2024 Cambridge IGCSE Physics (0625) 模擬試題連答案詳解

1. Find the maximum speed reached during acceleration:
\(v = u + at = 0 + (2.0\text{ m/s}^2 \times 6.0\text{ s}) = 12\text{ m/s}\).

2. Calculate the distance travelled in each of the three stages:
- Stage 1 (acceleration): \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6.0\text{ s} \times 12\text{ m/s} = 36\text{ m}\).
- Stage 2 (constant speed): \(d_2 = \text{speed} \times \text{time} = 12\text{ m/s} \times 10\text{ s} = 120\text{ m}\).
- Stage 3 (deceleration): \(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 12\text{ m/s} = 24\text{ m}\).

3. Sum the distances for the total distance:
\(d_{\text{total}} = 36\text{ m} + 120\text{ m} + 24\text{ m} = 180\text{ m}\).

Award 1 mark for the correct calculation of total distance (180 m).

1. Calculate the total volume of the 10 washers by water displacement:
\(V = 36.0\text{ cm}^3 - 30.0\text{ cm}^3 = 6.0\text{ cm}^3\).

2. Use the density formula:
\(\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{45.0\text{ g}}{6.0\text{ cm}^3} = 7.50\text{ g/cm}^3\).

Award 1 mark for the correct calculation of the density of the metal.

Brownian motion is the random, erratic movement of visible microscopic particles (such as pollen grains or smoke particles) suspended in a fluid (gas or liquid). It is caused by the continuous bombardment of these particles by much smaller, rapidly moving molecules of the fluid (which are too small to be seen directly).

Award 1 mark for selecting the correct explanation of Brownian motion.

1. Use the wave equation to find the speed of the wave in deep water:
\(v_{\text{deep}} = f \lambda_{\text{deep}} = 15\text{ Hz} \times 0.040\text{ m} = 0.60\text{ m/s}\).

2. When a wave changes medium (e.g., from deep to shallow water), its frequency \(f\) remains constant:
\(f = 15\text{ Hz}\).

3. Use the wave equation with the new speed in shallow water to find the new wavelength:
\(\lambda_{\text{shallow}} = \frac{v_{\text{shallow}}}{f} = \frac{0.45\text{ m/s}}{15\text{ Hz}} = 0.030\text{ m} = 3.0\text{ cm}\).

Award 1 mark for the correct calculation of the wavelength in shallow water.

1. The resistance \(R\) of a wire of length \(L\) and cross-sectional area \(A\) is given by:
\(R = \rho \frac{L}{A}\), where \(A = \pi r^2 = \pi \frac{d^2}{4}\).
So, \(R \propto \frac{L}{d^2}\).

2. For the second wire, \(L_2 = 2L\) and \(d_2 = \frac{1}{2}d\).

3. Substitute these into the proportionality:
\(R_2 \propto \frac{2L}{\left(\frac{1}{2}d\right)^2} = \frac{2L}{\frac{1}{4}d^2} = 8 \frac{L}{d^2}\).

Thus, \(R_2 = 8R\).

Award 1 mark for the correct factor of resistance change.

1. Dark, dull surfaces are excellent emitters of thermal radiation. Therefore, Canister 1 (dull black) will emit radiation at the highest rate, meaning it cools down the fastest.

2. Light, shiny surfaces are very poor emitters of thermal radiation (and excellent reflectors). Therefore, Canister 2 (shiny silver) will emit radiation at the lowest rate, meaning it cools down the slowest.

Award 1 mark for identifying Canister 1 as the fastest and Canister 2 as the slowest cooler.

1. Correct the detector readings by subtracting background radiation:
- Initial corrected count rate = \(216 - 24 = 192\text{ counts/minute}\).
- Corrected count rate after \(6.0\text{ hours}\) = \(48 - 24 = 24\text{ counts/minute}\).

2. Determine the fraction of activity remaining:
\(\frac{24}{192} = \frac{1}{8}\).

3. Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), this represents exactly 3 half-lives.

4. Calculate the half-life:
\(3 \times T_{1/2} = 6.0\text{ hours} \implies T_{1/2} = 2.0\text{ hours}\).

Award 1 mark for the correct calculation of half-life taking background radiation into account.

1. Use the orbital speed formula:
\(v = \frac{2 \pi r}{T}\).

2. Identify and convert the variables:
- Radius \(r = 8.0 \times 10^3\text{ km} = 8000\text{ km}\).
- Period \(T = 2.0\text{ hours} = 2.0 \times 3600\text{ s} = 7200\text{ s}\).

3. Calculate the orbital speed:
\(v = \frac{2 \pi \times 8000\text{ km}}{7200\text{ s}} \approx \frac{50265\text{ km}}{7200\text{ s}} \approx 6.98\text{ km/s}\).

Rounded to 2 significant figures, this is \(7.0\text{ km/s}\).

Award 1 mark for the correct calculation of the average orbital speed.

The resistance of a wire is given by the formula \(R = \rho \frac{L}{A}\), where \(A = \frac{\pi d^2}{4}\).

Thus, \(R \propto \frac{L}{d^2}\).

For the second wire:
- The length is doubled to \(2L\).
- The diameter is halved to \(0.5d\), which means the cross-sectional area decreases by a factor of \(2^2 = 4\).

Substituting these changes into the proportionality:
\(R_{new} \propto \frac{2L}{(0.5d)^2} = \frac{2L}{0.25 d^2} = 8 \left(\frac{L}{d^2}\right)\).

Therefore, the new resistance is \(8R\).

1 mark for identifying that resistance is proportional to length and inversely proportional to the square of the diameter, leading to the correct calculation of \(8R\).

Using the principle of conservation of momentum: total momentum before collision equals total momentum after collision.

Let the initial direction of the car be positive.

\(p_{\text{initial}} = m_{\text{car}} u_{\text{car}} + m_{\text{truck}} u_{\text{truck}}\)
\(p_{\text{initial}} = (0.40\text{ kg} \times 5.0\text{ m/s}) + (0.80\text{ kg} \times 0\text{ m/s}) = 2.0\text{ kg m/s}\)

Let the final speed of the truck be \(v_{\text{truck}}\). Since the car rebounds, its final velocity is \(-1.0\text{ m/s}\).

