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

The distance traveled during each phase is found from the area under the speed-time graph. First phase (acceleration): \(\text{distance}_1 = 0.5 \times 4.0\text{ s} \times 12\text{ m/s} = 24\text{ m}\). Second phase (constant speed): \(\text{distance}_2 = 6.0\text{ s} \times 12\text{ m/s} = 72\text{ m}\). Third phase (deceleration): \(\text{distance}_3 = 0.5 \times 2.0\text{ s} \times 12\text{ m/s} = 12\text{ m}\). Total distance = \(24\text{ m} + 72\text{ m} + 12\text{ m} = 108\text{ m}\). Total time = \(4.0\text{ s} + 6.0\text{ s} + 2.0\text{ s} = 12.0\text{ s}\). Average speed = \(\text{total distance} / \text{total time} = 108\text{ m} / 12.0\text{ s} = 9.0\text{ m/s}\).

[1 mark] C - Correct calculation of total distance (108 m) and dividing by total time (12 s) to find 9.0 m/s.

The useful work done in lifting the load is \(W = mgh = 15\text{ kg} \times 9.8\text{ m/s}^2 \times 4.0\text{ m} = 588\text{ J}\). The useful power output of the motor is \(P_{\text{out}} = W / t = 588\text{ J} / 5.0\text{ s} = 117.6\text{ W}\). The efficiency of the motor is given by \(\text{efficiency} = P_{\text{out}} / P_{\text{in}}\). Therefore, the power input is \(P_{\text{in}} = P_{\text{out}} / 0.60 = 117.6\text{ W} / 0.60 = 196\text{ W}\).

[1 mark] C - Correctly calculates useful power output (117.6 W) and divides by efficiency (0.60) to obtain 196 W.

The difference in pressure \(\Delta P\) between the top and bottom of the cylinder depends only on the difference in depth \(\Delta h\), which is the height of the cylinder itself (\(0.50\text{ m}\)). Using the pressure formula: \(\Delta P = \rho g \Delta h = 1200\text{ kg/m}^3 \times 9.8\text{ m/s}^2 \times 0.50\text{ m} = 5880\text{ Pa} = 5.88\text{ kPa}\), which rounds to \(5.9\text{ kPa}\).

[1 mark] B - Correctly applies pressure difference formula with depth difference equal to the cylinder height (0.50 m).

For a fixed mass of gas at constant temperature, Boyle's Law applies: \(P_1 V_1 = P_2 V_2\). Substituting the given values: \(1.2 \times 10^5\text{ Pa} \times 60\text{ cm}^3 = P_2 \times 24\text{ cm}^3\). Solving for \(P_2\) gives: \(P_2 = \frac{1.2 \times 10^5 \times 60}{24} = 3.0 \times 10^5\text{ Pa}\).

[1 mark] C - Correctly applies Boyle's law equation to find the new pressure.

When a wave crosses a boundary and changes speed, its frequency \(f\) remains constant. First, find the frequency of the wave in deep water: \(f = v_{\text{deep}} / \lambda_{\text{deep}} = 12\text{ cm/s} / 2.4\text{ cm} = 5.0\text{ Hz}\). In the shallow region, the frequency is still \(5.0\text{ Hz}\). Using the wave equation: \(\lambda_{\text{shallow}} = v_{\text{shallow}} / f = 8.0\text{ cm/s} / 5.0\text{ Hz} = 1.6\text{ cm}\).

[1 mark] A - Correctly calculates the constant frequency (5.0 Hz) and uses it to find the shallow wavelength.

First, calculate the refractive index \(n\) of the glass using Snell's Law: \(n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(42^\circ)}{\sin(26^\circ)} \approx \frac{0.6691}{0.4384} \approx 1.526\). The critical angle \(c\) is related to the refractive index by \(\sin(c) = \frac{1}{n}\). Thus, \(\sin(c) = \frac{1}{1.526} \approx 0.6552\), which gives \(c = \arcsin(0.6552) \approx 40.9^\circ\), rounding to \(41^\circ\).

[1 mark] B - Correct calculation of refractive index (1.53) followed by correct calculation of the critical angle (41 degrees).

In bright light: total resistance \(R_1 = 6.0\ \Omega + 2.0\ \Omega = 8.0\ \Omega\). The current is \(I_1 = 12\text{ V} / 8.0\ \Omega = 1.5\text{ A}\). The voltage across the fixed resistor is \(V_1 = I_1 \times R = 1.5\text{ A} \times 6.0\ \Omega = 9.0\text{ V}\). In the dark: total resistance \(R_2 = 6.0\ \Omega + 18\ \Omega = 24\ \Omega\). The current is \(I_2 = 12\text{ V} / 24\ \Omega = 0.5\text{ A}\). The voltage across the fixed resistor is \(V_2 = I_2 \times R = 0.5\text{ A} \times 6.0\ \Omega = 3.0\text{ V}\). The change in the voltmeter reading is a decrease of \(9.0\text{ V} - 3.0\text{ V} = 6.0\text{ V}\).

[1 mark] B - Correct calculation of the two voltmeter readings (9.0 V and 3.0 V) and identifying the 6.0 V decrease.

Redshift indicates that the light from a galaxy has been shifted towards the red end of the spectrum, which means its observed wavelength is longer than the wavelength emitted by the galaxy. This occurs because the galaxy is moving away from the Earth (receding), which is evidence for the expansion of the Universe.

[1 mark] D - Correct identification of galaxy motion (moving away) and wavelength comparison (longer observed wavelength).

To find the total distance, we can calculate the distance travelled in each of the three phases:
1. Acceleration phase: The car accelerates from rest (\(u = 0\)) at \(a = 2.0\text{ m/s}^2\) for \(t_1 = 3.0\text{ s}\). The final speed reached is \(v = u + a t_1 = 0 + 2.0 \times 3.0 = 6.0\text{ m/s}\). The distance travelled is \(d_1 = \frac{1}{2} \times v \times t_1 = \frac{1}{2} \times 6.0 \times 3.0 = 9.0\text{ m}\).
2. Constant speed phase: The car travels at \(6.0\text{ m/s}\) for \(t_2 = 4.0\text{ s}\). The distance travelled is \(d_2 = v \times t_2 = 6.0 \times 4.0 = 24.0\text{ m}\).
3. Deceleration phase: The car decelerates from \(6.0\text{ m/s}\) to rest in \(t_3 = 2.0\text{ s}\). The distance travelled is \(d_3 = \frac{1}{2} \times v \times t_3 = \frac{1}{2} \times 6.0 \times 2.0 = 6.0\text{ m}\).
Total distance = \(d_1 + d_2 + d_3 = 9.0 + 24.0 + 6.0 = 39.0\text{ m}\).

