2018 AQA IGCSE Physics (9203) 模擬試題連答案詳解

An NTC (negative temperature coefficient) thermistor has a resistance that decreases as its temperature increases. This happens because the increased temperature provides energy to release more charge carriers (free electrons) into the conduction band, allowing a larger current to flow for a given potential difference.

1 mark for identifying the correct relationship and physical cause (B).

Stars with a mass much greater than the Sun follow a different life cycle path after the main sequence. They expand into a red supergiant, undergo a violent supernova explosion, and leave behind either a highly dense neutron star or, if the mass is sufficiently high, a black hole.

1 mark for the correct sequence for a high-mass star (B).

Using Snell's Law: \(n_1 \sin \theta_1 = n_2 \sin \theta_2\). Given \(n_1 = 1.33\), \(\theta_1 = 35^\circ\), and \(n_2 = 1.52\): \(1.33 \times \sin(35^\circ) = 1.52 \times \sin \theta_2\) which simplifies to \(1.33 \times 0.5736 = 1.52 \times \sin \theta_2\). Solving for \(\sin \theta_2\) gives \(0.5019\), and finding the inverse sine gives \(\theta_2 \approx 30.1^\circ\).

1 mark for the correct calculation resulting in 30.1 degrees (A).

For a converging lens to produce a virtual, upright, and magnified image (acting as a simple magnifier), the object must be placed between the lens and its principal focus (i.e., at a distance less than the focal length \(f\)).

1 mark for identifying the correct object distance relative to the focal length (C).

We find the number of half-lives that have elapsed by halving the activity: \(320 \rightarrow 160 \rightarrow 80 \rightarrow 40 \rightarrow 20\). This requires 4 half-lives. Since the total time is \(24\text{ hours}\), the length of one half-life is: \(t_{1/2} = 24\text{ hours} / 4 = 6\text{ hours}\).

1 mark for the correct half-life calculation (B).

Efficiency is calculated as (Useful energy output / Total energy input) x 100%, which is \((960 \text{ J} / 2400 \text{ J}) \times 100\% = 40\%\). Wasted energy is the difference between total input and useful output: \(2400 \text{ J} - 960 \text{ J} = 1440 \text{ J}\).

1 mark for the correct calculation of efficiency and wasted energy (A).

First, calculate the current flowing through Resistor A using Ohm's Law: \(I = \frac{V_A}{R_A} = \frac{4.5\text{ V}}{15\text{ }\Omega} = 0.3\text{ A}\). Since the resistors are connected in series, the same current of 0.3 A flows through Resistor B. Next, calculate the resistance of Resistor B: \(R_B = \frac{V_B}{I} = \frac{7.5\text{ V}}{0.3\text{ A}} = 25\text{ }\Omega\).

1 mark for calculating current: \(I = \frac{4.5}{15} = 0.3\text{ A}\). 1 mark for applying the same current to Resistor B using \(R = \frac{V}{I}\). 1 mark for final resistance of 25 \(\Omega\).

First, calculate the total initial momentum: \(p_{\text{initial}} = (0.8\text{ kg} \times 3.5\text{ m/s}) + (1.2\text{ kg} \times 0\text{ m/s}) = 2.8\text{ kg m/s}\). By the conservation of momentum, total final momentum must also be 2.8 kg m/s. The combined mass after the collision is \(0.8\text{ kg} + 1.2\text{ kg} = 2.0\text{ kg}\). Therefore, the final velocity is \(v = \frac{p}{m_{\text{total}}} = \frac{2.8\text{ kg m/s}}{2.0\text{ kg}} = 1.4\text{ m/s}\).

1 mark for calculating initial momentum as 2.8 kg m/s. 1 mark for identifying the total mass of the combined system as 2.0 kg. 1 mark for calculating the final velocity of 1.4 m/s.

First, find the electrical current: \(I = \frac{P}{V} = \frac{2300\text{ W}}{230\text{ V}} = 10\text{ A}\). Next, convert the time to seconds: \(t = 3.0\text{ minutes} \times 60\text{ s/minute} = 180\text{ s}\). Finally, calculate the charge using \(Q = I \times t = 10\text{ A} \times 180\text{ s} = 1800\text{ C}\).

1 mark for calculating current: \(I = \frac{P}{V} = \frac{2300}{230} = 10\text{ A}\). 1 mark for converting time to seconds: \(3.0\text{ mins} = 180\text{ s}\). 1 mark for final charge of 1800 C.

First, calculate the refractive index (n) of the plastic using Snell's Law: \(n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(50^\circ)}{\sin(31^\circ)} = \frac{0.7660}{0.5150} \approx 1.487\). Next, use the critical angle equation: \(\sin(c) = \frac{1}{n} = \frac{1}{1.487} \approx 0.6725\). Solving for critical angle (c): \(c = \arcsin(0.6725) \approx 42.3^\circ\).

1 mark for calculating the refractive index as 1.49 (or 1.5). 1 mark for setting up the equation \(\sin(c) = \frac{1}{n}\). 1 mark for calculating critical angle in range 42.1 to 42.5 degrees.