\(p_{\text{final}} = m_{\text{car}} v_{\text{car}} + m_{\text{truck}} v_{\text{truck}}\)
\(p_{\text{final}} = (0.40\text{ kg} \times -1.0\text{ m/s}) + (0.80\text{ kg} \times v_{\text{truck}}) = -0.40 + 0.80 v_{\text{truck}}\)

Setting \(p_{\text{initial}} = p_{\text{final}}\):
\(2.0 = -0.40 + 0.80 v_{\text{truck}}\)
\(2.4 = 0.80 v_{\text{truck}}\)
\(v_{\text{truck}} = 3.0\text{ m/s}\)

1 mark for applying the conservation of momentum equation correctly, accounting for the direction of the rebounding car, to find the speed of \(3.0\text{ m/s}\).

First, calculate the frequency of the waves in deep water using \(v = f \lambda\):
\(f = \frac{v}{\lambda} = \frac{30\text{ cm/s}}{4.0\text{ cm}} = 7.5\text{ Hz}\).

When a wave crosses a boundary between two media, its frequency remains unchanged. Therefore, the frequency in shallow water is still \(7.5\text{ Hz}\).

Next, calculate the wavelength in shallow water using the new speed:
\(\lambda_{\text{shallow}} = \frac{v_{\text{shallow}}}{f} = \frac{15\text{ cm/s}}{7.5\text{ Hz}} = 2.0\text{ cm}\).

1 mark for recognizing that frequency does not change across boundaries, and using the wave equation to correctly find \(f = 7.5\text{ Hz}\) and \(\lambda = 2.0\text{ cm}\).

To find the best estimate of the total time for 20 oscillations, calculate the mean of the two readings:
\(\text{Mean time} = \frac{34.8\text{ s} + 35.2\text{ s}}{2} = 35.0\text{ s}\).

The period \(T\) is the time for one complete oscillation:
\(T = \frac{35.0\text{ s}}{20} = 1.75\text{ s}\).

To improve the accuracy of the period, the student should measure a larger number of oscillations (e.g., 50 oscillations instead of 20). This reduces the percentage error introduced by human reaction time when starting and stopping the stopwatch.

1 mark for calculating the correct period of \(1.75\text{ s}\) and identifying that counting more oscillations reduces the effect of human reaction time errors.

According to the kinetic theory of gases, temperature is a measure of the average kinetic energy (and thus speed) of the gas molecules. Since temperature is constant, the molecules do not speed up or hit the walls with more force on average.

However, halving the volume means the same number of molecules are packed into half the space. This increases the density of the molecules, causing them to collide with the cylinder walls more frequently per unit area. This increase in collision rate causes an increase in pressure.

1 mark for identifying that the pressure increase is due to the increased frequency of collisions of the molecules with the walls at constant temperature.

1. Find the angle of refraction \(r\) using Snell's Law:
\(n = \frac{\sin i}{\sin r}\)
\(1.50 = \frac{\sin(45^\circ)}{\sin r}\)
\(\sin r = \frac{\sin(45^\circ)}{1.50} = \frac{0.707}{1.50} \approx 0.471\)
\(r = \sin^{-1}(0.471) \approx 28^\circ\).

2. Find the speed of light in glass \(v\):
\(n = \frac{c}{v}\)
\(v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.50} = 2.0 \times 10^8\text{ m/s}\).

1 mark for calculating both the correct angle of refraction (approx \(28^\circ\)) and the correct speed of light in glass (\(2.0 \times 10^8\text{ m/s}\)).

1. Deduct the background radiation to find the initial count rate of the source alone:
\(\text{Source rate}_{t=0} = 140\text{ cpm} - 20\text{ cpm} = 120\text{ cpm}\).

2. Determine the number of half-lives that have elapsed in \(45\text{ minutes}\):
\(N = \frac{45\text{ minutes}}{15\text{ minutes}} = 3\text{ half-lives}\).

3. Calculate the remaining activity of the source after 3 half-lives:
- After 1 half-life (15 min): \(120 / 2 = 60\text{ cpm}\)
- After 2 half-lives (30 min): \(60 / 2 = 30\text{ cpm}\)
- After 3 half-lives (45 min): \(30 / 2 = 15\text{ cpm}\).

4. Add the background count rate back to find the new total measured count rate:
\(\text{Measured rate}_{t=45} = 15\text{ cpm} + 20\text{ cpm} = 35\text{ cpm}\).

1 mark for subtracting background radiation first, dividing the remaining source activity by 8 (3 half-lives), and then adding back the background radiation to get \(35\text{ cpm}\).

First, calculate the operating current of the heater:
\(I = \frac{P}{V} = \frac{1500\text{ W}}{230\text{ V}} \approx 6.52\text{ A}\).

A fuse must have a rating slightly higher than the normal operating current of the appliance. Therefore, a \(13\text{ A}\) fuse is appropriate (a \(5\text{ A}\) fuse would blow immediately under normal operation).

The earth wire is connected to the metal casing. If a fault causes the live wire to touch the metal casing, a very large current flows through the earth wire to the ground, which instantly blows the fuse and prevents a user touching the casing from receiving a fatal electric shock.

1 mark for calculating the operating current to select the correct \(13\text{ A}\) fuse and stating the correct safety role of the earth wire.

The uncorrected reading from the micrometer is: \(\text{Main scale reading} + \text{Thimble reading} = 4.5\text{ mm} + (24 \times 0.01\text{ mm}) = 4.74\text{ mm}\). To find the actual diameter, we subtract the positive zero error: \(\text{Actual diameter} = \text{Uncorrected reading} - \text{Zero error} = 4.74\text{ mm} - 0.03\text{ mm} = 4.71\text{ mm}\).

1 mark for the correct calculation and subtraction of the zero error, leading to 4.71 mm.

First, find the total distance travelled by finding the area under the velocity-time graph. For the first part, distance is \(0.5 \times 5.0\text{ s} \times 12\text{ m/s} = 30\text{ m}\). For the second part, distance is \(10\text{ s} \times 12\text{ m/s} = 120\text{ m}\). For the third part, distance is \(0.5 \times 3.0\text{ s} \times 12\text{ m/s} = 18\text{ m}\). Total distance is \(30 + 120 + 18 = 168\text{ m}\). Total time is \(18.0\text{ s}\). Average speed is \(\frac{\text{Total distance}}{\text{Total time}} = \frac{168\text{ m}}{18.0\text{ s}} \approx 9.3\text{ m/s}\).

1 mark for calculating the total distance as 168 m and dividing by 18 s to obtain 9.3 m/s.

