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(a) As the magnet falls, its magnetic field lines cut through the turns of the coil. This creates a change in magnetic flux linkage, which induces an electromotive force (voltage) across the coil. Since the coil forms a complete circuit, an induced current flows.

(b) When the magnet enters the coil, the magnetic flux increases, inducing a voltage in one direction (first peak). When the magnet leaves the coil, the magnetic flux decreases, inducing a voltage in the opposite direction (second peak). Because the magnet is accelerating due to gravity, it moves faster when exiting the coil than when entering. This greater rate of change of flux results in a taller peak, and the shorter time interval results in a narrower peak.

(c) Dropping the magnet from a greater height means it will be travelling faster both when entering and leaving the coil. Therefore, both peaks will be taller (greater induced voltage due to faster flux cutting) and narrower (the magnet spends less time passing through the coil).

(a) 1 mark for mentioning magnetic field lines cutting the coil. 1 mark for change in magnetic flux linkage. 1 mark for inducing an electromotive force / voltage.

(b) 1 mark for identifying that entering/leaving causes opposite change in flux. 1 mark for explaining opposite peaks are due to opposite directions of motion/flux change. 1 mark for noting the magnet accelerates / moves faster at the bottom. 1 mark for linking higher speed to greater rate of change of flux (taller peak). 1 mark for linking higher speed to shorter duration of flux change (narrower peak).

(c) 1 mark for stating peaks are taller. 1 mark for stating peaks are narrower / occur earlier.

(a) The diagram should show a semi-circular glass block. A ray of light should enter the curved face along the normal (radius) so it does not bend. The ray must strike the midpoint of the flat face at the critical angle, with the refracted ray travelling along the flat boundary of the glass block at \(90^\circ\) to the normal.

(b) (i) \(n = \frac{1}{\sin(c)}\)
(ii) \(n = \frac{1}{\sin(42^\circ)} = \frac{1}{0.669} \approx 1.49\)

(c) (i) Since \(35^\circ\) is less than the critical angle of \(42^\circ\), the light will refract out into the air. Using Snell's law: \(n \sin(i) = 1 \cdot \sin(r)\) where the ray is going from glass to air, so \(\sin(r) = n \sin(i) = 1.49 \times
\sin(35^\circ) = 1.49 \times 0.574 = 0.855\). Therefore, \(r = \arcsin(0.855) \approx 59^\circ\).
(ii) Since \(45^\circ\) is greater than the critical angle of \(42^\circ\), the ray will undergo total internal reflection (TIR) and reflect back into the glass block at an angle of \(45^\circ\).

(a) 1 mark for ray entering along the normal of the curved face. 1 mark for ray striking the center of the flat boundary. 1 mark for refracted ray directed along the boundary at 90 degrees to the normal.

(b)(i) 1 mark for correct formula: n = 1 / sin(c).
(ii) 1 mark for correct substitution: n = 1 / sin(42). 1 mark for correct evaluation to 1.49 (or 1.5).

(c)(i) 1 mark for stating that the ray refracts out of the block. 1 mark for correct use of Snell's law equation. 1 mark for calculating r as 59 degrees (accept 58 to 60 degrees).
(ii) 1 mark for stating total internal reflection occurs.

(a) In a beta decay, the nucleon number remains unchanged, so \(a = 0\). The proton number increases by 1, which means the beta particle (electron) must have a charge of \(-1\), so \(b = -1\).

(b) (i) Half-life is the time taken for the activity or the number of radioactive nuclei in a sample to decrease to half of its initial value.
(ii) The activity halves as follows: \(800\text{ Bq} \rightarrow 400\text{ Bq} \rightarrow 200\text{ Bq} \rightarrow 100\text{ Bq}\). This process requires 3 half-lives. Since 3 half-lives equal 24 days, 1 half-life is \(24 / 3 = 8\text{ days}\).

(c) (i) Inside the body, beta radiation travels only a short distance (a few millimeters in tissue), meaning it can target and destroy the localized tumor cells without causing extensive damage to healthy tissues further away. However, if placed far outside the body, beta radiation would be absorbed by the air or the outer layers of dead skin cells and would not reach the deep-seated tumor.
(ii) Safety precautions include: 1. Keep sources at a distance using tongs. 2. Store radioactive sources in lead-lined containers when not in use.

