2023 OCR GCSE Twenty First Century Science - Physics B - J259 模擬試題連答案詳解

a) First, find the extension of the spring: \(x = 0.18\text{ m} - 0.12\text{ m} = 0.06\text{ m}\).
Using Hooke's Law: \(F = kx\).
Rearranging for spring constant \(k\):
\(k = \frac{F}{x} = \frac{4.5\text{ N}}{0.06\text{ m}} = 75\text{ N/m}\).

b) The energy stored in the elastic store is given by:
\(E = \frac{1}{2} F x\) or \(E = \frac{1}{2} k x^2\).
Using \(E = \frac{1}{2} F x\):
\(E = 0.5 \times 4.5\text{ N} \times 0.06\text{ m} = 0.135\text{ J}\).

c) In elastic deformation, the object returns to its original length and shape once the stretching force is removed. In plastic deformation, the object does not return to its original shape; it remains permanently deformed.

a) [1 mark] calculating the extension of 0.06 m. [1 mark] correctly recalling/rearranging the formula \(F = kx\). [1 mark] correct calculation of 75 N/m.
b) [1 mark] recalling/using \(E = \frac{1}{2} F x\) or \(E = \frac{1}{2} k x^2\). [1 mark] correct calculation of 0.135 J.
c) [1 mark] defining elastic deformation (returns to original shape). [1 mark] defining plastic deformation (permanent change in shape).

a) Temperature change \(\Delta \theta = 95\text{ °C} - 15\text{ °C} = 80\text{ °C}\).
Thermal energy required: \(E = m c \Delta \theta = 0.80\text{ kg} \times 4200\text{ J/kg°C} \times 80\text{ °C} = 268,800\text{ J}\).

b) Power \(P = 2.2\text{ kW} = 2200\text{ W}\).
Since \(P = \frac{E}{t}\), time \(t = \frac{E}{P} = \frac{268800\text{ J}}{2200\text{ W}} = 122.18\text{ s} \approx 122\text{ s}\).

c) As temperature increases, the thermal energy is transferred to the kinetic store of the water particles, making them move and vibrate faster. This increased motion causes them to collide more forcefully, pushing them slightly further apart on average and increasing the overall volume.

a) [1 mark] determining temperature difference (80 °C). [1 mark] substitution of values into \(E = m c \Delta \theta\). [1 mark] correct final value (268800 J).
b) [1 mark] converting kW to W (2200 W) and rearranging \(t = E/P\). [1 mark] correct final time (122 s, accept 122.18 s).
c) [1 mark] stating particles gain kinetic energy/move faster. [1 mark] stating that particles push each other further apart on average.

a) First, find the equivalent resistance of the parallel branch, \(R_p\):
\(\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{4.0} = \frac{2}{4.0} = \frac{1}{2.0}\), so \(R_p = 2.0\\ \Omega\).
Then, add the series resistor to find total resistance:
\(R_{\text{total}} = R_s + R_p = 6.0\\ \Omega + 2.0\\ \Omega = 8.0\\ \Omega\).

b) Total current \(I\) is:
\(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\\ \Omega} = 1.5\text{ A}\).

c) Because the parallel branches have equal resistance (each is \(4.0\\ \Omega\)), the current splits equally between them. Therefore, the current through each of the parallel resistors is half of the total current (\(0.75\text{ A}\)).

a) [1 mark] calculating parallel resistance (2.0 \(\Omega\)). [1 mark] adding series resistance (6.0 + 2.0). [1 mark] correct total resistance (8.0 \(\Omega\)).
b) [1 mark] recalling \(I = V/R\). [1 mark] correct calculation (1.5 A).
c) [1 mark] stating current is half of total/0.75 A. [1 mark] explaining that parallel branches are identical so current divides equally.

a) Momentum is given by \(p = m v\).
Initial momentum: \(p_i = 0.50\text{ kg} \times 4.0\text{ m/s} = 2.0\text{ kg m/s}\).
Final momentum: \(p_f = 0\text{ kg m/s}\).
Change in momentum: \(\Delta p = p_f - p_i = 0 - 2.0 = -2.0\text{ kg m/s}\).

b) Force is the rate of change of momentum: \(F = \frac{\Delta p}{t}\).
\(F = \frac{-2.0\text{ kg m/s}}{0.25\text{ s}} = -8.0\text{ N}\). (The magnitude of the average force is 8.0 N).

c) If the time interval increases, the rate of change of momentum decreases. Because force is proportional to the rate of change of momentum (\(F = \frac{\Delta p}{\Delta t}\)), the average force acting on the trolley would be smaller.

a) [1 mark] correct initial momentum calculation (2.0). [1 mark] correct change in momentum (-2.0 or 2.0). [1 mark] correct unit (kg m/s or N s).
b) [1 mark] recalling \(F = \Delta p / t\). [1 mark] correct force calculation (-8.0 N or 8.0 N).
c) [1 mark] stating the force would decrease. [1 mark] explanation linked to increased collision time reducing rate of change of momentum.