Using the principle of moments: Anticlockwise moment = Clockwise moment. Weight of the first box: \(F_1 = 12\text{ kg} \times 9.8\text{ N/kg} = 117.6\text{ N}\). Anticlockwise moment: \(M_{\text{anticlockwise}} = 117.6\text{ N} \times 1.5\text{ m} = 176.4\text{ Nm}\). This must equal the clockwise moment: \(176.4\text{ Nm} = F_2 \times 1.8\text{ m} \Rightarrow F_2 = 98\text{ N}\). Finally, convert this force back to mass: \(M = \frac{98\text{ N}}{9.8\text{ N/kg}} = 10\text{ kg}\).

1 mark for calculating the moment of the first box: \(117.6\text{ N} \times 1.5\text{ m} = 176.4\text{ Nm}\) (or equivalent ratio \(12 \times 1.5 = M \times 1.8\)). 1 mark for equating moments to find \(F_2 = 98\text{ N}\). 1 mark for the final mass of 10 kg.

First, calculate the total distance traveled by the ultrasound pulse: \(\text{Distance} = \text{Speed} \times \text{Time} = 1500\text{ m/s} \times 0.080\text{ s} = 120\text{ m}\). Since the ultrasound pulse travels to the bottom of the lake and back up to the receiver, the depth is half of the total distance traveled: \(\text{Depth} = \frac{120\text{ m}}{2} = 60\text{ m}\).

1 mark for substituting speed and time into the equation \(d = v \times t\). 1 mark for calculating total round-trip distance as 120 m (or using 0.040 s as the one-way travel time). 1 mark for calculating the correct depth of 60 m.

First, calculate the total work done by the pulling force: \(W = F \times d = 85\text{ N} \times 12\text{ m} = 1020\text{ J}\). The total energy supplied is 1020 J. The useful energy transferred is the gravitational potential energy gained, which is 720 J. The energy wasted as heat is the difference: \(1020\text{ J} - 720\text{ J} = 300\text{ J}\).

1 mark for using the formula \(W = F \times d\). 1 mark for calculating total work done as 1020 J. 1 mark for subtracting the useful energy to get the wasted energy of 300 J.

Use Einstein's mass-energy equivalence equation: \(E = m \times c^2\). Substitute the given values: \(E = (3.6 \times 10^{-29}\text{ kg}) \times (3.0 \times 10^8\text{ m/s})^2 = 3.6 \times 10^{-29} \times 9.0 \times 10^{16} = 3.24 \times 10^{-12}\text{ J}\).

1 mark for recalling \(E = m \times c^2\). 1 mark for correct substitution including squaring the speed of light: \(3.6 \times 10^{-29} \times (3.0 \times 10^8)^2\). 1 mark for correct final calculation yielding \(3.24 \times 10^{-12}\text{ J}\) (or \(3.2 \times 10^{-12}\text{ J}\)).

First, convert the time from minutes to seconds: \(t = 2.0\text{ minutes} = 2.0 \times 60 = 120\text{ s}\). Next, calculate the charge flow (\(Q\)): \(Q = I \times t = 0.60\text{ A} \times 120\text{ s} = 72\text{ C}\). Finally, calculate the energy transferred (\(E\)): \(E = V \times Q = 4.5\text{ V} \times 72\text{ C} = 324\text{ J}\). Alternatively, calculate power (\(P\)) first: \(P = I \times V = 0.60\text{ A} \times 4.5\text{ V} = 2.7\text{ W}\). Then, calculate energy: \(E = P \times t = 2.7\text{ W} \times 120\text{ s} = 324\text{ J}\).

- 1 mark for converting time to seconds: \(120\text{ s}\) (or for calculating Power \(P = 2.7\text{ W}\)). - 1 mark for correct calculation of charge: \(72\text{ C}\) (or for showing \(E = 2.7 \times 120\)). - 1 mark for correct final value with unit: \(324\text{ J}\) (allow \(324\text{ Joules}\)).

1. Calculate the stored elastic potential energy (\(E_e\)): \(E_e = \frac{1}{2} k e^2 = \frac{1}{2} \times 200\text{ N/m} \times (0.050\text{ m})^2 = 100 \times 0.0025 = 0.25\text{ J}\). 2. Equate the elastic potential energy to kinetic energy (\(E_k\)) and solve for speed (\(v\)): \(E_k = \frac{1}{2} m v^2 = 0.25\text{ J}\), so \ \frac{1}{2} \times 0.080\text{ kg} \times v^2 = 0.25\text{ J}\), giving \(0.040 \times v^2 = 0.25\), so \(v^2 = 6.25\). Thus, \(v = \sqrt{6.25} = 2.5\text{ m/s}\).

- 1 mark for calculating stored elastic potential energy: \(0.25\text{ J}\). - 1 mark for correct algebraic rearrangement or substitution to find speed: \(v^2 = 6.25\). - 1 mark for the correct speed with unit: \(2.5\text{ m/s}\).