(a) 1 mark for a = 0. 1 mark for b = -1.

(b)(i) 1 mark for defining half-life in terms of activity or remaining nuclei.
(ii) 1 mark for identifying that 3 half-lives have elapsed. 1 mark for setting up the equation 3 * T_half = 24. 1 mark for final value of 8 days.

(c)(i) 1 mark for explaining that inside the body it limits damage to healthy tissue due to its short range. 1 mark for explaining that outside the body it cannot penetrate to reach the tumor.
(ii) 1 mark per valid safety precaution (e.g., using tongs, storing in lead containers, wearing film badges, pointing sources away). Max 2 marks.

(a) Acceleration is given by: \(a = \frac{v - u}{t}\). Substituting the values: \(a = \frac{9.0 - 0}{6.0} = 1.5\text{ m/s}^2\).

(b) The graph should have velocity in \(\text{m/s}\) on the vertical axis (y-axis) and time in \(\text{s}\) on the horizontal axis (x-axis). The shape should be:
1. A straight line starting from the origin \((0,0)\) up to the point \((6.0, 9.0)\).
2. A horizontal line from \(t = 6.0\text{ s}\) to \(t = 18.0\text{ s}\) at a constant velocity of \(9.0\text{ m/s}\).
3. A straight line descending from \((18.0, 9.0)\) to \((22.0, 0)\).

(c) The total distance is the area under the velocity-time graph. The area can be split into three parts:
- First section (triangle): \(\text{Area}_1 = \frac{1}{2} \times 6.0 \times 9.0 = 27.0\text{ m}\)
- Second section (rectangle): \(\text{Area}_2 = 12.0 \times 9.0 = 108.0\text{ m}\)
- Third section (triangle): \(\text{Area}_3 = \frac{1}{2} \times 4.0 \times 9.0 = 18.0\text{ m}\)

Total distance = \(27.0 + 108.0 + 18.0 = 153\text{ m}\).

(a) 1 mark for formula: a = (v - u) / t. 1 mark for calculation showing 1.5. 1 mark for unit: m/s^2 (or m s^-2).

(b) 1 mark for correctly labelled axes with units. 1 mark for correct shape (acceleration, constant velocity, deceleration). 1 mark for key points labelled with values (6.0 s, 18.0 s, 22.0 s, and 9.0 m/s).

(c) 1 mark for stating that distance equals area under the graph. 1 mark for calculating area of both triangles. 1 mark for calculating area of the rectangle. 1 mark for summing the areas to obtain 153 m.

(a) The principle of conservation of momentum states that in the absence of external forces, the total momentum of a system before a collision is equal to the total momentum after the collision.

(b) (i) Momentum is given by: \(p = m \times v\). For car \(A\): \(p = 0.50 \times 2.4 = 1.2\text{ kg m/s}\).
(ii) By conservation of momentum: \(\text{Total momentum before} = \text{Total momentum after}\). The initial momentum is \(1.2\text{ kg m/s}\). The combined mass after the collision is \(0.50 + 0.30 = 0.80\text{ kg}\). Thus: \(1.2 = 0.80 \times v\) which gives \(v = \frac{1.2}{0.80} = 1.5\text{ m/s}\).

(c) An elastic collision is one where kinetic energy is conserved. Let's calculate the kinetic energy before and after:
- Initial kinetic energy: \(KE_{\text{initial}} = \frac{1}{2} m_A v_A^2 = \frac{1}{2} \times 0.50 \times (2.4)^2 = 1.44\text{ J}\).
- Final kinetic energy: \(KE_{\text{final}} = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} \times 0.80 \times (1.5)^2 = 0.90\text{ J}\).
Since the kinetic energy decreases (from \(1.44\text{ J}\) to \(0.90\text{ J}\)), kinetic energy is not conserved, meaning the collision is inelastic.

(a) 1 mark for total momentum before = total momentum after. 1 mark for specifying 'provided no external forces act'.