a) In alpha decay, the mass number decreases by 4 and the atomic number decreases by 2.
Mass number: \(226 - 4 = 222\), so \(A = 222\).
Atomic number: \(88 - 2 = 86\), so \(Z = 86\).

b) Determine the number of half-lives that have passed:
Number of half-lives \(n = \frac{4800\text{ years}}{1600\text{ years}} = 3\).
After 1 half-life: \(8.0 \times 10^4\).
After 2 half-lives: \(4.0 \times 10^4\).
After 3 half-lives: \(2.0 \times 10^4\) (or 20,000).

c) Alpha particles have a relatively large mass and double positive charge (+2), making them highly ionizing compared to beta particles. However, because they interact strongly with matter, they lose energy quickly and have very low penetrating power (stopped by a sheet of paper or a few cm of air), whereas beta radiation is moderately penetrating.

a) [1 mark] identifying A = 222. [1 mark] identifying Z = 86.
b) [1 mark] calculating number of half-lives (3). [1 mark] dividing initial value 3 times. [1 mark] correct answer (20,000 or \(2.0 \times 10^4\)).
c) [1 mark] correct comparison of ionizing power (alpha is more ionizing). [1 mark] correct comparison of penetrating power (alpha is less penetrating).

a) Use the formula for magnetic force on a conductor:
\(F = B I l\).
\(F = 0.40\text{ T} \times 3.5\text{ A} \times 0.20\text{ m} = 0.28\text{ N}\).

b) The rule is Fleming's Left-Hand Rule.
- The thumb represents the direction of the force (or motion).
- The first finger represents the direction of the magnetic field (North to South).
- The second finger represents the direction of the conventional current (positive to negative).

c) To reverse the force, you can either reverse the direction of the current (by swapping the connections of the power supply) or reverse the direction of the magnetic field (by flipping the magnets around).

a) [1 mark] recalling formula \(F = B I l\). [1 mark] substituting values correctly. [1 mark] correct answer (0.28 N).
b) [1 mark] naming Fleming's Left-Hand Rule. [1 mark] linking thumb to force/motion. [1 mark] linking first finger to field and second finger to current.
c) [1 mark] stating either reversing current OR reversing magnetic field direction.

a) Hydrostatic pressure is calculated using:
\(p = h \rho g\).
\(p = 120\text{ m} \times 1030\text{ kg/m}^3 \times 10\text{ N/kg} = 1,236,000\text{ Pa}\) (or \(1.236 \times 10^6\text{ Pa}\)).

b) Total pressure is the sum of fluid pressure and atmospheric pressure:
\(P_{\text{total}} = P_{\text{fluid}} + P_{\text{atm}}\).
\(P_{\text{total}} = 1,236,000\text{ Pa} + 1.0 \times 10^5\text{ Pa} = 1,336,000\text{ Pa}\) (or \(1.336 \times 10^6\text{ Pa}\)).

c) As the container is lowered, the surrounding liquid pressure increases. By Boyle's Law, volume is inversely proportional to pressure at a constant temperature. Therefore, the increasing external pressure compresses the gas inside, causing the volume of the flexible container to decrease.

a) [1 mark] recalling \(p = h \rho g\). [1 mark] substitution of values. [1 mark] correct calculation (1,236,000 Pa or \(1.236 \times 10^6\text{ Pa}\)).
b) [1 mark] adding atmospheric pressure. [1 mark] correct total (1,336,000 Pa or \(1.336 \times 10^6\text{ Pa}\)).
c) [1 mark] stating volume decreases. [1 mark] explanation linked to increased hydrostatic/external pressure compressing the gas.

a) Using the wave equation:
\(v = f \lambda\).
Rearranging to find wavelength:
\(\lambda = \frac{v}{f} = \frac{3.0 \times 10^8\text{ m/s}}{1.2 \times 10^{15}\text{ Hz}} = 2.5 \times 10^{-7}\text{ m}\).

b) Hazard: UV is highly energetic and ionizing, which can damage skin cells leading to sunburn or skin cancer, or cause cataracts in the eyes. Application: Used to detect counterfeit bank notes (using fluorescence) or to sterilize drinking water or medical equipment.

c) The frequency of the wave is determined by the source and does not change as it passes through different mediums. Since the wave speed \(v\) decreases in glass and \(v = f \lambda\), the wavelength \(\lambda\) must decrease proportionally to maintain this relationship.

a) [1 mark] recalling \(v = f \lambda\). [1 mark] correct substitution. [1 mark] correct calculation of wavelength (\(2.5 \times 10^{-7}\text{ m}\)).
b) [1 mark] any correct hazard. [1 mark] any correct application.
c) [1 mark] stating frequency remains constant. [1 mark] explaining wavelength decreases due to decrease in speed.

**Part (a)**
- Elastic deformation: The material returns to its original length and shape once the stretching force is removed.
- Plastic deformation: The material is permanently deformed and does not return to its original shape or length when the force is removed.