When first dropped, the downward force of gravity (weight) is greater than the upward resistive force (drag), causing the ball to accelerate downwards. As the speed of the ball increases, the upward drag force also increases. Eventually, the upward drag force becomes equal to the downward weight of the ball. At this point, the resultant force acting on the ball is zero, so it ceases to accelerate and continues to fall at a constant terminal velocity.

1 mark: For stating that the upward drag/resistive force increases as the speed of the ball increases.
1 mark: For explaining that the drag force eventually becomes equal to the weight of the ball.
1 mark: For concluding that when forces are balanced, the resultant force is zero, so the ball falls at a constant (terminal) velocity.

Nuclear fusion involves joining two lighter atomic nuclei to form a heavier nucleus. Since nuclei are positively charged, they experience a strong electrostatic repulsion force as they approach each other. Extremely high temperatures are needed to give the nuclei enough kinetic energy to overcome this electrostatic repulsion. High pressures are required to compress the nuclei close together, increasing the rate of collisions so that fusion can take place.

1 mark: For identifying that nuclei are positively charged and repel each other electrostatically.
1 mark: For explaining that high temperature provides high kinetic energy to overcome this electrostatic repulsion.
1 mark: For explaining that high pressure increases collision frequency by keeping nuclei close together.

An ultrasound transmitter at the surface of the water emits a short pulse of high-frequency sound waves. This pulse travels down through the water, reflects off the solid bottom of the reservoir, and travels back to a detector at the surface. The detector measures the total time delay between emitting the pulse and receiving the echo. Since the wave has traveled to the bottom and back, the depth is calculated by multiplying the speed of sound in water by half of the measured time delay.

1 mark: For stating that a pulse is transmitted through the water, reflects off the bottom, and is detected at the surface.
1 mark: For stating that the time delay between transmission and detection of the echo is measured.
1 mark: For explaining that depth is calculated using speed multiplied by half the measured time delay because the wave travels a double distance.

The useful energy transfer is the chemical energy stored in the drill's battery being transferred electrically to the kinetic energy store of the rotating drill bit. The wasted energy transfers occur as energy is dissipated to the thermal energy store of the drill motor, the drill bit, and the wooden block due to friction and electrical resistance. Additionally, some energy is transferred away as sound waves to the surroundings.

1 mark: For identifying the useful energy pathway (chemical energy store of battery transferred electrically to kinetic energy of the drill bit).
1 mark: For identifying that energy is wasted/dissipated as thermal energy (due to friction or electrical resistance in the motor).
1 mark: For identifying sound energy as another wasted pathway dissipated to the surroundings.

As water waves travel into shallower water, the friction with the seabed causes their speed to decrease. According to the wave equation, wave speed is equal to frequency multiplied by wavelength (v = f * lambda). Since the frequency of the waves remains constant, a decrease in wave speed must result in a proportional decrease in wavelength. Therefore, the wavelength of the water waves decreases.

1 mark: For stating that the speed of the water waves decreases in shallower water.
1 mark: For referencing the wave equation (v = f * lambda) or stating that speed is directly proportional to wavelength when frequency is constant.
1 mark: For concluding that the wavelength of the waves decreases.

In a parallel circuit, the potential difference across each branch is identical, so both Resistor A and Resistor B experience the same potential difference. Current is defined as the rate of flow of charge. Since Resistor A has a lower resistance than Resistor B, it offers less opposition to the flow of charge. Consequently, more charge flows through Resistor A per second under the same potential difference, resulting in a larger current.

1 mark: For stating that the potential difference across both resistors is the same because they are connected in parallel.
1 mark: For stating that current is the rate of flow of charge, and that lower resistance offers less opposition to charge flow.
1 mark: For concluding that more charge flows per second through Resistor A, resulting in a higher current.

When passengers occupy the upper deck, they increase the mass at a higher position, which raises the overall centre of mass of the double-decker bus. As the bus is tilted, a higher centre of mass means that the vertical line of action of the bus's weight will fall outside the support base of its wheels at a smaller tilt angle. Once the line of action of the weight falls outside this base, it creates a resultant moment that rotates the bus further, causing it to topple over.

1 mark: For explaining that loading the upper deck raises the overall position of the centre of mass of the bus.
1 mark: For stating that tilting causes the vertical line of action of the weight to fall outside the base of support (wheels).
1 mark: For explaining that this creates a resultant/unbalanced moment that causes the bus to topple.

When the angle of incidence inside the glass block at the flat glass-air boundary is increased beyond the critical angle, the light ray cannot refract out into the air. Instead, it undergoes total internal reflection. All of the light is reflected back inside the glass block. The light obeys the law of reflection, meaning the angle of reflection within the glass is exactly equal to the angle of incidence.

1 mark: For stating that the ray undergoes total internal reflection.
1 mark: For explaining that no light is transmitted/refracted out into the air.
1 mark: For stating that the light behaves as if it hit a mirror, with the angle of reflection being equal to the angle of incidence inside the glass.