Award 1 mark for the correct option C.

First, calculate the useful work output done in lifting the mass: \(W = mgh = 15\text{ kg} \times 9.8\text{ m/s}^2 \times 8.0\text{ m} = 1176\text{ J}\).
Next, calculate the useful power output: \(P_{\text{out}} = \frac{W}{t} = \frac{1176\text{ J}}{6.0\text{ s}} = 196\text{ W}\).
Finally, calculate the efficiency: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{196\text{ W}}{280\text{ W}} \times 100\% = 70\%\).

Award 1 mark for the correct option C.

Let the direction to the right be positive.
Initial momentum of the system: \(p_i = m_X u_X + m_Y u_Y = (2.0 \times 6.0) + (4.0 \times 0) = 12.0\text{ kg m/s}\).
After the collision, trolley \(X\) travels to the left, so its velocity \(v_X = -1.5\text{ m/s}\).
By conservation of momentum: \(p_i = p_f \implies 12.0 = m_X v_X + m_Y v_Y \implies 12.0 = (2.0 \times -1.5) + (4.0 \times v_Y) \implies 12.0 = -3.0 + 4.0 v_Y \implies 15.0 = 4.0 v_Y \implies v_Y = 3.75\text{ m/s}\). Since this value is positive, the velocity is directed to the right.

Award 1 mark for the correct option C.

Since temperature is constant, the average kinetic energy and average speed of the gas molecules do not change. Therefore, the force of each individual collision remains the same. However, because the volume is reduced, the molecules are closer together and collide with the cylinder walls more frequently per unit area, resulting in increased pressure.

Award 1 mark for the correct option C.

Using Snell's Law: \(n = \frac{\sin i}{\sin r} \implies 1.50 = \frac{\sin i}{\sin 25^\circ} \implies \sin i = 1.50 \times \sin 25^\circ \approx 1.50 \times 0.4226 = 0.6339 \implies i \approx 39^\circ\).
The speed of light in glass is \(v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.50} = 2.0 \times 10^8\text{ m/s}\).

Award 1 mark for the correct option C.

Electrical energy transferred: \(E = V Q = 12\text{ V} \times 360\text{ C} = 4320\text{ J}\).
Current: \(I = \frac{Q}{t}\). Since time must be in seconds: \(t = 5.0 \times 60 = 300\text{ s}\).
Therefore, \(I = \frac{360\text{ C}}{300\text{ s}} = 1.2\text{ A}\).

Award 1 mark for the correct option A.

The equivalent resistance of the two parallel resistors is \(R_{\text{p}} = \frac{R \times R}{R + R} = 0.5R\).
The total resistance of the network is \(R_{\text{total}} = R_{\text{p}} + R = 0.5R + R = 1.5R\).
The circuit current is \(I = \frac{V}{1.5R} = \frac{2V}{3R}\).
The potential difference across the third resistor (in series) is \(V_3 = I \times R = \left(\frac{2V}{3R}\right) \times R = \frac{2}{3}V\).

Award 1 mark for the correct option C.

Stars with a mass much greater than the Sun follow this sequence: they form from a interstellar cloud of gas and dust (nebula), become stable main sequence stars, expand into red supergiants, explode in a supernova, and leave behind either a neutron star or a black hole.

Award 1 mark for the correct option B.

The deflection of a very small fraction of \(\alpha\)-particles through large angles (greater than \(90^\circ\)) indicates that they encountered a highly concentrated positive charge and mass. This concentration of mass and charge is the nucleus. Option A explains why most pass straight through, but not why some are deflected at large angles. Option B is incorrect because electrons have far too little mass to cause large deflections of heavy \(\alpha\)-particles. Option D is factually incorrect as some \(\alpha\)-particles are indeed deflected at large angles.

1 mark: C - Correct statement identifying that the concentrated mass and positive charge in the nucleus cause the large-angle deflection of a small fraction of alpha particles.

First, determine the number of half-lives that have elapsed in 24 hours: \(n = \frac{24\text{ hours}}{6.0\text{ hours}} = 4\). The number of active nuclei remaining is halved 4 times: \(N = N_0 \times \left(\frac{1}{2}\right)^4 = 8.0 \times 10^{12} \times \frac{1}{16} = 0.50 \times 10^{12} = 5.0 \times 10^{11}\).

1 mark: A - Correct calculation of the number of active nuclei remaining after 4 half-lives.

The unstretched length of the spring is \(L_0 = 15.0\text{ cm}\). For a load of \(4.0\text{ N}\), the extension \(e_1\) is \(17.0\text{ cm} - 15.0\text{ cm} = 2.0\text{ cm}\). Since the spring is within its limit of proportionality, Hooke's Law applies, so the extension is directly proportional to the load: \(e = \frac{F}{k}\). Since a load of \(4.0\text{ N}\) causes a \(2.0\text{ cm}\) extension, a load of \(8.0\text{ N}\) (which is twice as large) will cause twice the extension: \(e_2 = 2 \times 2.0\text{ cm} = 4.0\text{ cm}\). The total length of the spring with an \(8.0\text{ N}\) load is \(15.0\text{ cm} + 4.0\text{ cm} = 19.0\text{ cm}\).

1 mark: B - Correct application of Hooke's law to determine that an 8.0 N load produces a 4.0 cm extension, resulting in a total length of 19.0 cm.

The total pressure at depth \(h\) in a liquid is given by: \(P_{\text{total}} = P_{\text{atm}} + \rho g h\). Substituting the given values: \(1.3 \times 10^5\text{ Pa} = 1.0 \times 10^5\text{ Pa} + (1200\text{ kg/m}^3 \times 10\text{ m/s}^2 \times h)\). Simplifying this yields: \(0.3 \times 10^5\text{ Pa} = 12000 \times h\), which gives \(30000 = 12000 \times h\). Therefore, \(h = \frac{30000}{12000} = 2.5\text{ m}\).