**Part (b)**
- First, convert the extension from centimeters to meters:
\(x = 8.0\text{ cm} = 0.08\text{ m}\)
- Use the elastic potential energy formula:
\(E_e = \frac{1}{2} k x^2\)
- Substitute the values:
\(E_e = 0.5 \times 250\text{ N/m} \times (0.08\text{ m})^2\)
\(E_e = 125 \times 0.0064 = 0.8\text{ J}\)

**Part (c)**
- Polymers consist of long, tangled molecular chains. When a force is applied, these chains untangle and straighten out easily, allowing for a large extension without breaking the strong covalent bonds.
- In contrast, metals have a rigid crystalline lattice where stretching only slightly distorts the metallic bonds before layers of atoms permanently slide over one another.

**Part (a) [2 marks]**
- 1 mark: Clarifying that elastic deformation is temporary / returns to original shape.
- 1 mark: Clarifying that plastic deformation is permanent / does not return to original shape.

**Part (b) [3 marks]**
- 1 mark: Correct conversion of extension to meters: \(0.08\text{ m}\).
- 1 mark: Correct substitution into the formula \(E_e = \frac{1}{2} k x^2\) (or \(E_e = 0.5 \times 250 \times 0.08^2\)).
- 1 mark: Correct final answer: \(0.8\text{ J}\).

**Part (c) [2 marks]**
- 1 mark: Mentions long, tangled polymer chains that can slide, untangle, or straighten out.
- 1 mark: Contrasts with metals having a rigid structure or explains that untangling does not break the chemical bonds within the chains.

**Part (a)**
- First, find the equivalent resistance of the parallel branch (\(R_p\)) using:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\)
\(\frac{1}{R_p} = \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30}\)
\(R_p = \frac{30}{5} = 6\ \Omega\)
- Next, add the series resistor (\(4\ \Omega\)) to find total resistance:
\(R_{total} = R_p + R_{series} = 6\ \Omega + 4\ \Omega = 10\ \Omega\).

**Part (b)**
- Use Ohm's Law:
\(I_{total} = \frac{V}{R_{total}} = \frac{12\text{ V}}{10\ \Omega} = 1.2\text{ A}\).

**Part (c)**
- Calculate the potential difference across the parallel combination:
\(V_p = I_{total} \times R_p = 1.2\text{ A} \times 6\ \Omega = 7.2\text{ V}\).
- Calculate the current through the \(15\ \Omega\) resistor:
\(I_{15} = \frac{V_p}{15\ \Omega} = \frac{7.2\text{ V}}{15\ \Omega} = 0.48\text{ A}\).

**Part (a) [3 marks]**
- 1 mark: Correct formula or setup for parallel resistors (\(\frac{1}{10} + \frac{1}{15}\)).
- 1 mark: Correctly calculating the parallel branch resistance as \(6\ \Omega\).
- 1 mark: Adding the series resistor to get the total resistance of \(10\ \Omega\).

**Part (b) [2 marks]**
- 1 mark: Correct equation used (\(I = V/R\)).
- 1 mark: Correct calculation to give \(1.2\text{ A}\).

**Part (c) [2 marks]**
- 1 mark: Correctly determining the voltage across the parallel group as \(7.2\text{ V}\) (or finding the current ratio \(10 / (10+15) \times 1.2\text{ A}\)).
- 1 mark: Correct final current calculation of \(0.48\text{ A}\).

**Part (a)**
- Calculate the downward force of gravity (weight, \(W\)):
\(W = m \times g = 65\text{ kg} \times 10\text{ N/kg} = 650\text{ N}\).
- Calculate the resultant force (\(F_{res}\)):
\(F_{res} = \text{Weight} - \text{Air Resistance} = 650\text{ N} - 260\text{ N} = 390\text{ N}\) (downwards).
- Calculate the acceleration (\(a\)):
\(a = \frac{F_{res}}{m} = \frac{390\text{ N}}{65\text{ kg}} = 6.0\text{ m/s}^2\).

**Part (b)**
- Initially, as the skydiver speeds up, the air resistance increases.
- This reduces the resultant downward force on the skydiver, so the acceleration decreases.
- Eventually, the upward air resistance increases until it is equal in magnitude to the downward weight of the skydiver.
- The resultant force becomes zero, so according to Newton's First Law, the skydiver continues at a constant velocity (terminal velocity).

**Part (a) [4 marks]**
- 1 mark: Correct calculation of weight as \(650\text{ N}\).
- 1 mark: Correct calculation of resultant force as \(390\text{ N}\).
- 1 mark: Recall and arrangement of \(F = ma\) to \(a = F/m\).
- 1 mark: Correct acceleration value of \(6.0\text{ m/s}^2\) (accept \(6\text{ m/s}^2\)).

**Part (b) [3 marks]**
- 1 mark: Mentions that air resistance increases as velocity increases.
- 1 mark: States that air resistance becomes equal to weight, leading to zero resultant force.
- 1 mark: Mentions Newton's First Law / no resultant force means constant speed/terminal velocity.

**Part (a)**
- Calculate the number of half-lives that have elapsed:
\(\text{Number of half-lives} = \frac{24\text{ hours}}{6.0\text{ hours}} = 4\).
- Calculate the activity after each half-life:
- 1 half-life: \(400\text{ Bq}\)
- 2 half-lives: \(200\text{ Bq}\)
- 3 half-lives: \(100\text{ Bq}\)
- 4 half-lives: \(50\text{ Bq}\).
- Thus, the activity remaining is \(50\text{ Bq}\).