Initially, the only force acting on the ball bearing is its weight acting downwards, giving it maximum acceleration. As the ball bearing speeds up, the upward resistive force (drag) from the oil increases. The resultant force (weight minus drag) acting downwards therefore decreases, which causes the acceleration to decrease until the drag equals the weight and the acceleration becomes zero.

1 mark: Acceleration decreases.
1 mark: Drag / resistive force increases as the velocity increases.
1 mark: The resultant force decreases, so the acceleration decreases (Newton's second law, \(F = ma\)).

The fission of a uranium-235 nucleus is triggered by absorbing a neutron, which splits the nucleus and releases energy along with two or three fast-moving neutrons. These released neutrons are absorbed by other uranium-235 nuclei, triggering further fission events. Control rods are adjusted to absorb excess neutrons so that, on average, only one neutron from each fission goes on to cause another fission, maintaining a steady and controlled rate.

1 mark: Fission of the first nucleus releases two or three neutrons.
1 mark: These neutrons are absorbed by other uranium-235 nuclei, causing them to split (chain reaction).
1 mark: Control rods absorb excess neutrons to ensure only one neutron per fission causes further fission, keeping the reaction rate constant.

The glass core has a higher refractive index than the surrounding outer cladding layer. When light travels through the core and meets the boundary with the cladding, it strikes the boundary at an angle of incidence that is greater than the critical angle. Because of this, total internal reflection occurs, and all the light is reflected back into the core rather than refracting out.

1 mark: Core has a higher refractive index than the cladding.
1 mark: Light strikes the boundary at an angle of incidence greater than the critical angle.
1 mark: Total internal reflection occurs, preventing light from escaping the core.

An ultrasound transducer emits high-frequency sound waves into the mother's body. When these waves reach a boundary between different media (such as muscle, fluid, or bone), some of the wave is reflected back to the transducer while some is transmitted. The transducer detects these reflections, and by measuring the time delay of each echo, a computer calculates the depth of each boundary to construct a detailed image.

1 mark: Ultrasound waves are partially reflected at boundaries between different tissues/materials.
1 mark: The time taken for the reflected waves (echoes) to return to the detector is measured.
1 mark: The depth of each boundary is calculated (using speed and time), which is processed by a computer to form an image.

A decrease in temperature causes the resistance of the thermistor to increase. In a potential divider circuit, the total voltage is shared in proportion to the resistances of the components. Since the thermistor's resistance has increased relative to the fixed resistor, it takes a larger share of the input potential difference. The voltage across the thermistor rises until it reaches the threshold required to activate the buzzer.

1 mark: Resistance of the thermistor increases as temperature decreases.
1 mark: The potential difference (voltage) across the thermistor increases (or takes a larger share of the total voltage).
1 mark: The increased voltage reaches the threshold required to turn on the buzzer.

Once the hydrogen in the core is depleted, fusion stops and the outward radiation pressure decreases. The core contracts under gravity, which increases its temperature and pressure. This heating triggers hydrogen fusion in a shell surrounding the core, producing an outward pressure that causes the outer layers of the star to expand significantly and cool down, making the star appear red.

1 mark: The core contracts/collapses because core hydrogen fusion stops and radiation pressure drops.
1 mark: Hydrogen fusion begins in a shell around the core (due to heating from core contraction).
1 mark: The resulting outward pressure causes the outer layers of the star to expand and cool.

Energy is dissipated by heating up the plastic or metal body of the kettle itself, rather than the water. Energy is also wasted by thermal conduction and convection from the kettle's outer walls to the cooler surrounding air, as well as energy lost as steam escapes. A design improvement is to construct the kettle with double-walled vacuum insulation (or thick plastic walls) to reduce heat transfer to the surrounding air.

1 mark: Describes one dissipation pathway (e.g., heating the kettle body or heat lost as steam escapes).
1 mark: Describes a second dissipation pathway (e.g., thermal conduction/convection to the surrounding air).
1 mark: Suggests a valid design improvement (e.g., insulating walls / sealing the lid tightly) linked to reducing one of the described losses.

According to Newton's third law, when the skater exerts a forward force on the medicine ball, the ball exerts an equal and opposite backward force on the skater. From the perspective of the conservation of momentum, the total momentum of the skater-ball system before the throw is zero. Since momentum must be conserved, the total momentum after the throw must also be zero; therefore, the skater must gain backward momentum to balance the forward momentum of the ball.

1 mark: Total momentum before the throw is zero.
1 mark: States that to conserve momentum, the skater must gain momentum in the opposite (backward) direction to the ball's forward momentum (or references Newton's third law: equal and opposite force on the skater).
1 mark: Since the skater has mass, this backward force/momentum causes them to move backwards.

An ultrasound transmitter on a ship sends a pulse of ultrasound waves downwards into the ocean. When the waves reach the boundary between the water and the ocean floor, they are reflected back towards the ship. A receiver on the ship detects the reflected waves (the echo) and records the total time taken. Since the speed of sound in seawater is known, the depth of the ocean trench is calculated using the formula: depth = \(\frac{\text{speed of sound} \times \text{time taken}}{2}\).