1 mark: B - Correct calculation using the liquid pressure formula including atmospheric pressure to solve for depth.

When water waves travel from deep to shallow water, their speed decreases because of the shallower depth. The frequency of the wave remains unchanged because it is determined solely by the source. Using the wave equation \(v = f \lambda\), since the speed \(v\) decreases and the frequency \(f\) remains constant, the wavelength \(\lambda\) must also decrease.

1 mark: B - Correctly identifies that frequency remains constant, and wavelength and speed both decrease in shallower water.

First, calculate the equivalent resistance of the two parallel resistors: \(R_{\text{parallel}} = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\). Next, calculate the total resistance of the series circuit: \(R_{\text{total}} = R_{\text{parallel}} + 3.0\ \Omega = 3.0\ \Omega + 3.0\ \Omega = 6.0\ \Omega\). Finally, use Ohm's Law to find the current: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\).

1 mark: B - Correct calculation of parallel equivalent resistance, total circuit resistance, and the resulting total current.

According to the laws of planetary motion, planets further from the Sun experience a weaker gravitational pull, resulting in lower average orbital speeds. Because the orbital path is longer (larger radius) and the orbital speed is slower, the orbital period of Mars is significantly longer than that of the Earth.

1 mark: C - Correctly identifies that Mars, being further from the Sun, has a longer orbital period and a lower average orbital speed.

Redshift occurs when light waves are stretched, indicating that the source is moving away from the observer (receding). Hubble's Law states that the recessional velocity of a galaxy is directly proportional to its distance from us. Therefore, more distant galaxies are receding faster, which results in a greater redshift.

1 mark: A - Correctly links redshift to motion away from Earth and states that redshift increases with increasing distance.

1. Determine the mass of the object from its weight on Earth:
\(m = \frac{W_{\text{Earth}}}{g_{\text{Earth}}} = \frac{60\text{ N}}{10\text{ N/kg}} = 6.0\text{ kg}\).

2. Since mass is a fundamental property that does not change with location, the mass of the object on Planet X is also \(6.0\text{ kg}\).

3. Calculate the gravitational field strength on Planet X:
\(g_X = \frac{W_X}{m} = \frac{24\text{ N}}{6.0\text{ kg}} = 4.0\text{ N/kg}\).

Award 1 mark for the correct combination of mass and gravitational field strength: Mass = 6.0 kg and Gravitational field strength = 4.0 N/kg.

1. Find the volume of the submerged metal sphere using its mass and density:
\(V_{\text{sphere}} = \frac{\text{mass}}{\text{density}} = \frac{135\text{ g}}{9.0\text{ g/cm}^3} = 15\text{ cm}^3\).

2. When the sphere is fully submerged in the liquid, the reading on the cylinder increases by the volume of the sphere:
\(V_{\text{new}} = V_{\text{liquid}} + V_{\text{sphere}} = 80\text{ cm}^3 + 15\text{ cm}^3 = 95\text{ cm}^3\).

Award 1 mark for calculating the correct final volume of 95 cm³.

1. Use conservation of momentum to find the common velocity \(v\):
\(p_{\text{initial}} = m_1 u_1 + m_2 u_2 = (2.0 \times 6.0) + (4.0 \times 0) = 12.0\text{ kg m/s}\).
\(p_{\text{final}} = (m_1 + m_2) v = (2.0 + 4.0) v = 6.0 v\).
\(6.0 v = 12.0 \implies v = 2.0\text{ m/s}\).

2. Calculate the initial and final kinetic energies:
\(E_{k\text{, initial}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 2.0 \times (6.0)^2 = 36\text{ J}\).
\(E_{k\text{, final}} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 6.0 \times (2.0)^2 = 12\text{ J}\).

3. Calculate the loss in kinetic energy:
\(\text{Loss} = E_{k\text{, initial}} - E_{k\text{, final}} = 36\text{ J} - 12\text{ J} = 24\text{ J}\).

Award 1 mark for the correct common velocity of 2.0 m/s and kinetic energy loss of 24 J.

1. Calculate the weight of the block:
\(F = W = m \times g = 12\text{ kg} \times 10\text{ N/kg} = 120\text{ N}\).

2. Identify the faces of minimum and maximum surface area:
- Minimum area \(A_{\text{min}} = 0.20\text{ m} \times 0.30\text{ m} = 0.06\text{ m}^2\).
- Maximum area \(A_{\text{max}} = 0.30\text{ m} \times 0.50\text{ m} = 0.15\text{ m}^2\).

3. Calculate the maximum and minimum pressures:
- Maximum pressure \(P_{\text{max}} = \frac{F}{A_{\text{min}}} = \frac{120\text{ N}}{0.06\text{ m}^2} = 2000\text{ Pa}\).
- Minimum pressure \(P_{\text{min}} = \frac{F}{A_{\text{max}}} = \frac{120\text{ N}}{0.15\text{ m}^2} = 800\text{ Pa}\).

4. Calculate the difference:
\(\Delta P = P_{\text{max}} - P_{\text{min}} = 2000\text{ Pa} - 800\text{ Pa} = 1200\text{ Pa}\).

Award 1 mark for the correct difference between maximum and minimum pressure of 1200 Pa.

1. Calculate total electrical energy supplied by the heater:
\(E = P \times t = 800\text{ W} \times (3.0 \times 60\text{ s}) = 144,000\text{ J} = 144\text{ kJ}\).

2. Subtract energy lost to find useful energy gained by the block:
\(Q = 144\text{ kJ} - 16\text{ kJ} = 128\text{ kJ} = 128,000\text{ J}\).

3. Use the thermal energy equation to find temperature change \(\Delta T\):
\(Q = m c \Delta T \implies 128,000\text{ J} = (2.0\text{ kg}) \times (400\text{ J/kg}^\circ\text{C}) \times \Delta T\).
\(128,000 = 800 \times \Delta T \implies \Delta T = 160^\circ\text{C}\).

4. Determine final temperature:
\(T_{\text{final}} = 20^\circ\text{C} + 160^\circ\text{C} = 180^\circ\text{C}\).

Award 1 mark for calculating the correct final temperature of 180 °C.

When a water wave enters shallow water from deep water:
- Its frequency remains unchanged because the frequency of a wave is determined solely by the source generating it.
- Its speed decreases due to the interaction of the wave with the shallower boundaries of the container.
- Since wave speed \(v = f \lambda\), a decrease in speed with a constant frequency results in a proportional decrease in wavelength \(\lambda\).