Find the mass of each liquid: Mass of X is \(1.2\text{ g/cm}^3 \times 200\text{ cm}^3 = 240\text{ g}\). Mass of Y is \(0.80\text{ g/cm}^3 \times 300\text{ cm}^3 = 240\text{ g}\). Total mass is \(240\text{ g} + 240\text{ g} = 480\text{ g}\). Total volume is \(200\text{ cm}^3 + 300\text{ cm}^3 = 500\text{ cm}^3\). Density is \(\frac{\text{Total mass}}{\text{Total volume}} = \frac{480\text{ g}}{500\text{ cm}^3} = 0.96\text{ g/cm}^3\).

1 mark for calculating total mass of 480 g, total volume of 500 cm³ and obtaining 0.96 g/cm³.

Calculate the useful work output: \(E_p = mgh = 80\text{ kg} \times 9.8\text{ m/s}^2 \times 15\text{ m} = 11\,760\text{ J}\). Calculate the total electrical energy input: \(E_{in} = \text{Power} \times \text{time} = 2000\text{ W} \times 10\text{ s} = 20\,000\text{ J}\). Calculate the efficiency: \(\text{Efficiency} = \frac{11\,760}{20\,000} \times 100\% = 58.8\%\), which rounds to \(59\%\).

1 mark for calculating useful work as 11,760 J, input energy as 20,000 J, and obtaining the efficiency of approximately 59%.

The relationship between refractive index \(n\) and critical angle \(c\) is given by: \(\sin c = \frac{1}{n} = \frac{1}{1.52} \approx 0.658\). Thus, \(c = \sin^{-1}(0.658) \approx 41^\circ\). If the angle of incidence is \(45^\circ\), which is greater than the critical angle (\(45^\circ > 41^\circ\)), the light ray undergoes total internal reflection.

1 mark for calculating the critical angle as 41° and identifying that since the angle of incidence (45°) is greater than the critical angle, total internal reflection occurs.

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

1 mark for correctly applying Boyle's law and calculating the new pressure as 3.0 × 10⁵ Pa.

First, calculate the equivalent resistance of the two resistors in parallel: \(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{3}{12.0}\), so \(R_p = 4.0\ \Omega\). Next, add the series resistor to get the total resistance: \(R_{\text{total}} = R_p + 4.0\ \Omega = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega\).

1 mark for calculating the parallel resistance as 4.0 Ω and adding the series 4.0 Ω resistor to obtain 8.0 Ω.

First, subtract the background count rate to find the initial corrected count rate: \(360 - 24 = 336\text{ counts per minute}\). Determine the number of half-lives that have passed in 12.0 hours: \(\frac{12.0\text{ hours}}{4.0\text{ hours}} = 3\). Find the corrected count rate after 3 half-lives: \(\frac{336}{2^3} = \frac{336}{8} = 42\text{ counts per minute}\). Finally, add the background count rate back: \(42 + 24 = 66\text{ counts per minute}\).

1 mark for subtracting background, halving source rate three times, and adding background back to get 66 counts per minute.

To find the average period of the pendulum, first find the mean time for 30 oscillations:

\(\text{Mean time} = \frac{45.3\text{ s} + 44.8\text{ s} + 45.5\text{ s}}{3} = \frac{135.6\text{ s}}{3} = 45.2\text{ s}\)

Next, calculate the period (the time for one single oscillation) by dividing the mean time by the number of oscillations (30):

\(\text{Period } T = \frac{45.2\text{ s}}{30} \approx 1.5067\text{ s}\)

Rounding to three significant figures gives \(1.51\text{ s}\).

1 mark: Correct calculation of the average time for 30 oscillations (45.2 s) and dividing by 30 to obtain 1.51 s.

The distance travelled in each stage of the motion can be calculated using the area under the speed-time graph:

1. During the acceleration phase, the distance \(d_1\) is the area of a triangle:
\(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times v = 0.5vt\)

2. During the constant speed phase, the distance \(d_2\) is the area of a rectangle:
\(d_2 = \text{base} \times \text{height} = 2t \times v = 2vt\)

3. During the deceleration phase, the distance \(d_3\) is the area of a triangle:
\(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times v = 0.5vt\)

Total distance \(D = d_1 + d_2 + d_3 = 0.5vt + 2vt + 0.5vt = 3vt\).

The fraction of the total distance travelled while accelerating is:
\(\frac{d_1}{D} = \frac{0.5vt}{3vt} = \frac{0.5}{3} = \frac{1}{6}\).

1 mark: Correct analysis of the three distance components and determining the ratio of the accelerating phase to the total distance as 1/6.

First, calculate the refractive index \(n\) of the plastic using Snell's Law:
\(n = \frac{\sin i}{\sin r} = \frac{\sin(45^\circ)}{\sin(28^\circ)} \approx \frac{0.7071}{0.4695} \approx 1.51\)

Next, the relationship between refractive index and critical angle \(c\) is:
\(\sin c = \frac{1}{n}\)
\(\sin c = \frac{1}{1.51} \approx 0.662\)
\(c = \sin^{-1}(0.662) \approx 41.5^\circ \approx 42^\circ\).

1 mark: Correct calculation of the refractive index and subsequently the critical angle to the nearest degree.

Matte black surfaces are highly effective emitters of infrared radiation, whereas shiny, light-colored surfaces are poor emitters (good reflectors). Since both cans contain hot water at the same temperature and are exposed to the same cooler environment, the matte black can loses thermal energy through radiation more rapidly than the shiny silver can, meaning the water inside it cools down more quickly.

1 mark: Correct identification that the black can cools more quickly because matte black surfaces are superior emitters of thermal radiation.

First, calculate the equivalent resistance \(R_p\) of the parallel combination:
\(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{2}{12.0} + \frac{1}{12.0} = \frac{3}{12.0}\)
\(R_p = 4.0\ \Omega\)

Now, calculate the total resistance \(R_{\text{total}}\) of the circuit:
\(R_{\text{total}} = R_p + R_{\text{series}} = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega\)

The total current \(I_{\text{total}}\) from the battery is:
\(I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\)

The potential difference across the parallel combination is:
\(V_p = I_{\text{total}} \times R_p = 1.5\text{ A} \times 4.0\ \Omega = 6.0\text{ V}\)

Finally, the current through the \(6.0\ \Omega\) resistor is:
\(I = \frac{V_p}{R} = \frac{6.0\text{ V}}{6.0\ \Omega} = 1.0\text{ A}\).

