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

The average speed is calculated as the total distance divided by the total time. First, find the distance for each section of the motion using the area under a speed-time graph:

1. Acceleration phase: \(\text{Distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 6.0\text{ m/s} = 12.0\text{ m}\).
2. Constant speed phase: \(\text{Distance}_2 = \text{time} \times \text{speed} = 10.0\text{ s} \times 6.0\text{ m/s} = 60.0\text{ m}\).
3. Deceleration phase: \(\text{Distance}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.0\text{ s} \times 6.0\text{ m/s} = 6.0\text{ m}\).

Total distance \(D = 12.0\text{ m} + 60.0\text{ m} + 6.0\text{ m} = 78.0\text{ m}\).
Total time \(T = 4.0\text{ s} + 10.0\text{ s} + 2.0\text{ s} = 16.0\text{ s}\).

Average speed \(v = \frac{D}{T} = \frac{78.0\text{ m}}{16.0\text{ s}} = 4.875\text{ m/s} \approx 4.9\text{ m/s}\).

B is correct [1 mark].
Award 1 mark for correct calculations of total distance and time leading to 4.9 m/s.

Force is defined as the rate of change of momentum:
\(F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}\)

Taking the initial direction as positive:
\(u = +15\text{ m/s}\)
\(v = -10\text{ m/s}\)

\(\Delta p = 0.20\text{ kg} \times (-10\text{ m/s} - 15\text{ m/s}) = 0.20 \times (-25) = -5.0\text{ kg m/s}\)

The magnitude of the change in momentum is \(5.0\text{ N s}\).

The average force is:
\(F = \frac{5.0\text{ N s}}{0.050\text{ s}} = 100\text{ N}\).

C is correct [1 mark].
Award 1 mark for correctly using change in momentum divided by contact time to get 100 N.

The total pressure at depth \(h\) is the sum of atmospheric pressure and hydrostatic pressure:
\(P = P_0 + \rho g h\)

\(P_{\text{hydrostatic}} = 1025\text{ kg/m}^3 \times 9.8\text{ m/s}^2 \times 80\text{ m} = 803,600\text{ Pa} \approx 8.04 \times 10^5\text{ Pa}\)

\(P = 1.0 \times 10^5\text{ Pa} + 8.04 \times 10^5\text{ Pa} = 9.04 \times 10^5\text{ Pa} \approx 9.0 \times 10^5\text{ Pa}\).

C is correct [1 mark].
Award 1 mark for adding atmospheric pressure to liquid pressure to obtain 9.0 x 10^5 Pa.

First, calculate the critical angle \(\theta_c\) for the glass-air boundary:
\(\sin(\theta_c) = \frac{1}{n} = \frac{1}{1.60} = 0.625\)
\(\theta_c = \arcsin(0.625) \approx 38.7^\circ\)

Since the angle of incidence (\(38^\circ\)) is less than the critical angle (\(38.7^\circ\)), the light ray undergoes refraction into the air rather than total internal reflection.

Now, use Snell's Law to find the angle of refraction \(r\):
\(n_{\text{glass}} \sin(i) = n_{\text{air}} \sin(r)\)
\(1.60 \times \sin(38^\circ) = 1.00 \times \sin(r)\)
\(\sin(r) = 1.60 \times 0.6157 = 0.985\)
\(r = \arcsin(0.985) \approx 80^\circ\).

C is correct [1 mark].
Award 1 mark for calculating critical angle and using Snell's Law to find the correct refraction angle.

1. Calculate the useful work output (gain in gravitational potential energy):
\(W_{\text{out}} = mgh = 50\text{ kg} \times 9.8\text{ m/s}^2 \times 12\text{ m} = 5880\text{ J}\)

2. Calculate the useful power output:
\(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{5880\text{ J}}{10\text{ s}} = 588\text{ W}\)

3. Calculate the electrical power input using the efficiency:
\(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\)
\(0.60 = \frac{588}{P_{\text{in}}}\)
\(P_{\text{in}} = \frac{588}{0.60} = 980\text{ W}\).

C is correct [1 mark].
Award 1 mark for calculating useful output power and dividing by efficiency to find 980 W.

In a series circuit, the total voltage is shared between the components:
\(V_{\text{total}} = V_{\text{resistor}} + V_{\text{thermistor}}\)
\(12\text{ V} = V_{\text{resistor}} + 4.0\text{ V}\)
\(V_{\text{resistor}} = 8.0\text{ V}\)

The ratio of the voltages across the series components equals the ratio of their resistances:
\(\frac{V_{\text{thermistor}}}{V_{\text{resistor}}} = \frac{R_{\text{thermistor}}}{R_{\text{resistor}}}\)
\(\frac{4.0\text{ V}}{8.0\text{ V}} = \frac{R_{\text{thermistor}}}{400\ \Omega}\)
\(R_{\text{thermistor}} = 400 \times 0.50 = 200\ \Omega\).

B is correct [1 mark].
Award 1 mark for finding the voltage across the resistor and using the potential divider ratio to get 200 ohms.

1. Subtract the background count rate to find the initial source activity:
\(\text{Initial source activity} = 820\text{ counts/s} - 20\text{ counts/s} = 800\text{ counts/s}\)

2. Determine the number of half-lives in \(1.0\text{ hour}\) (\(60\text{ minutes}\)):
\(\text{Number of half-lives} = \frac{60\text{ minutes}}{15\text{ minutes}} = 4\)

3. Calculate the source activity after 4 half-lives:
\(\text{Activity after 4 half-lives} = 800 \times \left(\frac{1}{2}\right)^4 = \frac{800}{16} = 50\text{ counts/s}\)

4. Add the background count rate back to find the measured count rate:
\(\text{Measured count rate} = 50\text{ counts/s} + 20\text{ counts/s} = 70\text{ counts/s}\).

C is correct [1 mark].
Award 1 mark for subtracting background, halving 4 times, and adding back background to get 70 counts/s.