Award 1 mark for selecting the correct row: Frequency is unchanged, Speed decreases, and Wavelength decreases.

1. Calculate the resistance of the parallel combination of two \(6.0\ \Omega\) resistors:
\(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\).

2. Calculate total circuit resistance, including the third resistor connected in series:
\(R_{\text{total}} = R_p + 6.0\ \Omega = 3.0\ \Omega + 6.0\ \Omega = 9.0\ \Omega\).

3. Calculate the total current drawn from the supply:
\(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{9.0\ \Omega} \approx 1.33\text{ A}\).

4. Calculate the potential difference across the parallel combination (and therefore across each of the parallel resistors):
\(V_p = I \times R_p = \frac{4}{3}\text{ A} \times 3.0\ \Omega = 4.0\text{ V}\).

Award 1 mark for identifying that the current is 1.33 A and potential difference across one parallel resistor is 4.0 V.

- Stars with a mass much larger than the Sun will eventually end their lives in a supernova explosion, leaving behind a neutron star or a black hole. (Stars with a mass similar to the Sun become red giants and then white dwarfs).
- The redshift of light from distant galaxies shows that they are moving away from us, providing evidence that the Universe is expanding.
- Therefore, option C is correct.

Award 1 mark for selecting the correct option describing a supernova for massive stars and redshift indicating an expanding universe.

First, find the total distance travelled by calculating the area under the speed-time graph. Distance during acceleration: \(d_1 = 0.5 \times 5.0\text{ s} \times 12.0\text{ m/s} = 30.0\text{ m}\). Distance during constant speed: \(d_2 = (15.0 - 5.0)\text{ s} \times 12.0\text{ m/s} = 120.0\text{ m}\). Distance during deceleration: \(d_3 = 0.5 \times (20.0 - 15.0)\text{ s} \times 12.0\text{ m/s} = 30.0\text{ m}\). Total distance travelled \(d = 30.0 + 120.0 + 30.0 = 180.0\text{ m}\). The average speed is given by \(\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{180.0\text{ m}}{20.0\text{ s}} = 9.0\text{ m/s}\).

1 mark for the correct total distance of 180 m and correct division by 20 s to obtain 9.0 m/s.

First, find the useful power output of the motor: \(\text{Useful power} = \text{Efficiency} \times \text{Input power} = 0.40 \times 1500\text{ W} = 600\text{ W}\). Next, calculate the work done in lifting the load: \(W = mgh = 50\text{ kg} \times 9.8\text{ m/s}^2 \times 12\text{ m} = 5880\text{ J}\). Finally, calculate the time taken: \(t = \frac{\text{Work done}}{\text{Useful power}} = \frac{5880\text{ J}}{600\text{ W}} = 9.8\text{ s}\).

1 mark for the correct determination of useful power (600 W), correct work done (5880 J), and the correct time of 9.8 s.

By the conservation of momentum, taking right as the positive direction: Initial momentum \(p_{\text{initial}} = (2.0\text{ kg} \times 6.0\text{ m/s}) + (3.0\text{ kg} \times (-2.0\text{ m/s})) = 12.0 - 6.0 = +6.0\text{ kg}\cdot\text{m/s}\). The combined mass after sticking together is \(M = 2.0\text{ kg} + 3.0\text{ kg} = 5.0\text{ kg}\). The common velocity \(v\) is given by \(v = \frac{p_{\text{initial}}}{M} = \frac{6.0\text{ kg}\cdot\text{m/s}}{5.0\text{ kg}} = 1.2\text{ m/s}\). Since the sign is positive, the direction is to the right.

1 mark for correct conservation of momentum equation including correct signs, and correct calculation of final velocity as 1.2 m/s to the right.

The pressure in the liquid is determined by the input piston: \(P = \frac{F_1}{A_1} = \frac{150\text{ N}}{8.0 \times 10^{-4}\text{ m}^2} = 1.875 \times 10^5\text{ Pa} \approx 1.9 \times 10^5\text{ Pa}\). The force exerted by the output piston is \(F_2 = F_1 \times \frac{A_2}{A_1} = 150\text{ N} \times \frac{200\text{ cm}^2}{8.0\text{ cm}^2} = 3750\text{ N}\).

1 mark for correctly calculating the pressure in the liquid (1.9 x 10^5 Pa) and the force exerted by the output piston (3750 N).

Using Boyle's Law at constant temperature: \(P_1 V_1 = P_2 V_2\). Substitute the given values: \(1.5 \times 10^5\text{ Pa} \times 120\text{ cm}^3 = P_2 \times 45\text{ cm}^3\). Solving for \(P_2\) gives \(P_2 = \frac{1.5 \times 10^5 \times 120}{45} = 4.0 \times 10^5\text{ Pa}\).

1 mark for the correct application of Boyle's Law and correct calculation to find 4.0 x 10^5 Pa.

Convert time to seconds: \(t = 4.0\text{ minutes} = 240\text{ s}\). Total charge transferred: \(Q = I \times t = 1.5\text{ A} \times 240\text{ s} = 360\text{ C}\). The chemical energy transferred is: \(E = V \times Q = 9.0\text{ V} \times 360\text{ C} = 3240\text{ J}\).

1 mark for converting minutes to seconds (240 s) and calculating the correct energy of 3240 J.

At room temperature, the potential divider rule gives the voltage across the thermistor: \(V_{\text{out}} = 12.0\text{ V} \times \frac{6.0\text{ k}\Omega}{3.0\text{ k}\Omega + 6.0\text{ k}\Omega} = 8.0\text{ V}\). As the temperature of the thermistor increases, its resistance decreases. Consequently, it takes a smaller fraction of the supply voltage, so the reading on the voltmeter decreases.

1 mark for calculating the correct initial voltage of 8.0 V and correctly identifying that the voltage across the thermistor decreases as temperature increases.

First, convert the orbital period into seconds: \(T = 96\text{ minutes} = 96 \times 60 = 5760\text{ s}\). The formula for orbital speed in a circular orbit is \(v = \frac{2\pi r}{T}\). Substitute \(r = 7.0 \times 10^3\text{ km}\) and \(T = 5760\text{ s}\): \(v = \frac{2 \times \pi \times 7.0 \times 10^3\text{ km}}{5760\text{ s}} \approx 7.6\text{ km/s}\).