1 mark: Correct step-by-step determination of circuit resistance, total current, parallel branch voltage, and final current through the 6.0 ohm resistor.

First, determine the secondary voltage \(V_s\) using the transformer equation:
\(\frac{V_s}{V_p} = \frac{N_s}{N_p}\)
\(V_s = 230\text{ V} \times \frac{900}{150} = 230\text{ V} \times 6 = 1380\text{ V}\)

Next, calculate the secondary current \(I_s\) through the resistor:
\(I_s = \frac{V_s}{R} = \frac{1380\text{ V}}{460\ \Omega} = 3.0\text{ A}\)

For an ideal transformer, input electrical power equals output electrical power:
\(V_p \times I_p = V_s \times I_s\)
\(I_p = I_s \times \frac{V_s}{V_p} = 3.0\text{ A} \times 6 = 18\text{ A}\).

1 mark: Correct calculation of secondary voltage, secondary current, and applying the conservation of electrical power to find the primary current of 18 A.

The orbital speed \(v\) is given by the formula:
\(v = \frac{2 \pi r}{T}\)

First, convert the period \(T\) from hours into seconds:
\(T = 2.5\text{ hours} \times 3600\text{ s/hour} = 9000\text{ s}\)

The radius of the orbit is \(r = 8.0 \times 10^3\text{ km}\).

Now, substitute these values into the speed formula:
\(v = \frac{2 \times \pi \times 8.0 \times 10^3\text{ km}}{9000\text{ s}} \approx \frac{50265.5\text{ km}}{9000\text{ s}} \approx 5.585\text{ km/s}\)

Rounding to two significant figures gives \(5.6\text{ km/s}\).

1 mark: Correct unit conversion of period and application of the orbital speed formula to obtain 5.6 km/s.

First, find the number of half-lives that have passed in \(24\text{ hours}\):
\(\text{Number of half-lives} = \frac{24\text{ hours}}{6.0\text{ hours}} = 4\)

After 4 half-lives, the number of active nuclei is halved four times:
\(\text{Remaining nuclei} = 1.6 \times 10^{20} \times \left(\frac{1}{2}\right)^4 = \frac{1.6 \times 10^{20}}{16} = 1.0 \times 10^{19}\).

1 mark: Correctly identifies 4 half-lives have elapsed and calculates the remaining active nuclei as 1.0 x 10^19.

The resistance \(R\) of a wire of length \(L\) and diameter \(d\) is given by the formula:

\(R = \rho \frac{L}{A} = \rho \frac{4L}{\pi d^2}

Thus, resistance is directly proportional to length and inversely proportional to the square of the diameter:

\)R \propto \frac{L}{d^2}

For the second wire:
- Length is doubled (\(L \to 2L\))
- Diameter is halved (\(d \to 0.5d\))

Substituting these changes:

\(R_{\text{new}} \propto \frac{2L}{(0.5d)^2} = \frac{2L}{0.25d^2} = 8 \times \frac{L}{d^2}

Therefore, the resistance of the second wire is \)8 \times 8.0\ \Omega = 64.0\ \Omega\).

1 mark for the correct calculation of the final resistance of 64.0 \(\Omega\).

First, find the initial extension \(x_1\) of the spring:
\(x_1 = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}

Since the spring remains within its limit of proportionality, the extension is directly proportional to the load applied (\)F = kx\)):

\(\frac{F_1}{x_1} = \frac{F_2}{x_2}

\)\frac{4.0\text{ N}}{3.0\text{ cm}} = \frac{10.0\text{ N}}{x_2}

\(x_2 = \frac{10.0 \times 3.0}{4.0} = 7.5\text{ cm}

Now, calculate the new total length of the spring:

\)\text{Total length} = \text{Original length} + x_2 = 12.0\text{ cm} + 7.5\text{ cm} = 19.5\text{ cm}\).

1 mark for the correct calculation of the new total length of 19.5 cm.

Dull, dark black surfaces are the best emitters of thermal radiation. Therefore, Can 1 (dull black) will radiate thermal energy at the fastest rate, cooling down the most and reaching the lowest temperature.

Shiny, light silver surfaces are the worst emitters of thermal radiation. Therefore, Can 4 (shiny silver) will radiate thermal energy at the slowest rate, keeping its heat the longest and having the highest temperature.

1 mark for correctly identifying Can 1 as having the lowest temperature and Can 4 as having the highest temperature.

The distance travelled by the probe in one complete circular orbit is equal to the circumference of the circle, \(2\pi r\).

Since speed is distance divided by time:

\(v = \frac{2\pi r}{T}

Rearranging this formula to solve for the orbital period \)T\) gives:

\(T = \frac{2\pi r}{v}

1 mark for identifying the correct rearranged formula for orbital period.

When water waves enter a shallower region:
1. The shallower water slows the wave down, so the speed decreases.
2. The frequency of a wave is determined solely by the source generating it, so the frequency remains constant.
3. Since the wave speed equation is \(v = f \lambda\), a decrease in speed \(v\) while frequency \(f\) remains constant results in a decrease in wavelength \(\lambda\).

1 mark for identifying that speed decreases, frequency remains constant, and wavelength decreases.

1. Subtract the background radiation to find the initial activity rate due only to the source at 12:00:

\(360\text{ cpm} - 40\text{ cpm} = 320\text{ cpm}

2. Calculate the time elapsed from 12:00 to 13:00:

\)\text{Elapsed time} = 60\text{ minutes}

3. Determine the number of half-lives that have passed:

\(\text{Number of half-lives} = \frac{60\text{ min}}{20\text{ min}} = 3

4. Calculate the source activity after 3 half-lives:

\)\text{Final source activity} = \frac{320}{2^3} = \frac{320}{8} = 40\text{ cpm}

5. Add the background radiation back to get the total detector reading:

\(\text{Total detector reading} = 40\text{ cpm (source)} + 40\text{ cpm (background)} = 80\text{ cpm}\).

1 mark for the correct calculation of 80 counts per minute after subtracting and adding back background radiation.

Using the principle of conservation of momentum:

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

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

Substitute the given values into the equation:

\((0.50\text{ kg} \times 6.0\text{ m/s}) + (1.0\text{ kg} \times 0) = (0.50\text{ kg} + 1.0\text{ kg}) \times v

\)3.0\text{ kg m/s} = 1.5\text{ kg} \times v

\(v = \frac{3.0}{1.5} = 2.0\text{ m/s}\)

1 mark for the correct calculation of the final common velocity (2.0 m/s).