(b)(i) 1 mark for formula: p = mv. 1 mark for calculation: 1.2. 1 mark for correct unit: kg m/s (or N s).
(ii) 1 mark for equating total initial and final momentum. 1 mark for correct total mass of 0.80 kg. 1 mark for calculating v = 1.5 m/s.

(c) 1 mark for calculating both initial and final kinetic energies correctly. 1 mark for stating that kinetic energy is not conserved and therefore the collision is inelastic.

(a) (i) \(p = h \times \rho \times g\) (pressure difference = depth \times density \times gravitational field strength)
(ii) \(p = 120 \times 1025 \times 10 = 1,230,000\text{ Pa}\) (or \(1.23\text{ MPa}\))
(iii) \(\text{Total Pressure} = \text{Pressure Increase} + \text{Atmospheric Pressure} = 1,230,000\text{ Pa} + 101,000\text{ Pa} = 1,331,000\text{ Pa}\) (or \(1.33\text{ MPa}\))

(b) First, find the area of the window: \(A = \pi \times r^2 = \pi \times (0.15)^2 \approx 0.0707\text{ m}^2\).
Using the formula relating pressure, force, and area: \(p = \frac{F}{A}\), we get \(F = p \times A = 1,230,000 \times 0.0707 \approx 86,950\text{ N}\) (or \(87,000\text{ N}\) to 2 s.f.).

(c) As depth increases, there is a greater mass (and therefore a greater weight) of liquid above a given level. Since pressure is force per unit area, and the downward gravitational force (weight) increases, the pressure also increases with depth.

(a)(i) 1 mark for correct formula: p = h * rho * g.
(ii) 1 mark for correct substitution. 1 mark for correct evaluation (1,230,000 Pa or 1.23 x 10^6 Pa or 1230 kPa).
(iii) 1 mark for adding atmospheric pressure to the hydrostatic pressure. 1 mark for correct final value of 1,331,000 Pa (or 1.33 x 10^6 Pa or 1331 kPa).

(b) 1 mark for calculating the area of the window (0.0707 m^2). 1 mark for recalling the formula F = p * A. 1 mark for correct evaluation of force (accept range 86,900 N to 87,000 N depending on rounding of pi/area).

(c) 1 mark for mentioning the weight of the water column above. 1 mark for linking this increased downward force to an increase in pressure (force per unit area).

(a) An alternating current (a.c.) in the primary coil produces a changing magnetic field inside the primary coil. This changing magnetic field is transferred by the soft iron core to the secondary coil. The changing magnetic field cuts through the secondary coil, inducing an alternating voltage (and thus an alternating current if there is a complete circuit) across the secondary coil.

(b) Use the transformer ratio equation:
$$\frac{V_p}{V_s} = \frac{N_p}{N_s}$$
Substitute the given values:
$$\frac{230}{V_s} = \frac{2400}{120}$$
$$V_s = 230 \times \frac{120}{2400}$$
$$V_s = 11.5\text{ V}$$

(c) For a 100% efficient transformer:
$$V_p \times I_p = V_s \times I_s$$
Substitute the known values:
$$230 \times I_p = 11.5 \times 1.5$$
$$230 \times I_p = 17.25$$
$$I_p = \frac{17.25}{230} = 0.075\text{ A}$$
(Alternatively, using $I_p = I_s \times \frac{N_s}{N_p} = 1.5 \times \frac{120}{2400} = 0.075\text{ A}$)

(a) [4 marks maximum]
- An alternating current in primary produces an alternating magnetic field [1]
- Iron core channels/transfers the magnetic field to the secondary coil [1]
- The magnetic field cuts through/links with the secondary coil [1]
- This induces an alternating voltage/current in the secondary coil [1]

(b) [3 marks]
- Correct equation: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$ [1]
- Correct substitution: $\frac{230}{V_s} = \frac{2400}{120}$ [1]
- Correct calculation: $11.5\text{ V}$ [1]

(c) [3 marks]
- Correct equation: $V_p I_p = V_s I_s$ or $I_p / I_s = N_s / N_p$ [1]
- Correct substitution: $230 \times I_p = 11.5 \times 1.5$ [1]
- Correct calculation: $0.075\text{ A}$ (or $75\text{ mA}$) [1]

(a) The formula is:
$$n = \frac{\sin i}{\sin r}$$

(b) Substitute $i = 45^\circ$ and $r = 28^\circ$ into the formula:
$$n = \frac{\sin(45^\circ)}{\sin(28^\circ)}$$
$$n = \frac{0.7071}{0.4695} \approx 1.506$$
Rounding to 2 decimal places gives $n = 1.51$.