1 mark for converting minutes to seconds (5760 s) and correctly using the orbital speed formula to obtain 7.6 km/s.

(a)(i) Using the equation of motion for acceleration: \(a = \frac{v - u}{t} = \frac{15\text{ m/s} - 0\text{ m/s}}{5.0\text{ s}} = 3.0\text{ m/s}^2\). (a)(ii) Total distance is the total area under the velocity-time graph. We can divide this into three sections: 1. Left triangle: \(\text{Area}_1 = \frac{1}{2} \times 5.0\text{ s} \times 15\text{ m/s} = 37.5\text{ m}\). 2. Center rectangle: \(\text{Area}_2 = 10\text{ s} \times 15\text{ m/s} = 150\text{ m}\). 3. Right triangle: \(\text{Area}_3 = \frac{1}{2} \times 3.0\text{ s} \times 15\text{ m/s} = 22.5\text{ m}\). Total distance = \(37.5\text{ m} + 150\text{ m} + 22.5\text{ m} = 210\text{ m}\). (b) When the car travels at a constant velocity, there is no acceleration. The forward driving force from the engine is exactly balanced by the opposing resistive forces (friction and air resistance). This means the resultant force acting on the car is zero, hence there is no acceleration according to Newton's first law of motion.

(a)(i) [1] for correct formula or substitution of values, [1] for correct final answer of 3.0 with unit m/s^2. (a)(ii) [1] for recognizing that distance is represented by the area under the velocity-time graph, [1] for correct calculation of any of the component areas, [1] for correct total distance of 210 m with unit. (b) [1] for stating that the driving force equals the resistive forces / air resistance, [1] for stating that the resultant force is zero, [1] for stating that zero resultant force results in zero acceleration / constant velocity.

(a) Useful work done is equal to the gain in gravitational potential energy: \(W = mgh = 85\text{ kg} \times 9.8\text{ N/kg} \times 12\text{ m} = 9996\text{ J}\) (or approximately \(10000\text{ J}\)). (b) Power output is the rate of doing useful work: \(P = \frac{W}{t} = \frac{9996\text{ J}}{6.0\text{ s}} = 1666\text{ W}\) (or approximately \(1670\text{ W}\) or \(1.67\text{ kW}\)). (c)(i) Efficiency = \(\frac{\text{Useful Power Output}}{\text{Total Power Input}} \times 100\% = \frac{1666\text{ W}}{2500\text{ W}} \times 100\% = 66.6\%\) (or \(67\%\)). (c)(ii) The wasted energy is dissipated to the motor components and the surrounding environment as thermal energy (heat) and sound energy.

(a) [1] for formula W = mgh or correct substitution, [1] for correct final value of 9996 J or 10000 J with unit. (b) [1] for formula P = W/t or correct substitution, [1] for correct final value of 1666 W or 1670 W with unit. (c)(i) [1] for correct substitution into efficiency equation, [1] for correct calculation of 66.6% or 67%. (c)(ii) [1] for stating that wasted energy is dissipated/lost to the surroundings, [1] for identifying the forms as thermal energy (heat) and/or sound.

(a) The pressure due to a liquid is given by \(p = \rho g h = 1025\text{ kg/m}^3 \times 9.8\text{ N/kg} \times 150\text{ m} = 1,506,750\text{ Pa}\) (which can be rounded to \(1.5 \times 10^6\text{ Pa}\)). (b) The total pressure is the sum of liquid pressure and atmospheric pressure: \(p_{\text{total}} = p_{\text{water}} + p_{\text{atm}} = 1,506,750\text{ Pa} + 100,000\text{ Pa} = 1,606,750\text{ Pa}\) (which is \(1.6 \times 10^6\text{ Pa}\)). (c)(i) Area \(A = \pi r^2 = \pi \times (0.18\text{ m})^2 = 0.1018\text{ m}^2\), which is approximately \(0.10\text{ m}^2\). (c)(ii) The net pressure acting on the window is the difference between the external total pressure and the internal pressure: \(\Delta p = p_{\text{total}} - p_{\text{inside}} = 1,606,750\text{ Pa} - 100,000\text{ Pa} = 1,506,750\text{ Pa}\) (this is exactly equal to the water pressure). Force \(F = \Delta p \times A = 1,506,750\text{ Pa} \times 0.1018\text{ m}^2 = 153,387\text{ N}\). If the rounded area of \(0.10\text{ m}^2\) is used, \(F = 1,506,750\text{ Pa} \times 0.10\text{ m}^2 = 150,675\text{ N}\). Accept any answer in the range \(1.5 \times 10^5\text{ N}\) to \(1.53 \times 10^5\text{ N}\).

(a) [1] for formula p = rho * g * h or correct substitution, [1] for correct final value of 1.5 * 10^6 Pa (or 1.51 * 10^6 Pa). (b) [1] for adding atmospheric pressure to the liquid pressure, [1] for correct final value of 1.6 * 10^6 Pa (or 1.61 * 10^6 Pa). (c)(i) [1] for substitution showing A = pi * 0.18^2 = 0.102 m^2. (c)(ii) [1] for identifying that pressure difference is 1.5 * 10^6 Pa, [1] for formula F = p * A, [1] for correct force value in range 1.5 * 10^5 N to 1.53 * 10^5 N with unit.

(a) Gas particles are in constant, rapid, and random motion. As they move, they collide with the inner walls of the cylinder. During each collision, a particle exerts a tiny force on the wall. The cumulative force of all these collisions acting per unit area of the walls results in the gas pressure. (b)(i) Since the temperature remains constant, the gas obeys Boyle's law: \(p_1 V_1 = p_2 V_2\). Rearranging to find the new pressure \(p_2\): \(p_2 = \frac{p_1 V_1}{V_2} = \frac{1.2 \times 10^5\text{ Pa} \times 4.5 \times 10^{-4}\text{ m}^3}{1.8 \times 10^{-4}\text{ m}^3} = 3.0 \times 10^5\text{ Pa}\). (b)(ii) When the volume of the gas is reduced, the gas particles are closer together (the density of the gas is greater). As a result, they collide with the walls of the cylinder more frequently (there are more collisions per unit area per second). Because the temperature is constant, the average speed of the particles does not change, but the increased frequency of collisions results in a higher overall pressure.