**Part (b)**
- Gamma radiation is highly penetrating and can easily pass out of the body to be detected by an external camera.
- Alpha radiation is highly ionizing and poorly penetrating; it would be completely absorbed by body tissues, causing severe cellular damage/mutation and failing to exit the body for detection.

**Part (c)**
- In a beta minus decay, a neutron turns into a proton, emitting a high-energy electron (\(\vphantom{}^{0}_{-1}\text{e}\) or \(\beta^-\)).
- To balance the mass and atomic numbers, the missing particle must have a mass number of \(0\) and an atomic number of \(-1\):
\(\vphantom{}^{0}_{-1}\text{e}\).

**Part (a) [3 marks]**
- 1 mark: Determines that \(4\) half-lives have elapsed (\(24 / 6\)).
- 1 mark: Evidence of successive halving or applying \(\frac{1}{2^4}\).
- 1 mark: Correct final answer: \(50\text{ Bq}\).

**Part (b) [2 marks]**
- 1 mark: Explains that gamma is highly penetrating / can escape the body to be detected.
- 1 mark: Explains that alpha is highly ionizing / highly damaging / cannot escape the body.

**Part (c) [2 marks]**
- 1 mark: Correctly identifies the particle as an electron or beta particle (symbol \(\text{e}\) or \(\beta\)).
- 1 mark: Correct mass number of \(0\) and atomic number of \(-1\).

**Part (a)**
- Use the formula for the force on a conductor:
\(F = B I L\)
- Rearrange the formula to solve for current (\(I\)):
\(I = \frac{F}{B L}\)
- Substitute the values:
\(I = \frac{0.090\text{ N}}{0.40\text{ T} \times 0.15\text{ m}}\)
\(I = \frac{0.090}{0.06} = 1.5\text{ A}\).

**Part (b)**
- Hold the thumb, first finger, and second finger of the left hand mutually at right angles (perpendicular to each other).
- The **F**irst finger points in the direction of the magnetic **F**ield (North to South).
- The se**C**ond finger points in the direction of the **C**urrent (positive to negative).
- The **T**humb then points in the direction of the **M**otion / force.

**Part (c)**
- Increase the current flowing through the coil.
- Use stronger permanent magnets (increase magnetic field strength).
- Increase the number of turns on the coil.
- Add an iron core inside the coil.

**Part (a) [3 marks]**
- 1 mark: Recall or use of formula \(F = BIL\) or rearrangement \(I = F / (BL)\).
- 1 mark: Correct substitution: \(\frac{0.090}{0.40 \times 0.15}\).
- 1 mark: Correct calculation of \(1.5\text{ A}\).

**Part (b) [2 marks]**
- 1 mark: States fingers are held at right angles / mutually perpendicular.
- 1 mark: Correctly links fingers to variables (First finger = Field, Second finger = Current, Thumb = Force/Motion; at least two correctly identified for the mark).

**Part (c) [2 marks]**
- 2 marks: Any two valid modifications from: increase current, use stronger magnets, increase number of turns in coil, add a soft iron core (1 mark per valid modification).

Part (a):
- Set up the wire horizontally clamped securely at one end, passing over a low-friction pulley at the edge of the bench with a weight hanger attached to the other end.
- Measure the original length of the wire under small initial tension using a metre ruler.
- Attach a small paper marker or tape to the wire and position a second metre ruler parallel to the wire to read the position of the marker.
- Add slotted masses systematically to the hanger, recording the new position of the marker each time to calculate extension (new length minus original length).
- To minimize parallax error, ensure the eye is directly above the marker when taking readings. Use a longer wire to achieve larger, more measurable extensions.
- Safety precaution: Wear safety goggles to protect eyes in case the wire snaps under high tension, and place a foam tray below the masses.

Part (b):
- Identify formula: \(k = \frac{F}{x}\)
- Convert extension from millimetres to metres: \(x = 1.2\text{ mm} = 1.2 \times 10^{-3}\text{ m}\)
- Substitute the values: \(k = \frac{25\text{ N}}{1.2 \times 10^{-3}\text{ m}}\)
- Calculate the final value: \(k \approx 20833\text{ N/m}\) (or \(2.1 \times 10^4\text{ N/m}\)).

Part (a) [6 marks]:
- 1 mark: Clamp one end and use a pulley/mass system to apply force.
- 1 mark: Measure original length and use a marker on the wire to measure movement against a ruler.
- 1 mark: Systematic addition of masses and calculation of extension.
- 1 mark: Method to reduce uncertainty (e.g. eye level reading / parallax reduction / long wire).
- 1 mark: Specific safety precaution (e.g. safety goggles / protective tray).
- 1 mark: Logical sequence of experimental steps.