Use the redshift equation:
\(\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\)

Where:
\(\Delta \lambda = 412\text{ nm} - 400\text{ nm} = 12\text{ nm}\)
\(\lambda = 400\text{ nm}\)

Calculate the fractional shift:
\(\frac{\Delta \lambda}{\lambda} = \frac{12}{400} = 0.030\)

Now find the recession speed \(v\):
\(v = 0.030 \times c = 0.030 \times 3.0 \times 10^8\text{ m/s} = 9.0 \times 10^6\text{ m/s}\).

B is correct [1 mark].
Award 1 mark for correctly using the redshift formula to solve for recession speed.

First, find the maximum velocity reached during the acceleration phase: \(v = u + at = 0 + 3.0 \times 4.0 = 12.0\text{ m/s}\). The distance travelled in each phase is: 1) Acceleration phase: \(d_1 = \frac{1}{2} \times v \times t_1 = \frac{1}{2} \times 12.0 \times 4.0 = 24.0\text{ m}\). 2) Constant velocity phase: \(d_2 = v \times t_2 = 12.0 \times 5.0 = 60.0\text{ m}\). 3) Deceleration phase: \(d_3 = \frac{1}{2} \times v \times t_3 = \frac{1}{2} \times 12.0 \times 3.0 = 18.0\text{ m}\). Total distance: \(d = 24.0 + 60.0 + 18.0 = 102.0\text{ m}\). Total time: \(t = 4.0 + 5.0 + 3.0 = 12.0\text{ s}\). Average speed: \(v_{\text{avg}} = \frac{102.0}{12.0} = 8.5\text{ m/s}\).

1 mark for the correct option B.

The useful work done per minute is the change in gravitational potential energy of the water: \(W = mgh = 300 \times 9.8 \times 12 = 35\,280\text{ J}\). The useful power output is: \(P_{\text{out}} = \frac{W}{t} = \frac{35\,280}{60} = 588\text{ W}\). Since the efficiency is \(70\%\), the power input is: \(P_{\text{in}} = \frac{P_{\text{out}}}{0.70} = \frac{588}{0.70} = 840\text{ W}\).

1 mark for the correct option D.

Taking the direction towards the wall as positive: initial velocity \(u = 25\text{ m/s}\) and final velocity \(v = -15\text{ m/s}\). The change in momentum is: \(\Delta p = m(v - u) = 0.060 \times (-15 - 25) = -2.4\text{ kg m/s}\). The magnitude of the change in momentum is \(2.4\text{ kg m/s}\). The average force is: \(F = \frac{\Delta p}{\Delta t} = \frac{2.4}{0.050} = 48\text{ N}\).

1 mark for the correct option D.

The total liquid pressure at the bottom is the sum of the pressures exerted by each layer: \(p = \rho_{\text{oil}} g h_{\text{oil}} + \rho_{\text{water}} g h_{\text{water}}\). \(p = (800 \times 9.8 \times 0.40) + (1000 \times 9.8 \times 0.60) = 3136 + 5880 = 9016\text{ Pa}\). This rounds to \(9000\text{ Pa}\).

1 mark for the correct option C.

Using Snell's Law for light entering the glass block from air: \(n = \frac{\sin i}{\sin r} \implies 1.52 = \frac{\sin 50^\circ}{\sin r}\). Therefore, \(\sin r = \frac{\sin 50^\circ}{1.52} = \frac{0.7660}{1.52} \approx 0.5040\). Thus, \(r = \sin^{-1}(0.5040) \approx 30.3^\circ \approx 30^\circ\).

1 mark for the correct option A.

First, convert time to seconds: \(t = 20 \times 60 = 1200\text{ s}\). Use the formula for electrical energy transferred: \(E = I V t\). Rearranging for potential difference: \(V = \frac{E}{I \times t} = \frac{16\,200}{1.5 \times 1200} = \frac{16\,200}{1800} = 9.0\text{ V}\).

1 mark for the correct option B.

First, find the combined resistance of the parallel pair: \(R_p = \frac{6.0 \times 12.0}{6.0 + 12.0} = \frac{72.0}{18.0} = 4.0\ \Omega\). Next, calculate the total resistance of the circuit: \(R_{\text{total}} = R_p + 8.0 = 4.0 + 8.0 = 12.0\ \Omega\). The total current supplied by the source is: \(I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{12.0\ \Omega} = 1.0\text{ A}\). The potential difference across the parallel combination is: \(V_p = I_{\text{total}} \times R_p = 1.0\text{ A} \times 4.0\ \Omega = 4.0\text{ V}\). Finally, the current in the \(12.0\ \Omega\) resistor is: \(I_{12} = \frac{V_p}{12.0\ \Omega} = \frac{4.0\text{ V}}{12.0\ \Omega} \approx 0.33\text{ A}\).

1 mark for the correct option A.

We must subtract the background count rate from the measured values to find the corrected count rates of the source. Initial corrected count rate = \(344 - 24 = 320\text{ counts/minute}\). Final corrected count rate = \(64 - 24 = 40\text{ counts/minute}\). The fraction of initial activity remaining is: \(\frac{40}{320} = \frac{1}{8}\). Since \(\frac{1}{8} = (\frac{1}{2})^3\), exactly 3 half-lives have elapsed in the \(6.0\text{ hours}\). Therefore, the half-life is: \(\frac{6.0\text{ hours}}{3} = 2.0\text{ hours}\).

1 mark for the correct option B.

The final velocity \(v\) of the probe after decelerating for \(4.0\text{ s}\) can be found using:
\(v = u + at\)
Here, the probe is decelerating, so \(a = -2.5\text{ m/s}^2\):
\(v = 15 + (-2.5 \times 4.0) = 5.0\text{ m/s}\)

The average speed during this time interval is:
\(v_{\text{avg}} = \frac{u + v}{2} = \frac{15 + 5.0}{2} = 10\text{ m/s}\)

The distance descended is:
\(s = v_{\text{avg}} \times t = 10 \times 4.0 = 40\text{ m}\)

B1 for correct answer (C)
- 1 mark for the correct option C.