- A standard ruler has millimeter subdivisions, which is suitable and accurate enough to measure a length of about \(5\text{ cm}\).
- A micrometer screw gauge is designed to measure very small distances with high precision (typically to within \(0.01\text{ mm}\)), making it the ideal instrument to measure a wire's diameter of about \(1.5\text{ mm}\).
- A measuring cylinder cannot directly measure a small diameter of \(1.5\text{ mm}\) with precision.

1 mark for selecting ruler for length and micrometer screw gauge for diameter.

(a) As the light intensity decreases, the resistance of the LDR increases.

(b) (i) First, calculate the total resistance of the series circuit: \(R_{\text{total}} = R_{\text{LDR}} + R_{\text{fixed}} = 9.0\text{ k}\Omega + 3.0\text{ k}\Omega = 12.0\text{ k}\Omega\). Use the potential divider formula to find the voltage across the LDR: \(V_{\text{LDR}} = (R_{\text{LDR}} / R_{\text{total}}) \times V_{\text{supply}} = (9.0 / 12.0) \times 12\text{ V} = 9.0\text{ V}\).

(ii) Calculate the circuit current using Ohm's law: \(I = V_{\text{supply}} / R_{\text{total}} = 12\text{ V} / 12.0 \times 10^3\ \Omega = 1.0 \times 10^{-3}\text{ A} = 1.0\text{ mA}\).

(a) State that resistance increases [1]. Correctly link the change directly to decreasing light intensity [1].
(b)(i) Determine total resistance is 12 k-ohms [1]. Apply potential divider formula or find current first [1]. State final voltage as 9.0 V [1].
(b)(ii) State Ohm's law formula I = V/R [1]. State correct final current as 1.0 mA with correct unit [1].

(a) Matte black surfaces are highly efficient absorbers of electromagnetic radiation. Painting the copper pipe matte black maximizes the rate at which infrared radiation from the Sun is absorbed.

(b) 1. Radiation: Thermal energy travels from the Sun through the vacuum of space and the glass lid to the pipe mainly as infrared radiation.
2. Conduction: The thermal energy is absorbed by the surface of the copper pipe and is conducted through the solid metal wall of the pipe to the inner surface. This occurs because free electrons and lattice vibrations transfer kinetic energy.
3. Convection: The water in contact with the inner wall of the copper pipe is heated by conduction, becomes less dense, and rises. This sets up a convection current that distributes the thermal energy throughout the water.

(a) Identify matte black as a good/excellent absorber [1] of infrared radiation / thermal energy [1].
(b) State that energy from the Sun reaches the pipe via infrared radiation [1]. State that energy transfers through the copper tube by conduction [1]. Mention free electrons or lattice vibrations in the conduction explanation [1]. State that energy is distributed in the water by convection [1]. Explain that hot water expands, becomes less dense, and rises [1].

(a) Immediately after jumping, the only force acting is weight, so the skydiver accelerates downwards at \(g\) (\(9.8\text{ m/s}^2\)). As their downward velocity increases, the upward air resistance increases. The resultant downward force decreases since \(F = W - R\). Consequently, the downward acceleration decreases. Eventually, air resistance increases to equal the weight, making the resultant force zero. At this point, acceleration becomes zero, and the skydiver falls at a constant terminal velocity.

(b) First, calculate weight: \(W = m \times g = 75\text{ kg} \times 9.8\text{ m/s}^2 = 735\text{ N}\). The resultant downward force is: \(F = W - R = 735\text{ N} - 350\text{ N} = 385\text{ N}\). Using Newton's second law: \(a = F / m = 385\text{ N} / 75\text{ kg} \approx 5.13\text{ m/s}^2\).

(a) Mention that initial acceleration is equal to g / weight is the only initial force [1]. State that air resistance increases as velocity increases [1]. State that the resultant force (and thus acceleration) decreases [1]. State that terminal velocity is reached when air resistance equals weight (resultant force is zero) [1].
(b) Calculate weight: 735 N [1]. Calculate resultant force: 385 N [1]. Calculate acceleration: 5.13 m/s² (allow 5.1 m/s²) with correct units [1].

(a) The refractive index \(n\) is given by Snell's law: \(n = \frac{\sin i}{\sin r} = \frac{\sin 40.0^\circ}{\sin 25.0^\circ} \approx \frac{0.6428}{0.4226} \approx 1.52\).

(b) The relationship between critical angle \(c\) and refractive index is: \(\sin c = \frac{1}{n} = \frac{1}{1.52} \approx 0.6579\). Thus, \(c = \sin^{-1}(0.6579) \approx 41.1^\circ\). Because the angle of incidence inside the glass (\(45.0^\circ\)) is greater than the critical angle (\(41.1^\circ\)), the light ray undergoes total internal reflection. The entire ray is reflected back into the glass block at an angle of reflection of \(45.0^\circ\).

(a) State Snell's law formula [1]. Correct substitution of values [1]. Calculate correct refractive index value of 1.52 (accept 1.5) [1].
(b) State formula sin(c) = 1/n [1]. Calculate critical angle as 41.1° (or 41.8° if using 1.5) [1]. State that total internal reflection occurs [1]. Explain that this happens because the angle of incidence exceeds the critical angle [1].

(a) Use the gravitational potential energy formula: \(\Delta E_p = mgh = 600\text{ kg} \times 9.8\text{ m/s}^2 \times 45\text{ m} = 264,600\text{ J}\) (or \(2.65 \times 10^5\text{ J}\)).

(b) If \(12\%\) of the energy is lost, the remaining energy converted to kinetic energy (\(E_k\)) is: \(E_k = (1.00 - 0.12) \times \Delta E_p = 0.88 \times 264,600\text{ J} = 232,848\text{ J}\). Now equate this to the kinetic energy formula: \(E_k = \frac{1}{2} m v^2 \implies 232,848 = \frac{1}{2} \times 600 \times v^2\). This simplifies to: \(300 v^2 = 232,848 \implies v^2 = 776.16\). Taking the square root gives: \(v \approx 27.9\text{ m/s}\).