Part (b) [4 marks]:
- 1 mark: Recalls or uses \(k = \frac{F}{x}\).
- 1 mark: Correct conversion of extension to metres: \(1.2 \times 10^{-3}\text{ m}\).
- 1 mark: Correct substitution: \(\frac{25}{1.2 \times 10^{-3}}\).
- 1 mark: Final value with correct unit: \(20833\text{ N/m}\) (accept \(2.1 \times 10^4\text{ N/m}\) or \(20.8\text{ kN/m}\)).

Part (a):
- From \(-10^\circ\text{C}\) to \(0^\circ\text{C}\), the ice is solid. Heating increases the kinetic energy of the particles, causing them to vibrate faster about fixed positions, which corresponds to a temperature rise.
- At \(0^\circ\text{C}\), the ice melts. The temperature remains constant because the energy supplied is used to overcome the intermolecular forces holding the ice lattice together, rather than increasing particle speed (kinetic energy).
- From \(0^\circ\text{C}\) to \(100^\circ\text{C}\), the water is liquid. Heating increases the kinetic energy of the particles, which move around each other, causing a temperature rise.
- At \(100^\circ\text{C}\), the water boils. The temperature is constant because the energy is used to break the remaining intermolecular bonds to free the particles into a gas.
- Above \(100^\circ\text{C}\), the steam particles gain kinetic energy, moving rapidly in all directions as temperature rises.

Part (b):
- Energy to melt the ice: \(E_1 = m \times L_f = 0.25\text{ kg} \times 3.34 \times 10^5\text{ J/kg} = 83500\text{ J}\)
- Energy to heat the water: \(E_2 = m \times c \times \Delta\theta = 0.25\text{ kg} \times 4200\text{ J/kg}^\circ\text{C} \times (20^\circ\text{C} - 0^\circ\text{C}) = 0.25 \times 4200 \times 20 = 21000\text{ J}\)
- Total energy: \(E_{\text{total}} = E_1 + E_2 = 83500\text{ J} + 21000\text{ J} = 104500\text{ J}\) (or \(104.5\text{ kJ}\)).

Part (a) [6 marks]:
- Level 3 (5-6 marks): Detailed explanation of both temperature rise (linked to particle kinetic energy) and constant temperature zones (linked to energy breaking/weakening intermolecular forces during state transitions of melting and boiling).
- Level 2 (3-4 marks): Clear description of states, temperature changes, and transitions, but with less detail on the mechanism of intermolecular forces/bond breaking vs. kinetic energy.
- Level 1 (1-2 marks): Simple description of heating phases, identifying that temperature stays constant during melting or boiling.

Part (b) [4 marks]:
- 1 mark: Calculation of melting energy: \(0.25 \times 3.34 \times 10^5 = 83500\text{ J}\).
- 1 mark: Calculation of heating energy: \(0.25 \times 4200 \times 20 = 21000\text{ J}\).
- 1 mark: Sum of both energy values: \(83500 + 21000\).
- 1 mark: Final correct value and unit: \(104500\text{ J}\) (or \(104.5\text{ kJ}\)).

Part (a):
- Step 1: Calculate parallel resistance \(R_p\) of the two \(8.0\ \Omega\) resistors:
\(\frac{1}{R_p} = \frac{1}{8.0} + \frac{1}{8.0} = \frac{2}{8.0} = \frac{1}{4.0}\)
Therefore, \(R_p = 4.0\ \Omega\).
- Step 2: Add the series resistor to find total equivalent resistance \(R_{\text{total}}\):
\(R_{\text{total}} = R_s + R_p = 6.0\ \Omega + 4.0\ \Omega = 10.0\ \Omega\).
- Step 3: Calculate current \(I\) using Ohm's Law:
\(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{10.0\ \Omega} = 1.2\text{ A}\).

Part (b):
- Disconnecting one of the \(8.0\ \Omega\) resistors changes the parallel section to a single \(8.0\ \Omega\) resistor.
- This increases the total resistance of the circuit from \(10.0\ \Omega\) to \(14.0\ \Omega\) (\(6.0\ \Omega + 8.0\ \Omega\)).
- An increase in total resistance decreases the total current drawn from the power supply.
- Since \(V = IR\) for the \(6.0\ \Omega\) resistor, and the current \(I\) decreases, the potential difference across it will decrease. The voltmeter reading will decrease.

Part (a) [6 marks]:
- 1 mark: Recalls or uses the parallel resistance formula.
- 1 mark: Correct calculation of parallel pair resistance: \(4.0\ \Omega\).
- 1 mark: Correctly sums the parallel section and the series resistor.
- 1 mark: Correct total resistance value: \(10.0\ \Omega\).
- 1 mark: Recalls or uses \(I = V/R\).
- 1 mark: Correct calculation of current: \(1.2\text{ A}\).

Part (b) [4 marks]:
- 1 mark: States that disconnecting one resistor increases total resistance (to \(14\ \Omega\)).
- 1 mark: States that total circuit current decreases.
- 1 mark: Explains that potential difference across the \(6.0\ \Omega\) resistor is proportional to current (\(V=IR\)).
- 1 mark: Concludes the voltmeter reading will decrease.