[1 mark] State that ultrasound pulses are transmitted downwards and are reflected from the ocean floor / seabed.
[1 mark] State that the time taken for the echo/pulse to return to the receiver is measured.
[1 mark] Explain that the depth is calculated using \(\text{depth} = \frac{\text{speed} \times \text{time}}{2}\) or that the total distance calculated must be halved because the wave travels to the bottom and back.

Initially, when the ball bearing is released, the downward force of gravity (weight) is much larger than the upward drag force, causing the ball bearing to accelerate downwards. As the velocity of the ball bearing increases, the upward resistive force (drag/friction) acting on it also increases. Eventually, the upward drag force increases until it is equal in magnitude to the downward force of gravity. At this point, the resultant force acting on the ball bearing becomes zero, so it stops accelerating and falls at a constant terminal velocity.

[1 mark] State that as the speed of the ball bearing increases, the upward resistive force / drag force increases.
[1 mark] State that eventually the upward drag force becomes equal (and opposite) to the downward force of gravity / weight.
[1 mark] Explain that the resultant force is now zero, meaning the acceleration is zero and the ball bearing travels at a constant velocity.

By conservation of momentum: Total momentum before = Total momentum after. \( (2.0 \text{ kg} \times 3.0 \text{ m/s}) + (1.0 \text{ kg} \times 0 \text{ m/s}) = (2.0 \text{ kg} + 1.0 \text{ kg}) \times v \). \( 6.0 \text{ kg m/s} = 3.0 \text{ kg} \times v \). Therefore, \( v = 2.0 \text{ m/s} \).

Award 1 mark for the correct answer C. No marks if working is incorrect but C is not chosen.

Using Snell's law: \( n = \frac{\sin(i)}{\sin(r)} \). Here \( i = 45^\circ \) and \( r = 28^\circ \). \( n = \frac{\sin(45^\circ)}{\sin(28^\circ)} \approx \frac{0.7071}{0.4695} \approx 1.51 \).

Award 1 mark for the correct answer B.

Using Ohm's law: \( I = \frac{V}{R} \). Substituting the given values: \( I = \frac{6.0\text{ V}}{12\ \Omega} = 0.50\text{ A} \).

Award 1 mark for the correct answer A.

First, calculate the initial corrected count rate: \( 800\text{ cpm} - 40\text{ cpm} = 760\text{ cpm} \). After 8 hours (which is exactly 2 half-lives), the corrected count rate will halve twice: \( \frac{760}{2} = 380\text{ cpm} \) after 4 hours, and \( \frac{380}{2} = 190\text{ cpm} \) after 8 hours.

Award 1 mark for the correct answer A.

Useful work done equals the gain in gravitational potential energy: \( E_p = m \times g \times h = 5.0\text{ kg} \times 9.8\text{ N/kg} \times 2.0\text{ m} = 98\text{ J} \). Useful power output is the rate of doing useful work: \( P = \frac{E_p}{t} = \frac{98\text{ J}}{4.0\text{ s}} = 24.5\text{ W} \).

Award 1 mark for the correct answer B.

The total distance traveled by the pulse to the ocean floor and back is given by \( d_{\text{total}} = v \times t = 1500\text{ m/s} \times 0.24\text{ s} = 360\text{ m} \). The depth is half of the total distance: \( \text{Depth} = \frac{360\text{ m}}{2} = 180\text{ m} \).

Award 1 mark for the correct answer B.

The elastic potential energy stored is given by: \( E_e = \frac{1}{2} F e \). Substituting the given values: \( E_e = 0.5 \times 15\text{ N} \times 0.30\text{ m} = 2.25\text{ J} \).

Award 1 mark for the correct answer B.

The pivot is at 50 cm. Distance from pivot to the 4.0 N weight = \( 50\text{ cm} - 20\text{ cm} = 30\text{ cm} = 0.30\text{ m} \). Anticlockwise moment = \( 4.0\text{ N} \times 0.30\text{ m} = 1.2\text{ Nm} \). Distance from pivot to the balancing weight at 90 cm = \( 90\text{ cm} - 50\text{ cm} = 40\text{ cm} = 0.40\text{ m} \). For horizontal balance, Clockwise moment = Anticlockwise moment: \( W \times 0.40\text{ m} = 1.2\text{ Nm} \implies W = \frac{1.2}{0.40} = 3.0\text{ N} \).

Award 1 mark for the correct answer B.

To find the refractive index \(n\) from the critical angle \(c\), we use the formula:

\[n = \frac{1}{\sin c}\]

Substitute \(c = 38^\circ\) into the formula:

\[\sin(38^\circ) \approx 0.6157\]

\[n = \frac{1}{0.6157} \approx 1.62\]

Therefore, the refractive index of the plastic block is 1.62.