First, calculate the useful work output (gravitational potential energy gained by the block):
\(E_{\text{out}} = mgh = 80\text{ kg} \times 9.8\text{ N/kg} \times 12\text{ m} = 9408\text{ J}\)

Next, calculate the total electrical energy input:
\(E_{\text{in}} = \text{Power} \times \text{time} = 2000\text{ W} \times 6.0\text{ s} = 12\,000\text{ J}\)

Finally, calculate the efficiency:
\(\text{Efficiency} = \frac{E_{\text{out}}}{E_{\text{in}}} \times 100\% = \frac{9408}{12\,000} \times 100\% = 78.4\% \approx 78\%\)

B1 for correct answer (C)
- 1 mark for the correct option C.

Taking motion to the right as positive:
- Initial momentum of the first trolley: \(p_1 = m_1 u_1 = 2.0 \times (+6.0) = +12.0\text{ kg m/s}\)
- Initial momentum of the second trolley: \(p_2 = m_2 u_2 = 3.0 \times (-2.0) = -6.0\text{ kg m/s}\)

Total initial momentum:
\(p_{\text{total}} = p_1 + p_2 = +12.0 - 6.0 = +6.0\text{ kg m/s}\)

Let \(v\) be the common velocity of the combined trolleys after collision. Since they stick together, the total mass is:
\(M = m_1 + m_2 = 2.0 + 3.0 = 5.0\text{ kg}\)

Using the conservation of momentum:
\(p_{\text{final}} = M v = 5.0 \times v\)
\(5.0 v = +6.0\)
\(v = +1.2\text{ m/s}\) (positive sign indicates to the right).

B1 for correct answer (B)
- 1 mark for the correct option B.

First, calculate the energy supplied by the heater:
\(Q = \text{Power} \times \text{time} = 60\text{ W} \times (5.0 \times 60)\text{ s} = 18\,000\text{ J}\)

The temperature increase of the block is:
\(\Delta T = 44^\circ\text{C} - 20^\circ\text{C} = 24^\circ\text{C}\)

Now, use the specific heat capacity formula \(Q = mc\Delta T\):
\(c = \frac{Q}{m\Delta T} = \frac{18\,000\text{ J}}{1.5\text{ kg} \times 24^\circ\text{C}} = \frac{18\,000}{36} = 500\text{ J / (kg }^\circ\text{C)}\)

B1 for correct answer (C)
- 1 mark for the correct option C.

First, find the critical angle \(c\) of the glass-air boundary:
\(\sin(c) = \frac{1}{n} = \frac{1}{1.50} \approx 0.667\)
\(c = \sin^{-1}(0.667) \approx 41.8^\circ\)

The angle of incidence is \(i = 45^\circ\).
Since \(i > c\) (\(45^\circ > 41.8^\circ\)) and the light is travelling from a optically denser medium (glass) to a less dense medium (air), the light ray undergoes total internal reflection.
By the law of reflection, the angle of reflection equals the angle of incidence, which is \(45^\circ\).

B1 for correct answer (D)
- 1 mark for the correct option D.

The electromotive force (e.m.f.) is defined as the work done per unit charge in driving charge around a complete circuit:
\(E = \frac{W}{Q}\)

Therefore, the energy converted is:
\(W = E \times Q = 9.0\text{ V} \times 150\text{ C} = 1350\text{ J}\)

B1 for correct answer (D)
- 1 mark for the correct option D.

First, determine the corrected count rates by subtracting the background count rate (\(24\text{ counts/minute}\)):
- Initial corrected count rate \(R_0 = 216 - 24 = 192\text{ counts/minute}\)
- Final corrected count rate \(R = 48 - 24 = 24\text{ counts/minute}\)

Next, determine the number of half-lives \(n\) that have elapsed:
\(192 \xrightarrow{\text{1st half-life}} 96 \xrightarrow{\text{2nd half-life}} 48 \xrightarrow{\text{3rd half-life}} 24\)

Exactly 3 half-lives have elapsed in the total time of \(3.0\text{ hours}\).
Therefore, the half-life \(T_{1/2}\) is:
\(T_{1/2} = \frac{3.0\text{ hours}}{3} = 1.0\text{ hour}\)

B1 for correct answer (B)
- 1 mark for the correct option B.

Cosmological redshift occurs when the light from a receding source is stretched, which increases its wavelength (shifting it toward the red end of the spectrum) and decreases its frequency. This observation indicates that the distant galaxy is moving away from the Earth, supporting the theory of the expansion of the universe.

B1 for correct answer (B)
- 1 mark for the correct option B.

We use the equation of motion: \(v^2 = u^2 + 2as\), where \(u = 25\text{ m/s}\) is the initial velocity, \(v = 15\text{ m/s}\) is the final velocity, and \(s = 80\text{ m}\) is the distance. Rearranging for deceleration \(a\): \(a = \frac{v^2 - u^2}{2s} = \frac{15^2 - 25^2}{2 \times 80} = \frac{225 - 625}{160} = \frac{-400}{160} = -2.5\text{ m/s}^2\). The magnitude of the deceleration is \(2.5\text{ m/s}^2\).

C is correct (1 mark).

1. Calculate the useful work done in lifting the water: \(W = mgh = 300\text{ kg} \times 9.8\text{ m/s}^2 \times 8.0\text{ m} = 23\,520\text{ J}\). 2. Calculate the useful output power: \(P_{\text{out}} = \frac{W}{t} = \frac{23\,520\text{ J}}{12\text{ s}} = 1960\text{ W}\). 3. Calculate the input power using efficiency: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies 0.65 = \frac{1960}{P_{\text{in}}} \implies P_{\text{in}} = \frac{1960}{0.65} \approx 3015\text{ W} \approx 3.0\text{ kW}\).

C is correct (1 mark).

1. Initial momentum of the system: \(p_i = m_1 u_1 + m_2 u_2 = 0.80 \times 3.0 + 0 = 2.4\text{ kg m/s}\). 2. By conservation of momentum, final velocity \(v\) of the combined mass \((m_1 + m_2 = 2.00\text{ kg})\) is: \(v = \frac{p_i}{m_1 + m_2} = \frac{2.4}{2.00} = 1.2\text{ m/s}\). 3. Initial kinetic energy: \(E_{ki} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 0.80 \times 3.0^2 = 3.6\text{ J}\). 4. Final kinetic energy: \(E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 2.00 \times 1.2^2 = 1.44\text{ J}\). 5. Loss of kinetic energy: \(\Delta E_k = E_{ki} - E_{kf} = 3.6 - 1.44 = 2.16\text{ J} \approx 2.2\text{ J}\).