(c) Use the critical angle formula:
$$\sin c = \frac{1}{n}$$
Using $n = 1.51$:
$$\sin c = \frac{1}{1.51} \approx 0.6623$$
$$c = \arcsin(0.6623) \approx 41.5^\circ$$
(If using unrounded $n = 1.506$, $c \approx 41.6^\circ$)

(d) When light is inside the glass (the denser medium) and meets the boundary with air (the less dense medium), we compare the angle of incidence to the critical angle. Since the angle of incidence ($50^\circ$) is greater than the critical angle ($41.5^\circ$), total internal reflection occurs. No light is refracted into the air; all light is reflected back inside the glass block at an angle of reflection equal to $50^\circ$.

(a) [1 mark]
- Correct equation: $n = \frac{\sin i}{\sin r}$ [1]

(b) [3 marks]
- Substitution: $n = \frac{\sin(45)}{\sin(28)}$ [1]
- Calculation: $\frac{0.7071}{0.4695}$ or $1.506$ [1]
- Final answer: $1.51$ (must be to 2 d.p.) [1]

(c) [3 marks]
- Correct equation: $\sin c = \frac{1}{n}$ [1]
- Correct substitution: $\sin c = \frac{1}{1.51}$ [1]
- Final answer: $41.5^\circ$ (accept $41.4^\circ$ to $41.8^\circ$ depending on rounding of $n$) [1]

(d) [3 marks]
- Statement that the angle of incidence is greater than the critical angle ($50^\circ > 41.5^\circ$) [1]
- Description that total internal reflection occurs [1]
- No light refracts/escapes into the air / all light reflects inside at $50^\circ$ [1]

(a) In a beta-minus decay, a neutron turns into a proton, emitting an electron (the beta particle). The mass number does not change, so $A = 0$. The proton number increases by 1, so the charge of the beta particle is represented by $Z = -1$.

(b) Determine the number of half-lives that have elapsed:
- Start: $4.0\ \mu\text{g}$
- 1 half-life: $2.0\ \mu\text{g}$
- 2 half-lives: $1.0\ \mu\text{g}$
- 3 half-lives: $0.50\ \mu\text{g}$
This is a total of 3 half-lives.

Calculate the total age:
$$\text{Age} = 3 \times 5730\text{ years} = 17190\text{ years}$$

(c) Radioactive decay is described as 'random' because we cannot predict exactly when a specific nucleus will decay, and each nucleus has an equal, constant probability of decaying per unit time, unaffected by external physical or chemical conditions.

(d) Two safety precautions when handling radioactive sources:
1. Use long-handled tongs to maximize distance from the source.
2. Keep the source pointed away from the body and other people.
3. Store sources in lead-lined containers when not in use.
4. Minimize the exposure time near the source.

(a) [2 marks]
- $A = 0$ [1]
- $Z = -1$ [1]

(b) [4 marks]
- Halving sequence shown or implied: $4.0 \rightarrow 2.0 \rightarrow 1.0 \rightarrow 0.50$ [1]
- Determination that 3 half-lives have elapsed [1]
- Multiplication of half-lives by half-life value: $3 \times 5730$ [1]
- Correct calculation: $17190\text{ years}$ [1]

(c) [2 marks]
- We cannot predict when a particular nucleus will decay [1]
- The decay is unaffected by external factors (such as temperature, pressure, chemical reactions) [1]

(d) [2 marks]
- Any two valid safety precautions: e.g. Use tongs / do not touch with bare hands [1], store in lead-lined containers [1], point source away from body/people [1], limit exposure time [1]

(a) Use the acceleration formula:
$$a = \frac{v - u}{t}$$
$$a = \frac{15 - 0}{6.0} = 2.5\text{ m/s}^2$$
The unit is $\text{m/s}^2$.