1 mark for selecting the correct option C (1.62).

To find the refractive index \(n\) from the critical angle \(c\), we use the formula:

\[n = \frac{1}{\sin c}\]

Substitute \(c = 38^\circ\) into the formula:

\[\sin(38^\circ) \approx 0.6157\]

\[n = \frac{1}{0.6157} \approx 1.62\]

Therefore, the refractive index of the plastic block is 1.62.

1 mark for selecting the correct option C (1.62).

1. Find the velocity of the combined system immediately after the collision using conservation of momentum:
\(m_1 u_1 = (m_1 + m_2) v\)
\(3.0 \times 8.0 = (3.0 + 5.0) \times v\)
\(24.0 = 8.0 \times v\)
\(v = 3.0 \text{ m/s}\)

2. Calculate the kinetic energy of the combined system immediately after the collision:
\(E_k = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 8.0 \times (3.0)^2 = 36.0 \text{ J}\)

3. Use the work-energy theorem (work done by the resistive force equals the loss in kinetic energy) to find the distance \(d\):
\(W = F \times d\)
\(36.0 = 6.0 \times d\)
\(d = 6.0 \text{ m}\)

- **1 mark**: Recalls and applies conservation of momentum to find the equation for the final velocity: \(3.0 \times 8.0 = 8.0 \times v\).
- **1 mark**: Calculates the final velocity of the system as \(3.0 \text{ m/s}\).
- **1 mark**: Calculates the kinetic energy after the collision as \(36.0 \text{ J}\) (or calculates deceleration as \(a = \frac{F}{m} = \frac{6.0}{8.0} = 0.75 \text{ m/s}^2\)).
- **1 mark**: Calculates the distance correctly as \(6.0 \text{ m}\).

1. Calculate the equivalent resistance of the two identical parallel resistors:
\(R_p = \frac{12.0 \times 12.0}{12.0 + 12.0} = 6.0 \ \Omega\)

2. Find the total resistance of the circuit:
\(R_T = 3.0 + 6.0 = 9.0 \ \Omega\)

3. Calculate the total current in the circuit:
\(I_T = \frac{V}{R_T} = \frac{18}{9.0} = 2.0 \text{ A}\)

4. Determine the current flowing through one of the parallel branches:
Since both resistors in parallel are identical, the current splits equally:
\(I_{branch} = \frac{2.0}{2} = 1.0 \text{ A}\)

5. Calculate the power dissipated in one of the parallel resistors:
\(P = I^2 R = (1.0)^2 \times 12.0 = 12.0 \text{ W}\)

- **1 mark**: Calculates equivalent resistance of parallel branch as \(6.0 \ \Omega\).
- **1 mark**: Calculates total current as \(2.0 \text{ A}\) using \(I = \frac{V}{R}\).
- **1 mark**: Calculates branch current as \(1.0 \text{ A}\) (or calculates voltage drop across parallel branch as \(V = 2.0 \times 6.0 = 12.0 \text{ V}\)).
- **1 mark**: Correctly calculates power as \(12.0 \text{ W}\).

1. Calculate the refractive index of the plastic block (\(n_p\)):
\(n_p = \frac{\sin i}{\sin r} = \frac{\sin 50.0^\circ}{\sin 30.0^\circ} = \frac{0.7660}{0.5000} = 1.532\)

2. Use the equation for the critical angle at the boundary between two media (from denser plastic to rarer water):
\(\sin \theta_c = \frac{n_{water}}{n_{plastic}} = \frac{1.33}{1.532} \approx 0.8681\)

3. Calculate the critical angle (\(\theta_c\)):
\(\theta_c = \sin^{-1}(0.8681) \approx 60.2^\circ\)

- **1 mark**: Recall and substitute into refractive index formula: \(n = \frac{\sin 50.0^\circ}{\sin 30.0^\circ}\).
- **1 mark**: Calculates refractive index of plastic as \(1.53\) (or \(1.532\)).
- **1 mark**: Recall and substitute into critical angle formula for boundaries between two media: \(\sin \theta_c = \frac{1.33}{1.532}\).
- **1 mark**: Calculates critical angle as \(60.2^\circ\) (accept range \(60.0^\circ\) to \(60.5^\circ\)).

1. Calculate the useful thermal energy needed to heat the water:
\(Q = m \cdot c \cdot \Delta \theta\)
\(Q = 1.5 \times 4200 \times (100 - 20) = 1.5 \times 4200 \times 80 = 504,000 \text{ J}\)

2. Use the efficiency formula to find the total electrical energy input required:
\(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}}\)
\(0.80 = \frac{504,000}{E_{in}} \implies E_{in} = \frac{504,000}{0.80} = 630,000 \text{ J}\)

3. Calculate the time required using \(P = \frac{E}{t}\), converting power from kW to W (\(2.4 \text{ kW} = 2400 \text{ W}\)):
\(t = \frac{E_{in}}{P} = \frac{630,000}{2400} = 262.5 \text{ s}\)

- **1 mark**: Calculates the useful heat energy needed as \(504,000 \text{ J}\).
- **1 mark**: Rearranges the efficiency equation and calculates total energy input needed as \(630,000 \text{ J}\).
- **1 mark**: Converses power correctly (\(2400 \text{ W}\)) and recalls \(t = \frac{E}{P}\).
- **1 mark**: Calculates time correctly as \(262.5 \text{ s}\) (accept \(263 \text{ s}\)).