C is correct (1 mark).

Since the temperature remains constant, the average speed and kinetic energy of the gas particles do not change, so the average force exerted during a single collision remains the same. However, reducing the volume increases the number of particles per unit volume, which causes the particles to collide with the walls of the container more frequently. This increases the total force per unit area, resulting in higher pressure.

C is correct (1 mark).

1. Using Snell's law for refraction from air to glass: \(n = \frac{\sin(i)}{\sin(r)} \implies 1.52 = \frac{\sin(35^\circ)}{\sin(r)} \implies \sin(r) = \frac{\sin(35^\circ)}{1.52} \approx \frac{0.5736}{1.52} \approx 0.3774 \implies r \approx 22^\circ\). 2. The formula for the critical angle \(c\) is: \(\sin(c) = \frac{1}{n} \implies \sin(c) = \frac{1}{1.52} \approx 0.6579 \implies c \approx 41^\circ\).

A is correct (1 mark).

The resistance of a wire is given by \(R = \rho \frac{L}{A}\). For the first wire, \(R_1 = \rho \frac{L}{A} = 16\ \Omega\). For the second wire, the length is \(2L\). Since the diameter is doubled, the radius is also doubled, meaning the cross-sectional area \(A_2\) is four times greater than the first wire's area, so \(A_2 = 4A\). The resistance of the second wire is: \(R_2 = \rho \frac{2L}{4A} = \frac{1}{2} \left(\rho \frac{L}{A}\right) = \frac{1}{2} R_1 = \frac{1}{2} \times 16\ \Omega = 8.0\ \Omega\).

B is correct (1 mark).

1. Using the transformer equation: \(\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 240 \times \frac{300}{1200} = 60\text{ V}\). 2. Find the secondary current using Ohm's law: \(I_s = \frac{V_s}{R} = \frac{60\text{ V}}{15\ \Omega} = 4.0\text{ A}\). 3. For an ideal transformer, input power equals output power: \(V_p I_p = V_s I_s \implies 240 \times I_p = 60 \times 4.0 \implies I_p = 1.0\text{ A}\).

B is correct (1 mark).

1. Calculate the initial corrected count rate of the source by subtracting the background count rate: \(355 - 35 = 320\text{ counts/minute}\). 2. Determine how many half-lives have passed: \(\frac{4.0\text{ hours}}{2.0\text{ hours}} = 2.0\text{ half-lives}\). 3. Calculate the corrected count rate after 2 half-lives: \(\frac{320}{2^2} = 80\text{ counts/minute}\). 4. Add back the background count rate to get the registered count rate: \(80 + 35 = 115\text{ counts/minute}\).

C is correct (1 mark).

We can divide the motion of the drone into three distinct phases to calculate the total height:

1. **First Phase (Uniform Acceleration):**
- From \(t = 0\) to \(t = 5.0\text{ s}\), acceleration \(a = 4.0\text{ m/s}^2\).
- Height gained: \(s_1 = \frac{1}{2} a t^2 = \frac{1}{2} \times 4.0 \times (5.0)^2 = 50\text{ m}\).
- Velocity reached at the end of this phase: \(v = a t = 4.0 \times 5.0 = 20\text{ m/s}\).

2. **Second Phase (Constant Velocity):**
- For the next \(10\text{ s}\) (from \(t = 5.0\text{ s}\) to \(t = 15.0\text{ s}\)), the drone travels at a constant velocity of \(20\text{ m/s}\).
- Height gained: \(s_2 = v \times t = 20 \times 10 = 200\text{ m}\).

3. **Third Phase (Free Fall after motors cut out):**
- At \(t = 15.0\text{ s}\), the initial upward velocity is \(u = 20\text{ m/s}\), and the acceleration is \(a = -g = -10\text{ m/s}^2\).
- The drone rises further until its velocity reaches zero: \(v^2 = u^2 + 2as_3 \implies 0 = 20^2 - 2 \times 10 \times s_3\).
- Height gained: \(s_3 = \frac{400}{20} = 20\text{ m}\).

**Total Maximum Height:**
\(H = s_1 + s_2 + s_3 = 50 + 200 + 20 = 270\text{ m}\).

Award 1 mark for the correct answer B.

First, calculate the useful work done by the motor in lifting the crate:
\(W_{\text{out}} = mgh = 120\text{ kg} \times 9.8\text{ m/s}^2 \times 15\text{ m} = 17\,640\text{ J}\).

Next, calculate the useful power output:
\(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{17\,640\text{ J}}{12\text{ s}} = 1470\text{ W}\).

Finally, use the efficiency formula to find the electrical power input:
\(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies 0.75 = \frac{1470\text{ W}}{P_{\text{in}}}\)
\(P_{\text{in}} = \frac{1470}{0.75} = 1960\text{ W} = 1.96\text{ kW}\).

Award 1 mark for the correct answer C.

The resistance of a cylindrical conductor is given by:
\(R = \rho \frac{L}{A}\), where \(A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\).

This gives: \(R = \frac{4\rho L}{\pi d^2}\).

For the second wire:
- Length \(L' = 3L\)
- Diameter \(d' = 2d\)

Substitute these into the resistance equation:
\(R' = \frac{4\rho L'}{\pi (d')^2} = \frac{4\rho (3L)}{\pi (2d)^2} = \frac{12\rho L}{4\pi d^2} = 3 \times \left(\frac{\rho L}{\pi d^2}\right)\).

Since \(\frac{\rho L}{\pi d^2} = \frac{R}{4}\):
\(R' = 3 \times \frac{R}{4} = 0.75R\).

Award 1 mark for the correct answer B.

Using the transformer turns-ratio equation to calculate the secondary voltage \(V_s\):
\(\frac{V_p}{V_s} = \frac{N_p}{N_s} \implies \frac{240\text{ V}}{V_s} = \frac{400}{100} = 4\)
\(V_s = 60\text{ V}\).