(b) The total distance is the area under the velocity-time graph. We can divide the motion into three segments:
- **Segment 1 (0 to 6 s - acceleration):**
$$\text{Distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6.0 \times 15 = 45\text{ m}$$
- **Segment 2 (6 to 18 s - constant velocity):**
$$\text{Distance}_2 = \text{base} \times \text{height} = 12 \times 15 = 180\text{ m}$$
- **Segment 3 (18 to 22 s - deceleration):**
$$\text{Distance}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \times 15 = 30\text{ m}$$

$$\text{Total Distance} = 45 + 180 + 30 = 255\text{ m}$$
(Alternatively, using the area of a trapezium: $\frac{1}{2} \times (a + b) \times h = \frac{1}{2} \times (12 + 22) \times 15 = 17 \times 15 = 255\text{ m}$).

(c) Calculate the deceleration during the last 4.0 s:
$$\text{Deceleration} = \frac{\text{change in velocity}}{\text{time}} = \frac{15 - 0}{4.0} = 3.75\text{ m/s}^2$$
Comparing this to the acceleration in the first 6.0 s ($2.5\text{ m/s}^2$):
The magnitude of the deceleration ($3.75\text{ m/s}^2$) is greater than the acceleration ($2.5\text{ m/s}^2$).

(a) [3 marks]
- Correct formula: $a = \frac{v - u}{t}$ [1]
- Correct calculation: $2.5$ [1]
- Correct unit: $\text{m/s}^2$ (or $\text{m s}^{-2}$) [1]

(b) [5 marks]
- Recognising that total distance is the area under the velocity-time graph [1]
- Correct calculation of the accelerating distance ($45\text{ m}$) [1]
- Correct calculation of the constant velocity distance ($180\text{ m}$) [1]
- Correct calculation of the decelerating distance ($30\text{ m}$) [1]
- Correct sum of distances: $255\text{ m}$ [1]
(Allow full marks if the trapezium method is shown correctly: $\frac{1}{2} \times (12 + 22) \times 15 = 255$)

(c) [2 marks]
- Calculation of deceleration magnitude as $3.75\text{ m/s}^2$ [1]
- Comparison stating that the deceleration is greater than the initial acceleration [1]

(a) The equation is:
$$\text{momentum} = \text{mass} \times \text{velocity}$$
$$p = mv$$

(b) Calculate the momentum of each skater before the collision:
- Skater 1: $p_1 = 60\text{ kg} \times 3.0\text{ m/s} = 180\text{ kg m/s}$
- Skater 2: $p_2 = 40\text{ kg} \times 0\text{ m/s} = 0\text{ kg m/s}$
$$\text{Total initial momentum} = 180 + 0 = 180\text{ kg m/s}$$

(c) By the conservation of momentum, total momentum after the collision equals total momentum before the collision:
$$\text{Total initial momentum} = \text{Total final momentum}$$
$$180 = (m_1 + m_2) \times v$$
$$180 = (60 + 40) \times v$$
$$180 = 100 \times v$$
$$v = \frac{180}{100} = 1.8\text{ m/s}$$

(d) Use the relationship between force, change in momentum, and time:
$$F = \frac{\Delta p}{t}$$
Calculate the change in momentum of the 40 kg skater:
$$\Delta p = m_2 v - m_2 u$$
$$\Delta p = (40 \times 1.8) - (40 \times 0) = 72\text{ kg m/s}$$
Now, calculate the average force:
$$F = \frac{72}{0.80} = 90\text{ N}$$
(Alternatively, using acceleration: $a = \frac{1.8 - 0}{0.80} = 2.25\text{ m/s}^2$; then $F = m \times a = 40 \times 2.25 = 90\text{ N}$)

(a) [1 mark]
- Correct equation: $\text{momentum} = \text{mass} \times \text{velocity}$ [1]