(a) State GPE formula mgh [1]. Calculate energy as 2.6 x 10^5 J or 2.65 x 10^5 J [1].
(b) Identify that 88% of the GPE is converted to KE [1]. Calculate kinetic energy value as 232,848 J [1]. State KE formula 0.5 * m * v^2 [1]. Equate KE to remaining energy and rearrange for v [1]. Calculate final speed as 27.9 m/s (allow 28 m/s) with correct unit [1].

(a) An alternating current (a.c.) in the primary coil creates a continuously changing magnetic field in the soft iron core. This changing magnetic field is guided by the core through the secondary coil. As the changing magnetic field cuts across the secondary coil, it induces an alternating electromotive force (e.m.f.) across its terminals. Because there are fewer turns on the secondary coil than on the primary coil, the output voltage is stepped down.

(b) The output power \(P_{\text{out}}\) of the secondary coil is \(24\text{ W}\). Since the transformer is \(90\%\) efficient: \(P_{\text{in}} = P_{\text{out}} / 0.90 = 24\text{ W} / 0.90 \approx 26.67\text{ W}\). The power input is also given by \(P_{\text{in}} = V_p \times I_p\). Therefore: \(I_p = P_{\text{in}} / V_p = 26.67\text{ W} / 240\text{ V} \approx 0.111\text{ A}\).

(a) State that alternating current in primary produces a changing magnetic field [1]. State that the iron core channels this magnetic field to the secondary coil [1]. State that this changing magnetic field induces an e.m.f./voltage in the secondary coil [1]. Link fewer turns on the secondary to a stepped-down voltage [1].
(b) Correctly state relation between efficiency, input power, and output power [1]. Calculate input power as 26.7 W [1]. Calculate primary current as 0.11 A or 0.111 A with correct unit [1].

(a) During beta-minus decay, a neutron inside the nucleus converts into a proton and an electron (beta particle). The nucleon number remains \(24\), but the proton number increases by \(1\) to become \(12\). The equation is: \(_{11}^{24}\text{Na} \rightarrow _{12}^{24}\text{Mg} + _{-1}^{\ \ 0}\beta\) (or \(_{-1}^{\ \ 0}\text{e}\)).

(b) First, determine the number of half-lives that have elapsed: \(N = 60\text{ hours} / 15\text{ hours} = 4\text{ half-lives}\). Now, halve the initial activity four times:
1 half-life: \(320 / 2 = 160\text{ Bq}\)
2 half-lives: \(160 / 2 = 80\text{ Bq}\)
3 half-lives: \(80 / 2 = 40\text{ Bq}\)
4 half-lives: \(40 / 2 = 20\text{ Bq}\).

(a) State correct beta particle symbol with correct mass number 0 and atomic number -1 [1]. Correctly identify Magnesium nucleon number as 24 [1]. Correctly identify Magnesium proton number as 12 [1].
(b) Determine that 4 half-lives have passed [1]. Show clear steps of halving or correct use of radioactive decay formula [2]. State final activity as 20 Bq with correct unit [1].

(a) Speed is defined as distance divided by time. For a circular path, the distance travelled in one full orbit is equal to the circumference, which is \(2\pi r\). The time taken for one full orbit is the orbital period \(T\). Substituting these gives \(v = \frac{2\pi r}{T}\). Here, \(r\) is the orbital radius, which is the total distance from the center of the Earth to the satellite.

(b) First, calculate the orbital radius \(r\) in meters: \(r = R_{\text{Earth}} + \text{altitude} = 6400\text{ km} + 600\text{ km} = 7000\text{ km} = 7.0 \times 10^6\text{ m}\). Convert the orbital period \(T\) to seconds: \(T = 97\text{ minutes} = 97 \times 60 = 5820\text{ s}\). Now calculate the speed: \(v = \frac{2 \pi \times 7.0 \times 10^6\text{ m}}{5820\text{ s}} \approx 7557.5\text{ m/s}\) (or \(7.6 \times 10^3\text{ m/s}\)).

(a) State that distance is the circumference 2*pi*r [1]. State that time is the period T [1]. Define r as the orbital radius (or distance from the center of the Earth to the satellite) [1].
(b) Calculate total orbital radius as 7,000 km or 7.0 x 10^6 m [1]. Convert 97 minutes to 5,820 s [1]. Correctly substitute values into speed equation [1]. Calculate final speed as 7560 m/s (allow range 7550 - 7600 m/s) with correct unit [1].

(a) Orbital radius \( r = \text{Radius of Earth} + \text{altitude} = 6.40 \times 10^6 \text{ m} + 3.60 \times 10^7 \text{ m} = 0.64 \times 10^7 \text{ m} + 3.60 \times 10^7 \text{ m} = 4.24 \times 10^7 \text{ m} \).\
\
(b)(i) \( T = 24.0 \text{ hours} = 24.0 \times 3600 \text{ s} = 86400 \text{ s} \).\
\
(b)(ii) \( v = \frac{2 \pi r}{T} = \frac{2 \pi \times 4.24 \times 10^7 \text{ m}}{86400 \text{ s}} = 3083 \text{ m/s} \approx 3080 \text{ m/s} \) (or \( 3.08 \times 10^3 \text{ m/s} \)).\
\
(c) Geostationary orbit. It remains above a fixed position on the Earth's surface, so ground antennas and receiver dishes do not need to turn to track the satellite, ensuring an uninterrupted signal connection.

(a) [1 mark] Shows addition: \( 6.40 \times 10^6 + 3.60 \times 10^7 = 4.24 \times 10^7 \text{ m} \).\
\
(b)(i) [1 mark] \( T = 24.0 \times 3600 = 86400 \text{ s} \).\
\
(b)(ii) [3 marks]\
- [1 mark] Uses orbital speed formula: \( v = \frac{2 \pi r}{T} \)\
- [1 mark] Correct substitution: \( v = \frac{2 \pi \times 4.24 \times 10^7}{86400} \)\
- [1 mark] Correct final speed with units: \( 3080 \text{ m/s} \) (or \( 3.08 \times 10^3 \text{ m/s} \); accept 3100 m/s or 3083 m/s; ignore minor rounding errors).\
\
(c) [2.27 marks]\
- [1 mark] Identifies orbit as geostationary (or geosynchronous).\
- [1.27 marks] Explains that it remains above the same position on the Earth's surface / ground dishes do not need to track it.