Now, calculate the secondary current \(I_s\) using Ohm's Law:
\(I_s = \frac{V_s}{R_s} = \frac{60\text{ V}}{12\ \Omega} = 5.0\text{ A}\).

For an ideal transformer, input power equals output power:
\(P_{\text{in}} = P_{\text{out}} \implies V_p I_p = V_s I_s\)
\(240 \times I_p = 60 \times 5.0\)
\(240 I_p = 300 \implies I_p = 1.25\text{ A}\).

Award 1 mark for the correct answer B.

First, find the number of half-lives that have elapsed in \(16\text{ hours}\):
\(n = \frac{16\text{ hours}}{4.0\text{ hours}} = 4\text{ half-lives}\).

Each half-life halves the number of undecayed nuclei. After \(4\) half-lives, the fraction of remaining undecayed nuclei is:
\(\left(\frac{1}{2}\right)^4 = \frac{1}{16}\).

Calculate the remaining nuclei:

\(N = N_0 \times \frac{1}{16} = \frac{6.4 \times 10^{11}}{16} = 0.4 \times 10^{11} = 4.0 \times 10^{10}\).

Award 1 mark for the correct answer A.

First, calculate the redshift \(z\):
\(z = \frac{\Delta\lambda}{\lambda_0} = \frac{515\text{ nm} - 500\text{ nm}}{500\text{ nm}} = \frac{15}{500} = 0.030\).

Now, find the recession speed \(v\) of the galaxy using the redshift equation:
\(v = z \times c = 0.030 \times 3.0 \times 10^8\text{ m/s} = 9.0 \times 10^6\text{ m/s}\).

Using Hubble's law, \(v = H_0 d\), to calculate the distance \(d\):
\(d = \frac{v}{H_0} = \frac{9.0 \times 10^6\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} \approx 4.1 \times 10^{24}\text{ m}\).

Award 1 mark for the correct answer C.

We apply Snell's Law for refraction at the boundary from air to glass:
\(n_{\text{air}} \sin(\theta_i) = n_{\text{glass}} \sin(\theta_r)\)

Given:
- \(n_{\text{air}} = 1.00\)
- \(\theta_i = 50^\circ\)
- \(n_{\text{glass}} = 1.52\)

Substitute the values:
\(1.00 \times \sin(50^\circ) = 1.52 \times \sin(\theta_r)\)
\(\sin(\theta_r) = \frac{\sin(50^\circ)}{1.52} = \frac{0.7660}{1.52} \approx 0.5040\)

Calculate the angle of refraction:
\(\theta_r = \arcsin(0.5040) \approx 30.3^\circ\), which is closest to \(30^\circ\).

Award 1 mark for the correct answer A.

First, calculate the initial kinetic energy \(E_{ki}\):
\(E_{ki} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} (2000) (6.0)^2 + 0 = 36\,000\text{ J} = 36.0\text{ kJ}\).

Next, use the conservation of momentum to find the common velocity \(v\) of the coupled cars:
\(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\)
\(2000 \times 6.0 + 0 = (2000 + 3000) v\)
\(12\,000 = 5000 v \implies v = 2.4\text{ m/s}\).

Now, calculate the final kinetic energy \(E_{kf}\):
\(E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (5000) (2.4)^2 = 2500 \times 5.76 = 14\,400\text{ J} = 14.4\text{ kJ}\).

Finally, determine the kinetic energy lost:
\(\text{Energy Lost} = E_{ki} - E_{kf} = 36\,000\text{ J} - 14\,400\text{ J} = 21\,600\text{ J} = 21.6\text{ kJ}\).

Award 1 mark for the correct answer C.

(a) Speed is a scalar quantity (has magnitude only), whereas velocity is a vector quantity (has both magnitude and direction).

(b)(i) \(a = \frac{v - u}{t} = \frac{80 - 0}{4.0} = 20\text{ m/s}^2\)

(b)(ii) Height reached = area under the speed-time graph
\(\text{Height} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \times 80 = 160\text{ m}\)

(b)(iii) \(\Delta E_p = m g h = 0.65 \times 9.8 \times 160 = 62.4\text{ J}\) (or \(62\text{ J}\))

(a) Speed is scalar, velocity is vector [1]

(b)(i) Formula \(a = \frac{v - u}{t}\) or substitution \(\frac{80}{4.0}\) [1]
Correct answer with unit: \(20\text{ m/s}^2\) [1]

(b)(ii) Realisation that distance is area under graph [1]
Calculation: \(\frac{1}{2} \times 4.0 \times 80\) [1]
Correct answer with unit: \(160\text{ m}\) [1]

(b)(iii) Formula \(\Delta E_p = m g h\) or substitution \(0.65 \times 9.8 \times 160\) [1]
Correct answer with unit: \(62.4\text{ J}\) (allow \(62\text{ J}\)) [1]

(a) The total momentum of a closed/isolated system remains constant, provided no external forces act on it.

(b)(i) Using conservation of momentum:
\(m_A u_A + m_B u_B = (m_A + m_B) v\)
\((12000 \times 4.5) + (8000 \times 1.5) = (12000 + 8000) v\)
\(54000 + 12000 = 20000 v\)
\(66000 = 20000 v \implies v = 3.3\text{ m/s}\)

(b)(ii) \(\text{Impulse} = \Delta p_B = m_B (v - u_B) = 8000 \times (3.3 - 1.5) = 8000 \times 1.8 = 14400\text{ N s}\) (or \(\text{kg m/s}\))

(a) Total momentum before collision = total momentum after collision [1]
In a closed system / no external forces [1]

(b)(i) Use of \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\) [1]
Correct substitution: \((12000 \times 4.5) + (8000 \times 1.5) = 20000 v\) [1]
Correct answer with unit: \(3.3\text{ m/s}\) [1]

(b)(ii) Formula for impulse: \(F t = \Delta p = m \Delta v\) [1]
Substitution: \(8000 \times (3.3 - 1.5)\) [1]
Correct answer with unit: \(14400\text{ N s}\) (or \(1.4 \times 10^4\text{ N s}\)) [1]

(a) Pressure is defined as force per unit area.