(b) [2 marks]
- Correct substitution: $60 \times 3.0$ [1]
- Correct calculation with units: $180\text{ kg m/s}$ (or $\text{N s}$) [1]

(c) [3 marks]
- Statement of conservation of momentum [1]
- Correct substitution: $180 = 100 \times v$ [1]
- Correct calculation: $1.8\text{ m/s}$ [1]

(d) [4 marks]
- Correct formula: $F = \frac{\Delta p}{t}$ or $F = ma$ [1]
- Correct calculation of change in momentum ($72\text{ kg m/s}$) or acceleration ($2.25\text{ m/s}^2$) [1]
- Correct substitution: $F = \frac{72}{0.80}$ or $F = 40 \times 2.25$ [1]
- Final calculation with units: $90\text{ N}$ [1]

(a) As the magnet falls, there is relative motion between the magnetic field of the magnet and the copper tube. The copper tube cuts the magnetic field lines of the magnet, which induces a voltage (e.m.f.) across the tube. Since copper is an electrical conductor and the tube forms a complete path, this induced voltage causes an electric current to flow around the tube. (b) The direction of the force experienced by the falling magnet is upwards. According to Lenz's law, the induced magnetic field opposes the change that created it (the downward motion of the magnet). This upward magnetic force acts against the weight of the magnet, decelerating the magnet or causing it to reach a slow, constant terminal velocity inside the tube. (c) Using the transformer relationship: \( \frac{V_p}{V_s} = \frac{N_p}{N_s} \), we rearrange to find \( V_s = V_p \times \frac{N_s}{N_p} \). Substituting the values: \( V_s = 12 \times \frac{3000}{120} = 12 \times 25 = 300 \text{ V} \).

(a) 1 mark: state that there is relative motion between magnet and copper tube (or magnet cuts magnetic field lines). 1 mark: state that a voltage / e.m.f. is induced. 1 mark: state that a current flows because copper is a conductor / complete path exists. (b) 1 mark: force is upwards. 1 mark: explain that the force opposes the motion of the magnet (Lenz's Law). 1 mark: state that the magnet slows down or falls at a constant speed lower than free-fall. (c) 1 mark: correct substitution into transformer formula \( V_s = 12 \times \frac{3000}{120} \). 1 mark: correct final voltage of 300 V.

(a) A slow-moving (thermal) neutron is absorbed by a Uranium-235 nucleus, making it highly unstable. The nucleus then splits into two smaller, lighter daughter nuclei, releasing two or three fast-moving neutrons and a large amount of energy (mostly as kinetic energy of the products). (b) The moderator (typically water or graphite) slows down the fast-moving neutrons produced in fission so that they are more easily absorbed by other Uranium-235 nuclei to sustain the reaction. Control rods (typically made of boron or cadmium) absorb excess neutrons. By adjusting their depth, the rate of fission can be maintained at a steady, controlled level. (c) Two differences: 1. Fission involves the splitting of a heavy nucleus into lighter nuclei, whereas fusion is the joining of two light nuclei to form a heavier nucleus. 2. Fusion requires extremely high temperatures and pressures to overcome electrostatic repulsion, whereas fission does not.

(a) 1 mark: slow-moving neutron is absorbed by a Uranium-235 nucleus. 1 mark: nucleus splits into two daughter nuclei and releases neutrons. 1 mark: energy is released. (b) 1 mark: moderator slows down neutrons to increase the rate of absorption. 1 mark: control rods absorb excess neutrons. 1 mark: control rods are adjusted (raised/lowered) to regulate the rate of the chain reaction. (c) 1 mark: fission is splitting of heavy nuclei, fusion is joining of light nuclei. 1 mark: fusion requires extreme temperature/pressure (or fusion releases more energy per unit mass than fission).