(a) [1] for referencing constant random motion of particles, [1] for stating particles collide with the walls, [1] for stating that the collision forces acting per unit area create the pressure. (b)(i) [1] for formula p1 * V1 = p2 * V2, [1] for correct substitution of values, [1] for correct final pressure of 3.0 * 10^5 Pa. (b)(ii) [1] for stating particles are closer together / more crowded / more concentrated, [1] for stating there is a higher frequency of collisions with the walls (collisions per unit area/time).

(a) The critical angle is defined as the angle of incidence in an optically denser medium for which the angle of refraction in the less dense medium (typically air) is \(90^\circ\). (b)(i) The speed of light in glass is calculated using the formula for refractive index: \(n = \frac{c}{v}\). Hence, \(v = \frac{c}{n} = \frac{3.00 \times 10^8\text{ m/s}}{1.52} = 1.97 \times 10^8\text{ m/s}\). (b)(ii) The critical angle \(c\) is calculated using the formula: \(\sin(c) = \frac{1}{n} = \frac{1}{1.52} = 0.6579\). Taking the inverse sine: \(c = \sin^{-1}(0.6579) = 41.1^\circ\). (c) Since the angle of incidence (\(45^\circ\)) is greater than the critical angle (\(41.1^\circ\)), the light cannot escape into the air. Instead, the ray is fully reflected back inside the glass block. This phenomenon is called total internal reflection.

(a) [1] for stating it is the angle of incidence in the denser medium, [1] for stating the angle of refraction is 90 degrees in the less dense medium. (b)(i) [1] for formula n = c/v or substitution, [1] for correct speed of 1.97 * 10^8 m/s with unit. (b)(ii) [1] for formula sin(c) = 1/n or substitution, [1] for correct angle of 41.1 degrees (or 41 degrees). (c) [1] for describing that the ray is reflected back inside the glass block (at an angle of 45 degrees), [1] for stating the name: total internal reflection.

(a) The power supply, the fixed resistor of \(1500\ \Omega\), and the LDR are connected in a single closed loop (in series). One terminal of the power supply connects to the fixed resistor, which connects to the LDR, which then connects back to the other terminal of the power supply. (b)(i) For components connected in series: \(R_{\text{total}} = R_1 + R_2 = 1500\ \Omega + 300\ \Omega = 1800\ \Omega\). (b)(ii) The current is calculated using Ohm's law: \(I = \frac{V}{R_{\text{total}}} = \frac{12.0\text{ V}}{1800\ \Omega} = 6.67 \times 10^{-3}\text{ A}\) (or \(6.7\text{ mA}\)). (b)(iii) The potential difference across the LDR is \(V_{\text{LDR}} = I \times R_{\text{LDR}} = 6.67 \times 10^{-3}\text{ A} \times 300\ \Omega = 2.0\text{ V}\). (c) When the light intensity decreases, the resistance of the LDR increases. The total resistance of the circuit increases, which reduces the overall current. However, because the LDR's resistance now represents a larger fraction of the total resistance, the potential difference across it increases (the LDR takes a larger share of the fixed supply voltage).

(a) [1] for stating all components are connected in series, [1] for clarifying a closed loop consisting of the battery/supply, fixed resistor, and LDR. (b)(i) [1] for 1800 ohms. (b)(ii) [1] for formula I = V/R or substitution, [1] for correct current of 6.7 mA or 6.67 * 10^-3 A with unit. (b)(iii) [1] for correct potential difference of 2.0 V. (c) [1] for stating that the potential difference across the LDR increases, [1] for explaining that the LDR's resistance is now a larger proportion of the total circuit resistance.

(a) A transformer relies on electromagnetic induction, which requires a changing magnetic field to induce an electromotive force (e.m.f.) in the secondary coil. A constant direct current (d.c.) in the primary coil produces a constant/steady magnetic field, so no e.m.f. is induced in the secondary. (b) Using the transformer turns ratio equation: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\). Therefore, the secondary voltage \(V_s = V_p \times \frac{N_s}{N_p} = 230\text{ V} \times \frac{2000}{400} = 230 \times 5 = 1150\text{ V}\). (c)(i) An ideal transformer is \(100\%\) efficient, meaning input power equals output power: \(P_{\text{in}} = P_{\text{out}} = 1150\text{ W}\). Using the power equation: \(P_{\text{in}} = I_p \times V_p\). Therefore, \(I_p = \frac{P_{\text{in}}}{V_p} = \frac{1150\text{ W}}{230\text{ V}} = 5.0\text{ A}\). (c)(ii) Two main reasons for energy losses in real transformers are: 1. Heating effect in the copper wires of the coils due to their electrical resistance. 2. Eddy currents induced in the iron core which cause resistive heating of the core. (Other correct reasons include magnetic hysteresis or magnetic flux leakage).

(a) [1] for stating d.c. produces a constant/steady magnetic field, [1] for explaining that a changing magnetic field/flux is needed to induce an e.m.f. (b) [1] for transformer equation or substitution, [1] for correct output voltage of 1150 V. (c)(i) [1] for recognizing that input power equals output power (1150 W) or for equating primary power to secondary power, [1] for correct primary current of 5.0 A with unit. (c)(ii) [1] each for any two valid reasons: thermal losses in coil resistance, thermal losses in the core due to eddy currents, magnetic flux leakage, or core hysteresis losses.

(a)(i) Redshift is the increase in the observed wavelength (or decrease in the observed frequency) of electromagnetic radiation emitted by a source that is moving away from the observer. (a)(ii) Redshift indicates that these distant galaxies are moving away from Earth (they are receding). (b)(i) \(v\) represents the recession velocity of the galaxy, and \(d\) represents the distance of the galaxy from Earth. (b)(ii) The age of the Universe \(t\) is approximately given by the reciprocal of the Hubble constant: \(t \approx \frac{1}{H_0}\). Therefore: \(t = \frac{1}{2.2 \times 10^{-18}\text{ s}^{-1}} = 4.545 \times 10^{17}\text{ s}\). To convert this into years: \(t = \frac{4.545 \times 10^{17}\text{ s}}{3.15 \times 10^7\text{ s/year}} = 1.44 \times 10^{10}\text{ years}\) (or \(1.4 \times 10^{10}\text{ years}\), which is 14 billion years).