(b)(i) \(F = m g = 150 \times 9.8 = 1470\text{ N}\)
\(A = 0.80 \times 0.60 = 0.48\text{ m}^2\)
\(P = \frac{F}{A} = \frac{1470}{0.48} = 3062.5\text{ Pa}\) (or \(3100\text{ Pa}\) to 2 s.f.)

(c)(i) \(P = \rho g h = 900 \times 9.8 \times 1.5 = 13230\text{ Pa}\) (or \(1.3 \times 10^4\text{ Pa}\))

(c)(ii) The pressure on the bottom surface of the block is greater than on the top surface.

(a) Force per unit area (or \(F/A\)) [1]

(b)(i) Weight calculation: \(150 \times 9.8 = 1470\text{ N}\) [1]
Area calculation: \(0.80 \times 0.60 = 0.48\text{ m}^2\) [1]
Correct pressure: \(3100\text{ Pa}\) (allow \(3062.5\text{ Pa}\)) [1]

(c)(i) Formula \(P = \rho g h\) [1]
Substitution: \(900 \times 9.8 \times 1.5\) [1]
Correct answer with unit: \(13200\text{ Pa}\) (allow \(13230\text{ Pa}\) or \(13000\text{ Pa}\)) [1]

(c)(ii) Bottom pressure is greater [1]

(a) Specific heat capacity is the thermal energy required per unit mass to increase the temperature of a substance by one degree Celsius (or Kelvin).

(b)(i) \(E = P \times t = 120\text{ W} \times (5.0 \times 60)\text{ s} = 120 \times 300 = 36000\text{ J}\)

(b)(ii) \(\Delta T = 48 - 20 = 28\,^\circ\text{C}\)
\(c = \frac{E}{m \Delta T} = \frac{36000}{2.5 \times 28} = \frac{36000}{70} = 514\text{ J/(kg }^\circ\text{C)}\) (or \(510\text{ J/(kg }^\circ\text{C)}\))

(b)(iii) The calculated value of the specific heat capacity would be larger than the true value.

(a) Thermal energy required to raise temperature of unit mass (1 kg) [1]
By 1 degree Celsius (or 1 K) [1]

(b)(i) Formula \(E = P \times t\) or conversion of time to seconds (300 s) [1]
Correct energy with unit: \(36000\text{ J}\) (or \(36\text{ kJ}\)) [1]

(b)(ii) Calculation of temp difference \(\Delta T = 28\,^\circ\text{C}\) [1]
Use of \(c = \frac{E}{m \Delta T}\) [1]
Correct answer with unit: \(510\text{ J/(kg }^\circ\text{C)}\) (allow \(514\text{ J/(kg }^\circ\text{C)}\)) [1]

(b)(iii) Calculated value is larger/greater than true value [1]

(a) 1. The light must travel from an optically denser medium to a less dense medium (e.g., glass to air).
2. The angle of incidence must be greater than the critical angle.

(b)(i) From Snell's Law:
\(n = \frac{\sin i}{\sin r} \implies 1.52 = \frac{\sin 45^\circ}{\sin r}\)
\(\sin r = \frac{\sin 45^\circ}{1.52} = \frac{0.7071}{1.52} = 0.4652\)
\(r = \arcsin(0.4652) = 27.7^\circ\) (or \(28^\circ\))

(b)(ii) Using critical angle formula:
\(\sin c = \frac{1}{n} = \frac{1}{1.52} = 0.6579\)
\(c = \arcsin(0.6579) = 41.1^\circ\) (or \(41^\circ\))

(a) Light goes from denser to rarer/less dense medium [1]
Angle of incidence > critical angle [1]

(b)(i) Formula \(n = \frac{\sin i}{\sin r}\) [1]
Substitution: \(\sin r = \frac{\sin 45^\circ}{1.52}\) [1]
Correct angle: \(28^\circ\) (allow \(27.7^\circ\)) [1]

(b)(ii) Formula \(\sin c = \frac{1}{n}\) [1]
Substitution: \(\sin c = \frac{1}{1.52}\) [1]
Correct angle with unit: \(41^\circ\) (allow \(41.1^\circ\)) [1]

(a)(i) A rectangle with a diagonal line through it that has a horizontal flat section at the bottom-left.
(a)(ii) A rectangle (or rectangle in a circle) with two arrows pointing towards it.

(b)(i) Using the potential divider formula:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{thermistor}}}{R_{\text{fixed}} + R_{\text{thermistor}}}\)
\(V_{\text{out}} = 12.0 \times \frac{1200}{800 + 1200} = 12.0 \times 0.60 = 7.2\text{ V}\)

(b)(ii) The voltmeter reading decreases. As temperature increases, the resistance of the thermistor decreases. Since the thermistor's resistance becomes a smaller share of the total circuit resistance, it receives a smaller share of the total potential difference (supply voltage).

(a)(i) Correct thermistor symbol [1]
(a)(ii) Correct LDR symbol [1]

(b)(i) Total resistance calculation: \(800 + 1200 = 2000\, \Omega\) [1]
Potential divider calculation: \(12.0 \times \frac{1200}{2000}\) [1]
Correct voltage with unit: \(7.2\text{ V}\) [1]

(b)(ii) Voltmeter reading decreases [1]
Resistance of thermistor decreases with temperature [1]
Thermistor takes a smaller fraction/share of total potential difference [1]

(a) Faraday's law states that the size of an induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage (or rate of cutting of magnetic field lines).

(b)(i) The needle of the millivoltmeter deflects to one side (e.g., to the right) and then returns to zero. Explanation: As the magnet moves, magnetic field lines cut the coil (flux changes), inducing an e.m.f. and causing a temporary current to flow.

(b)(ii) The reading is zero (0 mV).

(b)(iii) The needle deflects in the opposite direction (e.g., to the left) and has a larger peak deflection. Explanation: Pulling the magnet out reverses the direction of magnetic flux change (Lenz's Law), so the e.m.f. is reversed. Moving it rapidly increases the rate of cutting field lines, resulting in a larger induced e.m.f.