(a) Total internal reflection occurs only when: 1. The light travels from a medium with a higher refractive index (denser) towards a medium with a lower refractive index (less dense). 2. The angle of incidence in the denser medium is greater than the critical angle for that boundary. (b) The relationship is given by \( \sin(c) = \frac{1}{n} \). Substituting \( n = 1.48 \): \( \sin(c) = \frac{1}{1.48} \approx 0.6757 \). Therefore, \( c = \arcsin(0.6757) \approx 42.5^\circ \). (c) An endoscope contains two main bundles of optical fibres. The first bundle (the light guide) carries light down from an external source into the patient's body to illuminate the internal organ. The light travels through the fibres by total internal reflection. The second bundle (the image guide) collects the light reflected from the internal organ and carries the image back up to the doctor's eyes or a screen. The fibres in this bundle are arranged coherently so that the image is not distorted.

(a) 1 mark: light travels from high to low refractive index (denser to less dense). 1 mark: angle of incidence is greater than the critical angle. (b) 1 mark: correct formula used \( \sin(c) = \frac{1}{n} \). 1 mark: correct calculation of 42.5 degrees (accept range 42 to 43). (c) 1 mark: identifies two distinct bundles of optical fibres. 1 mark: describes the illumination bundle carrying light down. 1 mark: describes the imaging bundle carrying the reflected image back. 1 mark: mentions total internal reflection occurring continuously inside both bundles.

(a) The relationship is \( \text{Force} = \frac{\text{change in momentum}}{\text{time taken}} \) or \( F = \frac{mv - mu}{t} \). (b) According to the conservation of momentum, the total momentum before collision equals the total momentum after collision. Initial momentum = \( (1200 \times 15) + (1800 \times 0) = 18000 \text{ kg m/s} \). After the collision, the combined mass is \( 1200 + 1800 = 3000 \text{ kg} \). Therefore, \( 18000 = 3000 \times v \), which gives \( v = \frac{18000}{3000} = 6.0 \text{ m/s} \). (c) During a crash, the car must change its momentum to zero. A crumple zone is designed to deform and crush during the collision, which increases the time taken (\( t \)) for the vehicle to come to a stop. Since the change in momentum is constant, increasing the time taken decreases the rate of change of momentum. Since force is equal to the rate of change of momentum, the impact force experienced by the driver is significantly reduced, decreasing the risk of injury.

(a) 1 mark: force = change in momentum / time (written in words or symbols). (b) 1 mark: calculates initial momentum as 18000 kg m/s. 1 mark: equates initial momentum to total mass times final velocity \( 18000 = 3000 \times v \). 1 mark: correct final answer of 6.0 m/s with correct unit. (c) 1 mark: crumple zone deforms, increasing the duration of the impact (time). 1 mark: the change in momentum is fixed/same. 1 mark: rate of change of momentum is reduced. 1 mark: average force exerted on the driver is reduced.

(a) To transmit the same electrical power, a higher voltage requires a lower current because \(P = V \times I\). Since power loss in the transmission cables is given by \(P_{\text{loss}} = I^2 \times R\), a lower current significantly reduces the rate at which thermal energy is dissipated in the cables, making the transmission much more efficient.

(b) Using the transformer relationship:
$$\frac{V_p}{V_s} = \frac{N_p}{N_s}$$
Rearranging for \(V_s\):
$$V_s = V_p \times \frac{N_s}{N_p} = 250\text{ V} \times \frac{6000}{150} = 250 \times 40 = 10000\text{ V} = 10\text{ kV}$$

(c) For a 100% efficient transformer, the power output is equal to the power input:
$$P = 150\text{ kW} = 150000\text{ W}$$
Using the power equation:
$$P = V_s \times I_s$$
$$I_s = \frac{P}{V_s} = \frac{150000}{10000} = 15\text{ A}$$

(d) The rate at which thermal energy is wasted is the power loss in the cables:
$$P_{\text{loss}} = I^2 \times R = 15^2 \times 8.0 = 225 \times 8.0 = 1800\text{ W}$$

(a)
- Idea that high voltage results in a lower current for the same power (1)
- Recall/use of formula \(P = I^2 R\) (1)
- Idea that lower current leads to less thermal energy/heat wasted in the transmission cables (1)

(b)
- Selection and rearrangement of transformer equation: \(V_s = V_p \times \frac{N_s}{N_p}\) (1)
- Correct substitution leading to shown value of 10 kV (or 10,000 V) (1)

(c)
- Selection of equation: \(P = V I\) (1)
- Correct substitution: \(150000 / 10000\) (1)
- Correct calculation of current: 15 A (0.5)

(d)
- Selection of formula: \(P = I^2 R\) (1)
- Correct substitution and calculation: \(15^2 \times 8.0 = 1800\text{ W}\) (1)

(a) Acceptable sources of background radiation include cosmic rays, radon gas from rocks/ground, food/drink, medical sources (such as X-rays), or nuclear fallout.