(a) Momentum is the product of mass and velocity (or \( p = mv \)).\
\
(b)(i) Initial momentum: \( p = m \times v = 0.80 \text{ kg} \times 1.5 \text{ m/s} = 1.2 \text{ kg m/s} \) (or \( \text{N s} \)).\
\
(b)(ii) Total initial momentum = Total final momentum\
\( m_1 u_1 + 0 = (m_1 + m_2) v \)\
\( 1.2 = (0.80 + 0.40) v \)\
\( 1.2 = 1.20 v \)\
\( v = 1.0 \text{ m/s} \).\
\
(c) The carriage experiences an impulse which equals its change in momentum.\
\( \text{Change in momentum of carriage } \Delta p = m_2 v - 0 = 0.40 \text{ kg} \times 1.0 \text{ m/s} = 0.40 \text{ kg m/s} \).\
Using \( F \Delta t = \Delta p \):\
\( 6.0 \times \Delta t = 0.40 \)\
\( \Delta t = \frac{0.40}{6.0} = 0.0667 \text{ s} \approx 0.067 \text{ s} \).

(a) [1 mark] Momentum = mass \\times velocity (words or formula with symbols defined).\
\
(b)(i) [1.5 marks]\
- [1 mark] Correct calculation: \( 0.80 \times 1.5 = 1.2 \)\
- [0.5 mark] Correct unit: \( \text{kg m/s} \) or \( \text{N s} \).\
\
(b)(ii) [2.5 marks]\
- [1 mark] States or uses conservation of momentum: \( p_{\text{initial}} = p_{\text{final}} \)\
- [1 mark] Sets up equation correctly: \( 1.2 = 1.20 v \)\
- [0.5 mark] Correct value for \( v = 1.0 \text{ m/s} \) with unit.\
\
(c) [2.27 marks]\
- [1 mark] Calculates change in momentum of carriage: \( \Delta p = 0.40 \text{ kg m/s} \) (or uses impulse-momentum theorem on engine: \( \Delta p = 0.80 \times 1.5 - 0.80 \times 1.0 = 0.40 \text{ kg m/s} \)).\
- [1 mark] Recall and use of \( F \Delta t = \Delta p \) or \( \Delta t = \frac{\Delta p}{F} \).\
- [0.27 mark] Correct calculation of time: \( 0.067 \text{ s} \) (accept \( 0.0667 \text{ s} \)).

(a) When the metal base is heated, the metal atoms or ions at the bottom gain kinetic energy and vibrate more rapidly. They collide with neighboring atoms/ions, transferring thermal energy through the lattice. In addition, metals have free (delocalized) electrons. These free electrons gain kinetic energy, move rapidly through the metal lattice, and transfer energy efficiently by colliding with distant metal ions and other electrons.\
\
(b) Water near the bottom of the saucepan is heated, expands, and its density decreases. The warmer, less dense water rises to the top. Cooler, denser water at the top sinks to the bottom to take its place. This continuous circulation forms a convection current, heating all the water.\
\
(c) Shiny, polished surfaces are poor emitters of thermal (infrared) radiation. Therefore, the rate of thermal energy loss from the outside of the polished saucepan is significantly less than that from a dull black surface (which is a very good emitter of radiation).

(a) [3 marks]\
- [1 mark] Heating causes atoms/ions to vibrate more rapidly.\
- [1 mark] Vibration energy is transferred through collisions with neighboring atoms/ions (lattice vibration).\
- [1 mark] Free/delocalized electrons move/diffuse through the metal and transfer energy through collisions with distant ions.\
\
(b) [2.27 marks]\
- [1 mark] Water at the bottom is heated, expands, and becomes less dense.\
- [1 mark] Warmer, less dense water rises, and cooler, denser water sinks to replace it.\
- [0.27 mark] This forms a continuous circulation / convection current.\
\
(c) [2 marks]\
- [1 mark] Shiny/polished surfaces are poor/inefficient emitters (or good reflectors) of infrared/thermal radiation.\
- [1 mark] The rate of energy loss is reduced compared to a dull black surface (which is a good emitter).

(a) The thermometer reading is halfway between 84 and 85, which is \(84.5\ ^\circ\text{C}\).

(b) Calculate the temperature drop for each beaker:
- \(\Delta \theta_A = 84.5 - 69.0 = 15.5\ ^\circ\text{C}\)
- \(\Delta \theta_B = 84.5 - 74.0 = 10.5\ ^\circ\text{C}\)

(c) The aluminum foil lid is the better thermal insulator because Beaker B experienced a lower temperature drop (\(10.5\ ^\circ\text{C}\)) compared to Beaker A (\(15.5\ ^\circ\text{C}\)), meaning less thermal energy was lost to the surroundings.

(d) To make it a fair test, we must control parameters that affect the rate of cooling: the volume of hot water used in each beaker, the shape and size of the beaker, the initial starting temperature of the water, and the external room temperature (absence of drafts).

(e) Precautions for accurate thermometer readings include:
1. Stirring the water continuously before taking a reading to ensure uniform temperature throughout the beaker.
2. Reading the thermometer scale at eye level (perpendicularly) to prevent parallax error.
3. Making sure the thermometer bulb is fully submerged in the center of the liquid and does not contact the bottom or walls of the glass container.

Total: 10 Marks
- (a) [1 mark] State \(84.5\ ^\circ\text{C}\) (unit required).
- (b) [2 marks] \(\Delta \theta_A = 15.5\ ^\circ\text{C}\) [1 mark] and \(\Delta \theta_B = 10.5\ ^\circ\text{C}\) [1 mark].
- (c) [2 marks] State foil lid is a better insulator [1 mark] with reference to the smaller temperature drop [1 mark].
- (d) [2 marks] Identify any two control variables (e.g., volume of water, initial temperature, same beaker size, same room temperature) [1 mark each].
- (e) [2 marks] Identify any two precautions (e.g., stir before reading, eye level to avoid parallax, keep bulb away from bottom/sides) [1 mark each].

(a) Reading the scales:
- Voltmeter: \(1.0\text{ V} + 8 \times 0.1\text{ V} = 1.8\text{ V}\).
- Ammeter: \(0.30\text{ A} + 3 \times 0.02\text{ A} = 0.36\text{ A}\).

(b) Calculate \(R_1\):
\(R_1 = \frac{V}{I} = \frac{1.8}{0.36} = 5.0\ \Omega\).
Significant figures: 2 sig figs are appropriate because the input readings are given to 2 sig figs. Unit must be ohms (\(\Omega\)).

(c) Calculate \(R_2\):
\(R_2 = \frac{1.10}{0.88} = 1.25\ \Omega\).