(a) Induced e.m.f. / voltage [1]
Is proportional to the rate of cutting field lines / rate of change of flux linkage [1]

(b)(i) Deflection to one side (then returns to zero) [1]
Changing magnetic field/flux through solenoid [1]
Induces an e.m.f. / current [1]

(b)(ii) Zero / 0 [1]

(b)(iii) Deflection in opposite direction [1]
Larger peak/deflection (due to greater rate of cutting field lines) [1]

(a) Half-life is the time taken for the number of radioactive nuclei (or activity) in a sample to decrease to half of its initial value.

(b)(i) Number of half-lives: \(n = \frac{24\text{ days}}{8.0\text{ days}} = 3\).
After 3 half-lives, the activity is:
\(800 \times \left(\frac{1}{2}\right)^3 = 800 \times \frac{1}{8} = 100\text{ counts/s}\).

(b)(ii) Radioactive decay is a random process governed by probability, meaning we can never predict precisely when any individual nucleus will decay; there is always a non-zero probability that some nuclei remain undecayed.

(c) Safety precautions include:
1. Handling the source using long-handled tongs (to maintain distance).
2. Storing the source in a lead-lined container when not in use.
3. Keeping the source pointed away from body parts/people.

(a) Time taken for activity / count rate to halve [1]
OR time taken for half the radioactive nuclei to decay [1]

(b)(i) Calculation of number of half-lives (3) [1]
Activity calculation: \(800 \times \left(\frac{1}{2}\right)^3\) [1]
Correct answer with unit: \(100\text{ counts/s}\) [1]

(b)(ii) Decay is random / spontaneous [1]

(c) Any two precautions from:
- Use tongs / keep distance [1]
- Store in lead container [1]
- Minimize time of exposure [1]
- Point away from people [1]

(a) Momentum is defined as the product of mass and velocity. (b) Initial momentum = mass * velocity = 1500 kg * 120 m/s = 180,000 kg m/s. (c) The principle of conservation of momentum states that in a closed system with no external forces, the total momentum remains constant. (d) Using conservation of momentum: Initial momentum = Final momentum. 180,000 = (1450 kg * v) + (50 kg * -300 m/s). 180,000 = 1450 v - 15,000. 195,000 = 1450 v. v = 134.48 m/s, which rounds to 134 m/s.

(a) B1: product of mass and velocity (or p = mv with defined symbols). (b) C1: momentum = mass * velocity or 1500 * 120. A1: 180,000 kg m/s (or 1.8 * 10^5 kg m/s or N s). (c) B1: total momentum before equals total momentum after in a closed system / in the absence of external forces. (d) C1: use of momentum conservation equation. C1: substitution of values: 180,000 = 1450 * v + 50 * (-300). C1: rearrangement to 195,000 = 1450 * v. A1: 134 m/s (accept 134 m/s to 135 m/s).

(a) Weight of container = m * g = 450 * 9.8 = 4410 N. Base Area = 1.2 * 0.80 = 0.96 m2. Pressure = Force / Area = 4410 / 0.96 = 4593.75 Pa, which rounds to 4600 Pa. (b) Pressure = density * g * h = 1000 * 9.8 * 15 = 147,000 Pa (or 1.47 * 10^5 Pa). (c) Air molecules inside are in continuous, rapid, and random motion. They collide with the walls of the container. During each collision, there is a change in momentum of the molecules, which exerts a force on the wall. The sum of these forces over the surface area produces pressure.

(a) C1: Weight = 450 * 9.8 = 4410 N. C1: Area = 1.2 * 0.80 = 0.96 m2 and Pressure = Force / Area. A1: 4600 Pa (or 4590 Pa). (b) C1: P = density * g * h or 1000 * 9.8 * 15. A1: 147,000 Pa (or 1.5 * 10^5 Pa). (c) B1: molecules are in constant random motion. B1: molecules collide with the container walls. B1: collision / change in momentum produces a force (forces per unit area result in pressure).

(a) The thermometer shows the meniscus exactly halfway between 21 and 22, so $T_R = 21.5\text{ }^\circ\text{C}$.

(b) Over the first 60 seconds:
- For Container A: $\Delta\theta_A = 85.0 - 73.0 = 12.0\text{ }^\circ\text{C}$.
- For Container B: $\Delta\theta_B = 85.0 - 65.5 = 19.5\text{ }^\circ\text{C}$.

(c) Container A is a better thermal insulator because it had a smaller temperature drop over the 60-second period ($12.0\text{ }^\circ\text{C}$ compared to $19.5\text{ }^\circ\text{C}$ for Container B), which means it loses heat to the surroundings at a slower rate.

(d) Any two from:
- Volume of water used in each container.
- Initial temperature of the water (exactly $85.0\text{ }^\circ\text{C}$ for both).
- External surface area and dimensions of the containers.
- Ambient room temperature.
- Location (avoiding draft/air currents).

(e) Evaporation and convection from the open water surface account for a significant portion of the total heat loss. By using a lid, these pathways are minimized, meaning the measured heat loss is dominated by conduction through the container walls. This allows a more direct and accurate comparison of the thermal insulation of the container materials themselves.

- (a) 21.5 °C [1 mark]
- (b) Correct subtraction for A (12.0 °C) [1 mark]; Correct subtraction for B (19.5 °C) [1 mark]
- (c) Identifies Container A [1 mark]; Justified by referencing smaller temperature drop / slower rate of cooling from the table [1 mark]
- (d) Identifies two correct control variables (e.g., volume of water, initial temperature, room temperature, drafts) [2 marks, 1 mark each]
- (e) Mentions that a lid prevents/reduces evaporation or convection from the top surface [1 mark]; Mentions that this ensures heat loss occurs primarily through the container walls [1 mark]; Explains that this makes the comparison of container materials more valid [1 mark].

(a) The voltmeter symbol (circle with V) is drawn connected across the ends of the test wire.

(b) The needle is on the second tick mark past 0.4. Each subdivision represents 0.02, so $I = 0.40 + 2 \times 0.02 = 0.44\text{ A}$.

