- Ray parallel to principal axis refracts through the principal focus / focal point [B0.5]
- Image position located where the refracted rays intersect when extrapolated / extended backwards [B0.5]

(a) Using the potential divider formula:
\[ V_{\text{th}} = V_{\text{in}} \times \left( \frac{R_{\text{th}}}{R_1 + R_{\text{th}}} \right) \]
\[ V_{\text{th}} = 9.0\text{ V} \times \left( \frac{600\ \Omega}{1200\ \Omega + 600\ \Omega} \right) = 9.0 \times \frac{600}{1800} = 3.0\text{ V} \]

(b) As temperature increases, the resistance of the NTC thermistor decreases. Since total circuit resistance \(R_{\text{total}} = R_1 + R_{\text{th}}\) decreases, the circuit current \(I = \frac{V}{R_{\text{total}}}\) increases. The potential difference across the fixed resistor \(V_1 = I R_1\) therefore increases (or the fixed resistor accounts for a larger fraction of the total series resistance).

(a)
- Correct formula or substitution: \(V = 9.0 \times \frac{600}{1200 + 600}\) OR \(I = \frac{9.0}{1800} = 0.0050\text{ A}\) [C1]
- Correct value: \(3.0\text{ V}\) (allow unit mark) [A0.5]

(b)
- States that p.d. increases [B0.5]
- Explanation: resistance of thermistor decreases, so fixed resistor receives a greater fraction of total p.d. / circuit current increases [B0.5]

(a) Using the relation between critical angle and refractive index:
\[ \sin c = \frac{1}{n} = \frac{1}{1.60} = 0.625 \]
\[ c = \sin^{-1}(0.625) \approx 38.68^\circ = 38.7^\circ \]

(b) Because the angle of incidence \(i = 45^\circ\) is greater than the critical angle \(c = 38.7^\circ\) and the ray is travelling from a medium of higher refractive index (glass) towards a medium of lower refractive index (air), total internal reflection occurs and no light refracts into the air.

(a)
- Formula \(\sin c = \frac{1}{n}\) OR substitution \(\sin c = \frac{1}{1.60}\) [C1]
- Answer: \(38.7^\circ\) or \(39^\circ\) [A0.5]

(b)
- States: (total) internal reflection occurs / ray is reflected back into the glass [B0.5]
- Reason: angle of incidence (\(45^\circ\)) is greater than the critical angle (\(38.7^\circ\)) [B0.5]

(a) When the diode is forward-biased, it has zero resistance. The two branches are in parallel with resistances \(R_1 = 30\ \Omega\) and \(R_2 = 20\ \Omega\).
\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{30} + \frac{1}{20} = \frac{2 + 3}{60} = \frac{5}{60} = \frac{1}{12} \]
\[ R_{\text{total}} = 12\ \Omega \]

(b) When the diode is reversed, it becomes reverse-biased and has infinite resistance, so current in Branch 2 becomes zero. The total circuit resistance becomes simply \(30\ \Omega\). Because the total resistance increases from \(12\ \Omega\) to \(30\ \Omega\), the total current drawn from the \(12\text{ V}\) supply decreases from \(1.0\text{ A}\) to \(0.40\text{ A}\).

(a)
- Correct equation \(\frac{1}{R} = \frac{1}{30} + \frac{1}{20}\) OR \(R = \frac{30 \times 20}{30 + 20}\) [C1]
- \(12\ \Omega\) [A0.5]

(b)
- States that total current decreases [B0.5]
- Explains that the reversed diode has infinite resistance / blocks current in branch 2, so total resistance increases to \(30\ \Omega\) [B0.5]