(d) Theory check:
- Theoretical ratio: If resistance is inversely proportional to cross-sectional area, doubling the diameter increases the area by \(2^2 = 4\) times, reducing resistance to \(\frac{1}{4}\).
- Experimental ratio: \(\frac{R_1}{R_2} = \frac{5.0}{1.25} = 4.0\).
- Since the ratio is exactly 4.0, the results completely support the student's suggestion.

(e) Precaution against overheating: Keep the switch open (circuit turned off) when not taking readings, or use low currents by selecting a high-value series resistor/sliding rheostat setting.

Total: 10 Marks
- (a) [2 marks] \(V = 1.8\text{ V}\) [1 mark] and \(I = 0.36\text{ A}\) [1 mark].
- (b) [2 marks] Correct value of \(R_1 = 5.0\ \Omega\) [1 mark], correct unit and 2 or 3 sig figs [1 mark].
- (c) [1 mark] Correct calculation of \(R_2 = 1.25\ \Omega\) (allow ecf from b).
- (d) [3 marks] Show calculation of ratio \(R_1/R_2 = 4\) [1 mark], state clearly that this supports the theory [1 mark], and explain why (e.g., doubling diameter means area increases 4 times, so resistance drops to 1/4) [1 mark].
- (e) [2 marks] State any suitable precaution (e.g., switch off between readings [1 mark], use low current/high resistance rheostat [1 mark]).

(a) Since the scale is 1:5, the actual distance is:
\(u = 4.8\text{ cm} \times 5 = 24.0\text{ cm}\).

(b) Using \(u = 24.0\text{ cm}\) and \(v = 36.0\text{ cm}\):
\(f = \frac{24.0 \times 36.0}{24.0 + 36.0} = \frac{864}{60.0} = 14.4\text{ cm}\).

(c) For \(u = 30.0\text{ cm}\) and \(v = 28.1\text{ cm}\):
- \(u + v = 30.0 + 28.1 = 58.1\text{ cm}\)
- \(uv = 30.0 \times 28.1 = 843\text{ cm}^2\) (rounded to 3 significant figures).

(d) Typical practical difficulties in lens experiments:
1. Image focus is subjective; there is a small range of positions where the image looks equally sharp.
2. Measuring to the exact center of a thick lens is difficult with a standard ruler.

(e) Improvement: Move the screen slowly back and forth to locate the position where the image starts to become blurred in both directions, then take the midpoint. Alternatively, perform the experiment in a darkened room to increase image contrast.

(f) To protect lenses: Handle them by the rim/edge only to avoid finger grease/smudges, and place them on a soft cloth or lens holder when not in use.

Total: 10 Marks
- (a) [1 mark] \(u = 24.0\text{ cm}\) (value and unit needed).
- (b) [2 marks] \(f = 14.4\text{ cm}\) [1 mark for value, 1 mark for correct unit].
- (c) [2 marks] \(u + v = 58.1\text{ cm}\) [1 mark] and \(uv = 843\text{ cm}^2\) [1 mark, accept 843.0].
- (d) [2 marks] Two distinct difficulties (e.g., subjective sharp focus [1 mark], thickness of lens makes center hard to align [1 mark]).
- (e) [1 mark] Suitable improvement (e.g., mark boundaries of blur and take midpoint, or use a darkened room).
- (f) [2 marks] Lens handling precaution (hold by edges/rim [1 mark], do not touch optical surfaces [1 mark]).

Here is a complete experimental plan:

1. **Apparatus**:
- Paper cupcake cases (at least 5-6 of identical size and shape).
- Electronic balance (to measure the mass of the cases).
- Two tall retort stands and clamps to hold marker bands.
- Metre rule or tape measure (to measure the falling distance).
- Digital stopwatch (or light gates connected to a timer for better accuracy).

2. **Method**:
- Measure and record the mass \(m\) of a single paper cupcake case.
- Set up two horizontal markers (e.g., rubber bands on a tall retort stand) separated by a vertical distance \(d\) (e.g., \(1.0\text{ m}\)).
- Ensure the top marker is placed at least \(50\text{ cm}\) below the release point. This release height allows the light paper case to accelerate and reach its constant terminal velocity before passing the first marker.
- Drop the paper case from the release point. Start the stopwatch as it passes the top marker, and stop it as it passes the bottom marker. Record this time \(t\).
- Repeat this measurement twice to find an average time.
- Increase the mass of the falling object by nesting a second paper case inside the first one. Weigh the combined cases to find the new mass.
- Repeat the drop and time measurement.
- Continue adding nested cases one by one up to at least 5 cases.

3. **Control Variables**:
- The shape and cross-sectional area of the falling object (by nesting the paper cases carefully so they maintain the same outer profile).
- Avoid drafts in the room by closing doors and windows, as wind can affect the falling speed.

4. **Data Table**:
| Number of cases | Mass \(m\) / \text{g}\) | Distance \(d\) / \text{m}\) | Time \(t\) / \text{s}\) | Terminal Velocity \(v\) / \text{m/s}\) |
|---|---|---|---|---|

Where terminal velocity is calculated using \(v = \frac{d}{t}\).

5. **Analysis/Conclusion**:
- Plot a graph of Terminal Velocity \(v\) (y-axis) against Mass \(m\) (x-axis).
- Examine the shape of the curve/line. If the line is straight and passes through the origin, terminal velocity is directly proportional to mass. Otherwise, describe the relationship shown by the trend of the graph.

Total: 10 Marks
- [2 marks] Apparatus: Electronic balance [1 mark] and metre rule/tape measure/stopwatch [1 mark].
- [2 marks] Method: Drop nested cases to change mass [1 mark], and explain releasing them from a height above the start marker so they reach terminal velocity before timing begins [1 mark].
- [2 marks] Control variables: Mention keeping shape/surface area of cases constant (achieved by nesting) [1 mark], and conducting in a draft-free environment [1 mark].
- [2 marks] Data Table: Table drawn with clear headers and appropriate units (e.g., mass in \(\text{g}\) or \(\text{kg}\), distance in \(\text{m}\) or \(\text{cm}\), time in \(\text{s}\), velocity in \(\text{m/s}\)).
- [2 marks] Analysis: State how to calculate velocity (\(v = d/t\)) [1 mark] and suggest plotting a graph of \(v\) against \(m\) to establish the mathematical relationship [1 mark].