1. Calculate the useful work output (kinetic energy gained):
\(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200 \times (20)^2 = 240,000 \text{ J}\)

2. Calculate the total electrical energy input using power and time:
\(P = I \times V = 45 \times 400 = 18,000 \text{ W}\)
\(E_{in} = P \times t = 18,000 \times 15 = 270,000 \text{ J}\)

3. Calculate the efficiency:
\(\text{Efficiency} = \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\)
\(\text{Efficiency} = \frac{240,000}{270,000} \times 100 \approx 88.9\%\)

- **1 mark**: Calculates kinetic energy as \(240,000 \text{ J}\).
- **1 mark**: Calculates input electrical power as \(18,000 \text{ W}\).
- **1 mark**: Calculates total electrical energy input as \(270,000 \text{ J}\).
- **1 mark**: Calculates efficiency as \(88.9\%\) (accept \(0.889\) or \(89\%\)).

**Method 1 (Calculating speed first):**
1. Calculate total distance traveled to the bottom and back:
\(d_{bottom} = 2 \times 0.12 = 0.24 \text{ m}\)

2. Calculate the speed of ultrasound in the metal:
\(v = \frac{\text{distance}}{\text{time}} = \frac{0.24 \text{ m}}{80 \times 10^{-6} \text{ s}} = 3000 \text{ m/s}\)

3. Calculate distance traveled to the crack and back:
\(d_{crack} = v \times t_{crack} = 3000 \times (48 \times 10^{-6} \text{ s}) = 0.144 \text{ m}\)

4. Find the depth of the crack:
\(\text{depth} = \frac{d_{crack}}{2} = \frac{0.144}{2} = 0.072 \text{ m}\) (or \(7.2 \text{ cm}\))

**Method 2 (Using ratios):**
Since the speed is constant, depth is directly proportional to time:
\(\text{depth} = \text{thickness} \times \frac{t_{crack}}{t_{bottom}} = 0.12 \times \frac{48}{80} = 0.072 \text{ m}\)

- **1 mark**: Calculates total distance to the bottom and back as \(0.24 \text{ m}\).
- **1 mark**: Calculates speed of ultrasound as \(3000 \text{ m/s}\).
- **1 mark**: Calculates total distance to crack and back as \(0.144 \text{ m}\).
- **1 mark**: Correctly determines the depth of the crack as \(0.072 \text{ m}\) (or \(7.2 \text{ cm}\)).
*(Note: Full 4 marks if ratio method is used correctly: setting up ratio (2 marks), substitution (1 mark), correct final answer (1 mark).)*

1. Identify position of the center of mass:
Since the bar is uniform, its center of mass is at its midpoint, which is 1.0 m from the left end.

2. Calculate the perpendicular distances of the forces from the pivot (located 0.6 m from the left end):
- Distance to left weight (at 0.0 m) = \(0.6 \text{ m}\)
- Distance to center of mass of bar (at 1.0 m) = \(1.0 - 0.6 = 0.4 \text{ m}\) (to the right of the pivot)
- Distance to right mass (at 2.0 m) = \(2.0 - 0.6 = 1.4 \text{ m}\) (to the right of the pivot)

3. Apply the Principle of Moments (sum of clockwise moments = sum of anticlockwise moments):
\(M_{anticlockwise} = M_{clockwise}\)
\(m_{left} \times g \times 0.6 = (m_{bar} \times g \times 0.4) + (m_{right} \times g \times 1.4)\)
Dividing through by \(g\):
\(m_{left} \times 0.6 = (6.0 \times 0.4) + (1.5 \times 1.4)\)
\(m_{left} \times 0.6 = 2.4 + 2.1\)
\(m_{left} \times 0.6 = 4.5\)
\(m_{left} = \frac{4.5}{0.6} = 7.5 \text{ kg}\)

- **1 mark**: Identifies position of center of mass (1.0 m) and calculates distance from pivot to center of mass as \(0.4 \text{ m}\) AND distance from pivot to right end as \(1.4 \text{ m}\).
- **1 mark**: Recalls the Principle of Moments: sum of clockwise moments = sum of anticlockwise moments.
- **1 mark**: Substitutes values correctly into the moments equation: \(m_{left} \times 0.6 = (6.0 \times 0.4) + (1.5 \times 1.4)\).
- **1 mark**: Calculates mass correctly as \(7.5 \text{ kg}\).