Part (a):
- Setup: Tilt the ramp slightly until the trolley just rolls down at a constant speed when given a small push (friction-compensated ramp).
- Attach a string to the trolley, passing over a pulley with a mass hanger at the end to provide the accelerating force.
- Measure acceleration using two light gates connected to a data logger. A card of known length on the trolley interrupts the beams to record velocities \(u\) and \(v\) and the time interval \(t\), from which the software calculates acceleration.
- To vary force: Start with slotted masses on the trolley. Move masses one-by-one from the top of the trolley to the hanging mass holder. This increases the accelerating force (weight of hanging mass) while keeping the total mass of the system (trolley + hanger + masses) constant.
- Repeat the experiment for at least five different forces and calculate average accelerations.

Part (b):
- Step 1: Calculate the resultant force acting on the trolley:
\(F_{\text{res}} = F_{\text{applied}} - F_{\text{friction}} = 2.4\text{ N} - 0.60\text{ N} = 1.8\text{ N}\)
- Step 2: Recalls Newton's Second Law: \(F = m \times a\)
- Step 3: Rearrange to solve for acceleration:
\(a = \frac{F_{\text{res}}}{m} = \frac{1.8\text{ N}}{0.80\text{ kg}} = 2.25\text{ m/s}^2\).

Part (a) [6 marks]:
- 1 mark: Describes how to compensate for friction (tilt the runway).
- 1 mark: Mentions using hanging masses over a pulley to provide tension.
- 1 mark: Explains the use of light gates and a card to measure acceleration.
- 1 mark: Explicitly states that to keep system mass constant, masses are moved from the trolley to the hanger.
- 1 mark: Mentions repeating readings to calculate mean values.
- 1 mark: Logical structured description.

Part (b) [4 marks]:
- 1 mark: Calculates resultant force: \(1.8\text{ N}\).
- 1 mark: Recalls or uses \(a = F/m\).
- 1 mark: Substitutes values correctly: \(1.8 / 0.80\).
- 1 mark: Correct acceleration with units: \(2.25\text{ m/s}^2\) (accept \(2.3\text{ m/s}^2\)).

Part (a):
- Physical Nature: Alpha radiation consists of helium nuclei (2 protons, 2 neutrons); beta-minus consists of high-energy electrons; gamma consists of high-frequency electromagnetic waves.
- Ionizing Power: Alpha has the highest ionizing power due to its relatively large mass and +2 charge; beta has moderate ionizing power; gamma is weakly ionizing because it is uncharged.
- Penetrating Power: Alpha is stopped easily by a sheet of paper or a few cm of air; beta can penetrate paper but is stopped by a few mm of aluminium; gamma is highly penetrating and requires thick lead or metres of concrete to be significantly absorbed.
- Behavior in Electric Field: Alpha particles are positively charged and deflect slightly towards the negative plate; beta-minus particles are negatively charged, have a much smaller mass, and deflect strongly towards the positive plate; gamma rays are uncharged electromagnetic waves and pass straight through without deflection.

Part (b):
- Step 1: Determine the number of half-lives that have elapsed.
\(800\text{ Bq} \xrightarrow{\text{1 half-life}} 400\text{ Bq} \xrightarrow{\text{2 half-lives}} 200\text{ Bq} \xrightarrow{\text{3 half-lives}} 100\text{ Bq}\)
- Step 2: Set up equation for total time:
\(3 \times T_{1/2} = 15\text{ days}\)
- Step 3: Solve for the half-life:
\(T_{1/2} = \frac{15}{3} = 5\text{ days}\).

Part (a) [6 marks]:
- Level 3 (5-6 marks): Comprehensive comparison covering all four characteristics (nature, ionization, penetration, electric field deflection) for all three radiations with high accuracy.
- Level 2 (3-4 marks): Good comparison covering most characteristics for the three radiations, though some details or explanations may be less complete.
- Level 1 (1-2 marks): Basic identification of properties of some radiations with major omissions or lack of explanation.

Part (b) [4 marks]:
- 1 mark: Identifies that the activity halves 3 times to go from 800 Bq to 100 Bq.
- 1 mark: Shows clear working of halving sequence: 800 -> 400 -> 200 -> 100.
- 1 mark: Connects 3 half-lives to 15 days.
- 1 mark: Obtains correct half-life: 5 days (with correct unit).

Part (a):
- The permanent magnets establish a magnetic field across the coil.
- When current flows through the rectangular coil, the sides perpendicular to the field experience opposing forces due to the motor effect (Fleming's Left-Hand Rule), which creates a torque/turning effect that rotates the coil.
- As the coil reaches the vertical position, the forces on each side would act to bring the coil to a halt or reverse its direction if the current direction remained unchanged.
- The split-ring commutator rotates with the coil and changes contact from one carbon brush to the other every half turn (180 degrees).
- This reverses the direction of the current in the coil every half turn, which keeps the forces on each side acting in the same relative direction (e.g. left side always pushed up, right side always pushed down), maintaining continuous rotation in one direction.
- The carbon brushes provide a stable, sliding electrical connection from the stationary d.c. supply to the rotating commutator.