(a)(i) [1] for noting that the wavelength of light increases (or frequency decreases), [1] for noting this happens when the source moves away from the observer. (a)(ii) [1] for stating that the galaxies are moving away from Earth / us. (b)(i) [1] for v = recession velocity of the galaxy, [1] for d = distance of the galaxy from Earth. (b)(ii) [1] for using the relation t = 1/H0, [1] for calculating the time in seconds (4.5 * 10^17 s), [1] for correct conversion to years, giving approximately 1.4 * 10^10 years (or 1.44 * 10^10 years).

(a) Momentum is defined by the equation:
\(p = mv\)
where \(p\) is momentum, \(m\) is mass, and \(v\) is velocity.

(b) Using the principle of conservation of momentum:
Total momentum before collision = Total momentum after collision
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
Substitute the values:
\((0.50 \times 2.4) + (0.30 \times 0) = (0.50 \times 0.60) + (0.30 \times v_B)\)
\(1.20 = 0.30 + 0.30 v_B\)
\(0.30 v_B = 0.90\)
\(v_B = 3.0\text{ m/s}\)

(c) The average force \(F\) exerted on glider B is given by the rate of change of momentum:
\(F = \frac{\Delta p}{\Delta t} = \frac{m_B v_B - m_B u_B}{\Delta t}\)
\(F = \frac{0.30 \times 3.0 - 0}{0.12} = \frac{0.90}{0.12} = 7.5\text{ N}\)

(d) Calculate the total kinetic energy before the collision:
\(E_{ki} = \frac{1}{2} m_A u_A^2 = \frac{1}{2} (0.50) (2.4)^2 = 1.44\text{ J}\)

Calculate the total kinetic energy after the collision:
\(E_{kf} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = \frac{1}{2} (0.50) (0.60)^2 + \frac{1}{2} (0.30) (3.0)^2\)
\(E_{kf} = 0.09 + 1.35 = 1.44\text{ J}\)

Since the total kinetic energy before the collision is equal to the total kinetic energy after the collision, the collision is perfectly elastic.

(a) [1 mark]
- \(p = mv\) or word equation: \(\text{momentum} = \text{mass} \times \text{velocity}\).

(b) [3 marks]
- [1 mark] Correct statement of conservation of momentum formula or correct substitution: \(0.50 \times 2.4 = 0.50 \times 0.60 + 0.30 \times v_B\).
- [1 mark] Correct rearrangement: \(0.30 v_B = 0.90\).
- [1 mark] Correct final velocity value of \(3.0\) with correct units (\(\text{m/s}\) or \(\text{m s}^{-1}\)).

(c) [2 marks]
- [1 mark] Correct force formula: \(F = \frac{\Delta p}{t}\) or \(F = \frac{m(v-u)}{t}\) used with substitution.
- [1 mark] Correct value of \(7.5\) with unit \(\text{N}\).

(d) [2 marks]
- [1 mark] Correct calculation of both initial kinetic energy (\(1.44\text{ J}\)) and final kinetic energy (\(1.44\text{ J}\)).
- [1 mark] Correct statement that the collision is elastic because kinetic energy is conserved / remains unchanged.

(a) Faraday's law states that the magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate at which magnetic flux is cut (or the rate of change of magnetic flux linkage).

(b) As the magnet enters the coil, its magnetic field lines cut across the wires of the coil. This change in the magnetic field (or change in magnetic flux linkage) within the coil induces an electromotive force (e.m.f.) across the coil.

(c) (i) The maximum deflection of the voltmeter needle increases (larger deflection). Moving the magnet faster causes the magnetic field lines to be cut at a greater rate, which increases the induced e.m.f.
(ii) The voltmeter needle returns to zero (no deflection). When the magnet is stationary, there is no relative motion, meaning there is no change in magnetic flux linkage (magnetic field lines are not being cut) and therefore no induced e.m.f.

(a) [2 marks]
- [1 mark] Induced e.m.f./voltage is proportional to / depends on.
- [1 mark] Rate of change of magnetic flux linkage / rate of cutting of magnetic field lines.

(b) [2 marks]
- [1 mark] Magnetic field lines of the magnet pass through the coil.
- [1 mark] Movement of the magnet causes a change in magnetic flux linkage / cutting of magnetic field lines.

(c)(i) [2 marks]
- [1 mark] Deflection increases / needle moves further.
- [1 mark] Rate of change of magnetic flux increases / field lines cut faster.

(c)(ii) [2 marks]
- [1 mark] Deflection is zero / needle does not move.
- [1 mark] No change in magnetic flux linkage / field lines not being cut.

a) The object is at 10.5 cm and the lens is at 35.5 cm, so the distance is \(35.5\text{ cm} - 10.5\text{ cm} = 25.0\text{ cm}\). b) The lens is at 35.5 cm and the screen is at 73.0 cm, so the distance is \(73.0\text{ cm} - 35.5\text{ cm} = 37.5\text{ cm}\). c) Use the formula: \(f_1 = \frac{25.0 \times 37.5}{25.0 + 37.5} = \frac{937.5}{62.5} = 15.0\text{ cm}\). d) The average focal length is: \(f_{\text{avg}} = \frac{15.0 + 14.8 + 15.2}{3} = 15.0\text{ cm}\). e) The major difficulties are: (1) depth of focus, where the image appears sharp over a short range of positions rather than a single point; (2) aligning the optical centers horizontally. f) Improvements include: working in a darkened room, using an illuminated object with a distinct pattern (like grid lines), or taking the midpoint of the range of positions where the image is sharp.