(c) Resistance calculations:
- For $L = 60.0\text{ cm}$: $R_3 = 1.32 / 0.44 = 3.00\text{ }\Omega$
- For $L = 100.0\text{ cm}$: $R_5 = 2.20 / 0.44 = 5.00\text{ }\Omega$

(d) Yes, $R$ is directly proportional to $L$. This is justified because the ratio $R/L$ is constant for all data points:
- $1.00/20.0 = 0.05\text{ }\Omega/\text{cm}$
- $2.00/40.0 = 0.05\text{ }\Omega/\text{cm}$
- $3.00/60.0 = 0.05\text{ }\Omega/\text{cm}$
- $4.00/80.0 = 0.05\text{ }\Omega/\text{cm}$
- $5.00/100.0 = 0.05\text{ }\Omega/\text{cm}$
Since the ratio is constant within experimental limits, direct proportionality is confirmed.

(e) Current passing through the wire causes a heating effect. Since the resistance of metals increases with temperature, letting the wire warm up would change its resistance during the experiment, leading to systematic errors. Switching off the current allows the wire to cool back to room temperature between readings.

- (a) Voltmeter symbol (V in a circle) drawn correctly in parallel with the test wire [1 mark]
- (b) State current as 0.44 A [1 mark]
- (c) Calculate $R_3 = 3.00\text{ }\Omega$ [1 mark] and $R_5 = 5.00\text{ }\Omega$ [1 mark] (must keep decimal places consistent)
- (d) State "yes, directly proportional" [1 mark]; Calculate ratio $R/L$ or $L/R$ for at least two points to show it is constant ($0.05\text{ }\Omega/\text{cm}$) [2 marks]
- (e) State that current causes a temperature increase [1 mark]; Explain that a temperature rise increases the resistance of the wire, violating the assumption of a constant resistance wire [1 mark]

(a) Calculation of focal length:
$$f = \frac{30.0 \times 15.0}{30.0 + 15.0} = \frac{450.0}{45.0} = 10.0\text{ cm}$$

(b) Practical techniques to locate the sharpest image:
1. Move the screen slowly back and forth across the position of maximum sharpness, identifying the points where blurring just begins on either side, then place the screen exactly midway between these two positions.
2. Ensure the room is darkened so the image on the screen has high contrast and is easy to see.

(c) From the thin lens equation, we can rearrange to: $\frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$. Comparing this to $y = mx + c$ (where $y = \frac{1}{v}$ and $x = \frac{1}{u}$), the y-intercept $c$ is equal to $\frac{1}{f}$.

(d) Difficulties in measuring distances:
- Determining the exact position of the optical center of the lens inside the thick lens holder.
- Holding a meter ruler steady and parallel to the bench while measuring from the object to the lens.
- Parallax error when reading the ruler scale.

(e) Improvement:
- Use an optical bench with a built-in scale and clear index lines/pointers on the lens and screen holders to allow direct, parallax-free position readings.

- (a) Correct substitution and calculation to find $f = 10.0\text{ cm}$ (including units) [2 marks]
- (b) Two correct techniques described (e.g., move back and forth to find midpoint of sharpness [1 mark], perform in a darkened room [1 mark]) [2 marks]
- (c) Identifies the y-intercept as $\frac{1}{f}$ [2 marks]
- (d) Two correct difficulties in distance measurement (e.g., locating center of the lens, ruler alignment, parallax, depth of focus) [2 marks, 1 mark each]
- (e) One corresponding valid improvement (e.g., use an optical bench scale with aligned index markers/pointers, use a thin lens, or clamp a ruler alongside the apparatus) [2 marks]

- **Apparatus**:
- Wooden block (preferably with a hook)
- Spring balance (force meter) or a dual-range force sensor
- Selection of different surface materials (e.g., sandpaper, plastic, wood, carpet, glass)

- **Method**:
1. Place the wooden block on the first surface material.
2. Attach the spring balance horizontally to the hook of the wooden block.
3. Pull the spring balance horizontally with a gradually increasing force.
4. Record the force shown on the spring balance at the exact moment the block just begins to slide (this is the maximum static friction force).
5. Repeat the procedure for the same block on the other surface materials.

- **Control Variables**:
- The mass of the wooden block (and any added masses on top of it) must remain constant to ensure the normal contact force is the same.
- The same face of the wooden block (constant contact surface area) must be used on all surfaces.

- **Table Template**:
$$\begin{array}{|c|c|c|c|c|} \hline \text{Surface Material} & \text{Trial 1 Force } F_1 \text{ / N} & \text{Trial 2 Force } F_2 \text{ / N} & \text{Trial 3 Force } F_3 \text{ / N} & \text{Average Force } F_{\text{avg}} \text{ / N} \\ \hline & & & & \\ & & & & \\ & & & & \\ \hline \end{array}$$

- **Analysis**:
- Compare the average pulling force required to slide the block on each surface. The surface material that requires the largest pulling force has the highest friction, and the surface requiring the smallest force has the lowest friction.

- **Precaution / Accuracy Tip**:
- Pull the spring balance perfectly horizontally (parallel to the surface) to prevent any vertical component of the force from altering the normal force and the friction measurement.
- Alternatively, repeat the measurement at least three times on each surface and calculate the mean average to minimize random errors.

- **Apparatus** [1 mark]: List includes wooden block, spring balance/force sensor, and a selection of different surfaces.
- **Method** [2 marks]: Clear explanation of pulling the block slowly/gradually [1 mark]; describes measuring the maximum force right before the block starts to slide [1 mark].
- **Control Variables** [2 marks]: Identifies mass of the block/normal force [1 mark]; identifies contact area/same face of the block [1 mark].
- **Table Template** [2 marks]: Column for surface material [1 mark]; column for force (F) with unit Newton (N) [1 mark].
- **Analysis** [2 marks]: States that a higher pulling force indicates a higher friction force [1 mark]; compares the force values to rank the surfaces in order of friction [1 mark].
- **Precaution** [1 mark]: Pulling horizontally parallel to the surface, OR repeating trials and averaging the results to reduce experimental error.