1. Subtract background radiation to find the initial corrected count rate of the source:
\(360 - 40 = 320 \text{ counts per minute}\)

2. Determine the number of half-lives that have passed in 24 days:
\(\text{Number of half-lives} = \frac{24 \text{ days}}{8 \text{ days}} = 3 \text{ half-lives}\)

3. Calculate the corrected count rate after 3 half-lives:
\(\text{Corrected count rate} = 320 \times \left(\frac{1}{2}\right)^3 = 320 \times \frac{1}{8} = 40 \text{ counts per minute}\)

4. Add back the constant background count rate to get the final measured count rate:
\(\text{Final measured count rate} = 40 + 40 = 80 \text{ counts per minute}\)

- **1 mark**: Corrects initial count rate by subtracting background: \(360 - 40 = 320\).
- **1 mark**: Calculates the number of half-lives as 3.
- **1 mark**: Halves the corrected count rate three times to obtain \(40 \text{ counts per minute}\).
- **1 mark**: Adds background back to get the correct final answer of \(80 \text{ counts per minute}\).

### Indicative Content

**Circuit Setup:**
* Connect a power supply/cell, an ammeter, a switch, and the test wire in series.
* Connect a voltmeter in parallel across the length of the test wire being measured using crocodile clips.

**Measurements:**
1. Clip the crocodile clips to the test wire at a starting length of 10 cm (0.10 m), measured using a metre ruler.
2. Close the switch and record the potential difference ($V$) from the voltmeter and the current ($I$) from the ammeter.
3. Open the switch immediately after taking the readings.
4. Move one crocodile clip to increase the length of the wire in regular intervals (e.g., 20 cm, 30 cm, up to 100 cm) and repeat the measurements of potential difference and current for each length.

**Determining Resistance:**
* Calculate the resistance ($R$) for each length of wire using Ohm's Law formula:
$$R = \frac{V}{I}$$
* Plot a graph of resistance against length to show the relationship.

**Safety / Validity Precaution:**
* Open the switch between readings to disconnect the circuit. This prevents the wire from getting hot. A hot wire can cause burns (safety) and changes the resistivity of the metal, making the test unfair (validity).

**Level 3 (5–6 marks):**
A detailed and coherent method is described with logically ordered steps. It clearly details how to set up the circuit (with correct series/parallel connections for ammeter/voltmeter), explains how to vary length systematically, shows how to calculate resistance, and includes a relevant safety precaution regarding wire heating.

**Level 2 (3–4 marks):**
A coherent method is described with some detail. It explains how to set up the apparatus, vary wire length, measure voltage and current, and calculate resistance, but may omit the safety aspect or be less precise about systematic circuit configuration.

**Level 1 (1–2 marks):**
Simple statements showing some understanding of connecting a wire in a circuit to measure resistance. The response may lack a logical structure, fail to specify how to vary length systematically, or omit calculations.

**0 marks:**
No relevant content.

### Apparatus:
- Helical spring
- Clamp stand, clamps, and bosses
- Metre ruler
- Slotted masses and hanger (e.g., $100\text{ g}$ intervals)
- Pointer (attached to the bottom of the spring)

### Procedure:
1. Set up the clamp stand and suspend the spring vertically from the top clamp.
2. Position the metre ruler vertically alongside the spring, securing it in place.
3. Note the initial position of the bottom of the spring (or the pointer) on the ruler. This is the unstretched position ($L_0$).
4. Add a known mass (e.g., $100\text{ g}$ mass hanger) to the bottom of the spring. Convert this mass to force in Newtons ($W = m \times g$).
5. Record the new position of the pointer on the ruler ($L$).
6. Calculate the extension of the spring ($e = L - L_0$).
7. Add additional slotted masses one by one, recording the new pointer position and calculating the total extension for each load.
8. Repeat the measurements as masses are removed to check for hysteresis or permanent deformation.

### Accuracy Precautions:
- Ensure the metre ruler is perfectly vertical (use a set square or plumb line).
- Read the scale on the ruler at eye level with the pointer to avoid parallax error.
- Ensure the spring is completely stationary before taking a reading.

### Analysis:
- Plot a graph of applied force (in Newtons) on the y-axis against extension (in metres or millimetres) on the x-axis.
- Determine the spring constant ($k$) by calculating the gradient ($\text{gradient} = \frac{\Delta F}{\Delta e}$) of the linear portion of the graph where Hooke's Law is obeyed.

**Level 3 (5–6 marks):**
A detailed, coherent, and logically structured method is described. The equipment list is complete, and the procedure clearly explains how to obtain accurate measurements (e.g., avoiding parallax error, measuring unstretched length, and finding extension). A clear explanation of how to use a force-extension graph to determine the spring constant is provided.

**Level 2 (3–4 marks):**
A coherent method is described with some detail. It includes most steps of the procedure and lists the key measurements (force and extension). Some consideration of accuracy is mentioned, or there is a basic description of how to calculate the spring constant.

**Level 1 (1–2 marks):**
Simple statements are made showing some understanding of the investigation. The list of apparatus or the procedure is incomplete, with little or no explanation of how to ensure accuracy or analyze the results.

**0 marks:**
No relevant content.