Part (b):
- Recalls the formula: \(F = B \times I \times L\)
- Substitute the values: \(F = 0.25\text{ T} \times 4.0\text{ A} \times 0.15\text{ m}\)
- Calculate the value: \(F = 0.15\text{ N}\).

Part (a) [6 marks]:
- 1 mark: Mentions that current in magnetic field creates a force (motor effect).
- 1 mark: Explains that forces on opposite sides of the coil act in opposite directions, causing rotation.
- 1 mark: Identifies that without a commutator, the coil would stop or oscillate at the vertical position.
- 1 mark: Explains that the split-ring commutator reverses current direction every half-turn.
- 1 mark: Explains that reversing current maintains force direction on each side, ensuring continuous rotation.
- 1 mark: Explains role of carbon brushes as low-friction sliding electrical contacts.

Part (b) [4 marks]:
- 1 mark: Recalls or uses the equation \(F = BIL\).
- 1 mark: Correct substitution of values: \(0.25 \times 4.0 \times 0.15\).
- 1 mark: Correct calculation: \(0.15\).
- 1 mark: Correct unit: \(N\) or Newtons.

Part (a):
- Liquid pressure increases with depth because the pressure at any point is caused by the weight of the column of liquid directly above that point. Deeper down, there is a larger volume (and thus mass and weight) of water above, exerting a greater downward force over the area.
- Demonstration: Take a tall plastic bottle or cylinder and drill three small, identical holes at different heights along one side. Cover the holes with tape and fill the cylinder with water.
- Observation: Remove the tape from all three holes at the same time. Water will shoot out of all three holes.
- Analysis: The water jet from the lowest hole travels the furthest horizontally, while the jet from the top hole travels the shortest distance. Since a greater horizontal velocity indicates higher pressure pushing water out of the hole, this observation confirms that pressure is greatest at the bottom (deepest point).

Part (b):
- Step 1: Calculate pressure using \(p = h \times \rho \times g\):
\(p = 120\text{ m} \times 1025\text{ kg/m}^3 \times 10\text{ N/kg} = 1.23 \times 10^6\text{ Pa}\) (or \(1.23\text{ MPa}\)).
- Step 2: Calculate the area of the circular window:
\(A = \pi \times r^2 = \pi \times (0.15\text{ m})^2 = \pi \times 0.0225 \approx 0.0707\text{ m}^2\).
- Step 3: Calculate force using \(F = p \times A\):
\(F = 1.23 \times 10^6\text{ Pa} \times 0.0707\text{ m}^2 \approx 8.7 \times 10^4\text{ N}\) (accept values in range \(8.69 \times 10^4\text{ N}\) to \(8.70 \times 10^4\text{ N}\)).

Part (a) [6 marks]:
- 1 mark: Explains that pressure is due to the weight of the liquid column above.
- 1 mark: Explains that deeper points have more liquid weight above, increasing force per unit area.
- 1 mark: Describes the setup: cylinder with vertical holes at different heights filled with water.
- 1 mark: Describes releasing the water and observing the trajectories of the water jets.
- 1 mark: Mentions that the bottom jet travels the furthest horizontally.
- 1 mark: Connects the further distance of the bottom jet to a greater pressure at greater depth.

Part (b) [4 marks]:
- 1 mark: Recalls or uses \(p = h \rho g\) to get \(1.23 \times 10^6\text{ Pa}\).
- 1 mark: Calculates area of circular window: \(A = \pi r^2 \approx 0.0707\text{ m}^2\).
- 1 mark: Recalls or uses \(F = p \times A\).
- 1 mark: Calculates force correctly: \(8.7 \times 10^4\text{ N}\) (accept \(8.69 \times 10^4\) to \(8.70 \times 10^4\) with correct units).

Part (a):
- Place the rectangular glass block on a piece of white paper and trace around its perimeter with a pencil.
- Direct a narrow ray of light from a ray box at an angle into one of the long faces of the block.
- Mark the path of the incident ray, the emergent ray, and the entry and exit points on the block with small pencil crosses.
- Remove the glass block and use a ruler to connect the entry and exit points, representing the refracted ray inside the glass.
- Draw a normal line (90 degrees to the glass boundary) at the entry point. Use a protractor to measure the angle of incidence, \(i\) (between normal and incident ray), and the angle of refraction, \(r\) (between normal and refracted ray).
- To find the refractive index, repeat the measurements for several different angles of incidence. Plot a graph of \(\sin(i)\) against \(\sin(r)\). The gradient of the resulting straight line is equal to the refractive index of the glass (\(n = \frac{\sin i}{\sin r}\)).

Part (b):
- Step 1: Calculate the speed of light in glass:
\(v_{\text{glass}} = \frac{v_{\text{air}}}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.5} = 2.0 \times 10^8\text{ m/s}\).
- Step 2: Calculate the wavelength in glass:
\(\lambda_{\text{glass}} = \frac{\lambda_{\text{air}}}{n} = \frac{5.0 \times 10^{-7}\text{ m}}{1.5} \approx 3.33 \times 10^{-7}\text{ m}\).