- (a) [2 marks] 1 mark for correct subtraction showing working; 1 mark for accuracy (25.0 cm). - (b) [1 mark] Correct image distance (37.5 cm). - (c) [3 marks] 1 mark for correct calculation of 15.0; 1 mark for correct unit (cm); 1 mark for appropriate significant figures (2 or 3 s.f.). - (d) [1 mark] Correct calculation of average (15.0 cm). - (e) [2 marks] 1 mark for each valid difficulty (e.g., image is faint/room is too bright, depth of focus makes exact position hard to find, alignment issues). - (f) [1 mark] One valid experimental improvement linked to the stated difficulties (e.g., use a darkened room, use a grid on the object, take the midpoint of the sharp range).

a) A standard circuit diagram is drawn containing: a power source (cell or battery), an open or closed switch, an ammeter in series with the main loop, a resistance wire with a sliding contact, and a voltmeter in parallel across only the active length of the resistance wire. b) The voltmeter reading is \(1.5\text{ V} + 3 \times 0.1\text{ V} = 1.8\text{ V}\). The ammeter reading is halfway between 0.4 A and 0.5 A, which is 0.45 A. c) Use Ohm's Law: \(R_1 = \frac{1.8\text{ V}}{0.45\text{ A}} = 4.0\ \Omega\). d) Resistance per unit length: \(r_1 = \frac{4.0\ \Omega}{40.0\text{ cm}} = 0.10\ \Omega/\text{cm}\). e) The resistance at L = 80 cm is \(R_2 = \frac{2.4\text{ V}}{0.30\text{ A}} = 8.0\ \Omega\). f) At 40.0 cm, \(R_1 = 4.0\ \Omega\). At 80.0 cm, \(R_2 = 8.0\ \Omega\). Doubling the length from 40.0 cm to 80.0 cm exactly doubles the resistance from \(4.0\ \Omega\) to \(8.0\ \Omega\). Therefore, \(R \propto L\), and the results support the suggestion.

- (a) [3 marks] 1 mark for correct symbols for cell, switch, ammeter, and voltmeter; 1 mark for series layout of cell, switch, ammeter, and wire; 1 mark for voltmeter in parallel across length L of the wire. - (b) [2 marks] 1 mark for \(V_1 = 1.8\text{ V}\); 1 mark for \(I_1 = 0.45\text{ A}\). - (c) [1 mark] Correct calculation of \(R_1 = 4.0\ \Omega\) (accept 4). - (d) [1 mark] Correct calculation of \(r_1 = 0.10\) with correct unit \(\Omega/\text{cm}\) (or \(\Omega\text{ cm}^{-1}\)). - (e) [1 mark] Correct calculation of \(R_2 = 8.0\ \Omega\) (accept 8). - (f) [2 marks] 1 mark for state 'Yes'; 1 mark for showing ratios are equal (\(4/40 = 8/80 = 0.10\)) or explaining that doubling the length doubles the resistance.

a) The thermometer shows the level is exactly one division above 84.0 °C. Since each division is 0.5 °C, the temperature is \(84.5\text{ }^\circ\text{C}\). b) Insert 84.5 into the table for t = 0 in both columns. c) Beaker A temperature drop: \(84.5 - 66.0 = 18.5\text{ }^\circ\text{C}\). Beaker B temperature drop: \(84.5 - 56.0 = 28.5\text{ }^\circ\text{C}\). d) Rate of cooling: For Beaker A: \(R_A = 18.5 / 300 \approx 0.0617\text{ }^\circ\text{C/s}\) (or 0.062). For Beaker B: \(R_B = 28.5 / 300 = 0.095\text{ }^\circ\text{C/s}\). e) Beaker B cooled faster (larger temperature drop and cooling rate). The lid prevents hot air and steam from escaping, thereby reducing heat transfer via convection and evaporation. f) Control variables include keeping the volume of water the same, using identical beakers, and performing both trials under the same ambient room temperature.

- (a) [1 mark] Correct reading of 84.5 °C. - (b) [1 mark] 84.5 entered correctly in table for t = 0. - (c) [2 marks] 1 mark for \(\Delta \theta_A = 18.5\text{ }^\circ\text{C}\); 1 mark for \(\Delta \theta_B = 28.5\text{ }^\circ\text{C}\). - (d) [2 marks] 1 mark for correct calculations of rates (0.062 and 0.095); 1 mark for correct unit (\(^\circ\text{C/s}\) or \(^\circ\text{C}\text{ s}^{-1}\)). - (e) [2 marks] 1 mark for identifying Beaker B cooled faster; 1 mark for mentioning that the lid reduces convection or evaporation. - (f) [2 marks] 1 mark for each of any two valid control variables (same volume of water, same beaker size/material, same ambient temperature).

1. Additional Apparatus: A meter ruler to measure heights and distances, and a light gate connected to a digital timer (or a stopwatch to measure the time taken to travel a fixed distance d at the bottom). 2. Diagram: The diagram must show an inclined ramp with the toy car placed on it, a clear vertical line marked 'h' representing the height from the horizontal bench to the center of mass/base of the car, and a speed measuring system (such as a light gate) placed at the bottom flat section. 3. Method: Place the toy car at a chosen position on the ramp. Measure the vertical height h from the bench to this position using the meter ruler. Release the car from rest. Measure the time t taken for the car's card of width w to pass through the light gate, then calculate speed \(v = w/t\) (or measure time to traverse a short distance d at the bottom and use \(v = d/t\)). Repeat the process for at least 5 different release heights h. Repeat each measurement to take averages. 4. Control Variables: Keep the mass of the car constant, the surface and slope angle of the ramp constant, and ensure the car is always released from rest (without pushing). 5. Table of Results: Columns: Height \(h / \text{cm}\), Time 1 \(t_1 / \text{s}\), Time 2 \(t_2 / \text{s}\), Average Time \(t_{\text{avg}} / \text{s}\), Speed \(v / \text{m/s}\). 6. Analysis: Plot a graph of \(v^2\) against \(h\). A straight line through the origin would indicate that the speed squared is directly proportional to the release height (consistent with conservation of energy \(mgh = \frac{1}{2}mv^2\)).

- **Apparatus** [1 mark]: List includes meter ruler and stopwatch/light gates. - **Diagram** [2 marks]: 1 mark for drawing ramp with car and a clearly marked vertical release height h; 1 mark for showing a speed measurement method at the bottom of the slope. - **Method** [3 marks]: 1 mark for describing how to measure vertical height h with the ruler; 1 mark for explaining how to calculate speed (either \(v = d/t\) or via light gate); 1 mark for repeating for a minimum of 5 different heights. - **Control Variables** [2 marks]: 1 mark for keeping the same car/mass; 1 mark for keeping the ramp surface/angle constant, or releasing from rest. - **Table** [1 mark]: Correctly drawn headers with units (e.g., \(h / \text{cm}\), \(t / \text{s}\), \(v / \text{m/s}\)). - **Analysis** [1 mark]: Suggestion to plot a graph of speed \(v\) or \(v^2\) against height \(h\).