Part (a) [6 marks]:
- 1 mark: Traces block outline and marks incident/emergent rays.
- 1 mark: Draws normal line at the boundary.
- 1 mark: Correctly identifies/defines angle of incidence and angle of refraction (relative to normal).
- 1 mark: Measures multiple pairs of \(i\) and \(r\) across a range.
- 1 mark: Explains that refractive index \(n = \frac{\sin i}{\sin r}\).
- 1 mark: Mentions plotting \(\sin(i)\) vs \(\sin(r)\) to find gradient as \(n\) (or averaging calculations of \(n\)).

Part (b) [4 marks]:
- 1 mark: Recalls or uses \(v_{\text{glass}} = \frac{v}{n}\).
- 1 mark: Correct speed calculation: \(2.0 \times 10^8\text{ m/s}\).
- 1 mark: Recalls or uses \(\lambda_{\text{glass}} = \frac{\lambda}{n}\) (or \(v = f\lambda\)).
- 1 mark: Correct wavelength calculation: \(3.33 \times 10^{-7}\text{ m}\) (accept \(3.3 \times 10^{-7}\text{ m}\)).

### Part (a)(i)
1. Convert time into seconds:
\(t = 5.0\text{ minutes} = 5.0 \times 60 = 300\text{ s}\)

2. Calculate the electrical energy input (\(E\)):
\(E = V \times I \times t\)
\(E = 12.0\text{ V} \times 4.0\text{ A} \times 300\text{ s} = 14400\text{ J}\)

3. Calculate the change in temperature (\(\Delta \theta\)):
\(\Delta \theta = 50.0^\circ\text{C} - 22.0^\circ\text{C} = 28.0^\circ\text{C}\)

4. Use the specific heat capacity formula:
\(E = m c \Delta \theta \implies c = \frac{E}{m \Delta \theta}\)
\(c = \frac{14400\text{ J}}{1.0\text{ kg} \times 28.0^\circ\text{C}} \approx 514\text{ J/kg}^\circ\text{C}\) (or \(510\text{ J/kg}^\circ\text{C}\) to 2 significant figures).

### Part (a)(ii)
- Assumption: There is no heat loss from the metal block to the surroundings (or that 100% of the energy from the heater is transferred to the block, or that the heat capacity of the thermometer/heater is negligible).

### Part (b)*
**Indicative Scientific Content:**

**Modifications to the setup:**
- Wrap the metal block in an insulating material (such as bubble wrap, cotton wool, or foam) to reduce heat loss to the surrounding air by conduction and convection.
- Place the block on an insulating mat (e.g., cork or wood) rather than directly on a laboratory bench to prevent thermal conduction to the bench.
- Add a small amount of oil or petroleum jelly into the thermometer hole to improve thermal contact between the thermometer bulb and the block.

**Methodological improvements and Graph usage:**
- Rather than taking only initial and final temperature readings, measure and record the temperature at regular intervals (e.g., every 30 seconds) throughout the heating period.
- Plot a graph of temperature (on the y-axis) against time (on the x-axis).
- Identify the linear region of the graph where the heater is fully warmed up but heat losses are still low.
- Determine the gradient (\(\frac{\Delta \theta}{\Delta t}\)) of this linear section.
- Use the relationship: \(\text{Power } P = V I\) and \(P = m c \times \text{gradient}\). Rearranging this gives \(c = \frac{VI}{m \times \text{gradient}}\). Using the gradient from multiple data points averages out random errors and avoids errors due to initial lag or late heat loss.

### Part (a)(i) [3 marks]
- **1 mark**: Correct conversion of time to seconds (\(300\text{ s}\)) AND calculation of energy transferred: \(E = 12 \times 4 \times 300 = 14400\text{ J}\).
- **1 mark**: Correct calculation of temperature change: \(\Delta\theta = 28^\circ\text{C}\).
- **1 mark**: Correct substitution and calculation of specific heat capacity: \(c = 514\text{ J/kg}^\circ\text{C}\) (accept \(510\text{ J/kg}^\circ\text{C}\) due to significant figures or \(514.3\text{ J/kg}^\circ\text{C}\)).

### Part (a)(ii) [1 mark]
- **1 mark**: States that all thermal energy from the heater was transferred to the block (no heat loss to surroundings / negligible heat capacity of the thermometer/heater).

### Part (b)* [6 marks]
This is an extended writing question marked with an asterisk. Levels of response are used to award marks.

- **Level 3 (5–6 marks)**: The candidate provides a detailed, coherent, and logically structured response that addresses all three bullet points. They suggest practical improvements (insulation, oil/petroleum jelly), explain why these changes reduce systematic errors, and clearly explain how to use a temperature-time graph and its gradient to find a more accurate value for specific heat capacity.
- **Level 2 (3–4 marks)**: The candidate suggests some improvements (such as insulation or using oil) and makes a reasonable attempt to explain how to collect multiple readings or use a graph. The explanations are mostly clear but may lack full mathematical detail regarding the gradient of the graph.
- **Level 1 (1–2 marks)**: The candidate suggests at least one simple improvement (e.g., 'insulate the block' or 'take more readings') but does not fully explain how this improves accuracy or how to analyze the data using a graph.
- **0 marks**: No response of any relevance.