(b)(i) The corrected count rate is calculated by subtracting the background count rate from the total measured count rate:
- Corrected initial count rate = 344 - 24 = 320 counts per minute.
- Corrected final count rate = 64 - 24 = 40 counts per minute.

(b)(ii) Find the number of half-lives that have elapsed as the corrected count rate drops from 320 cpm to 40 cpm:
$$\text{320} \xrightarrow{\text{1st half-life}} \text{160} \xrightarrow{\text{2nd half-life}} \text{80} \xrightarrow{\text{3rd half-life}} \text{40}$$
This is exactly 3 half-lives.
Since 3 half-lives correspond to 15 hours:
$$\text{Half-life} = \frac{15\text{ hours}}{3} = 5\text{ hours}$$

(c) Handling the source with tongs increases the distance between the source and the body, which significantly reduces the intensity of radiation received by tissues. Wearing gloves prevents physical contamination (radioactive material sticking to the skin), which would lead to continuous localized exposure. Beta radiation is highly ionizing and can damage living cells and DNA.

(a)
- State any two valid sources of background radiation (1 mark per source, max 2 marks)

(b)(i)
- Subtracts background from initial total count: \(344 - 24 = 320\text{ cpm}\) (1)
- Subtracts background from final total count: \(64 - 24 = 40\text{ cpm}\) (1)

(b)(ii)
- Process of determining number of half-lives (e.g., showing 3 divisions by 2) (1)
- Equating 3 half-lives to 15 hours (1)
- Final value: 5 hours (1)

(c)
- Tongs increase distance to reduce radiation intensity/exposure (1)
- Gloves prevent transfer of radioactive material to skin / prevent contamination (1)
- Beta radiation is highly ionizing / causes cell damage (0.5)

(a) The formula is:
$$\sin(c) = \frac{1}{n}$$
where \(c\) is the critical angle and \(n\) is the refractive index.

(b)(i) Substitute \(n = 1.58\) into the formula:
$$\sin(c) = \frac{1}{1.58} \approx 0.6329$$
$$c = \arcsin(0.6329) \approx 39.3^\circ$$

(b)(ii) Since the angle of incidence (\(35^\circ\)) is less than the critical angle (\(39.3^\circ\)), the light ray will undergo refraction and exit the glass block into the air, bending away from the normal.
Using Snell's Law:
$$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$$
$$1.58 \times \sin(35^\circ) = 1.00 \times \sin(r)$$
$$\sin(r) = 1.58 \times 0.5736 \approx 0.9063$$
$$r = \arcsin(0.9063) \approx 65^\circ$$
The ray refracts into the air at an angle of refraction of \(65^\circ\).

(b)(iii) Since the angle of incidence (\(45^\circ\)) is greater than the critical angle (\(39.3^\circ\)), the light cannot refract out of the block. It undergoes total internal reflection (TIR) at the flat boundary and remains inside the glass block, reflecting at an angle of reflection of \(45^\circ\).

(a)
- Correct formula: \(\sin(c) = 1/n\) (1)

(b)(i)
- Substitution of values: \(\sin(c) = 1 / 1.58\) (1)
- Calculation of \(\sin(c) \approx 0.6329\) (1)
- Final value of \(c \approx 39.3^\circ\) (accept range 39.0 to 39.5) (1)

(b)(ii)
- Stating that light refracts / exits the block because the angle of incidence is less than the critical angle (1)
- Correct substitution into Snell's Law: \(1.58 \sin(35^\circ) = \sin(r)\) (1)
- Correct calculation of \(\sin(r) \approx 0.9063\) (1)
- Final angle of refraction: \(65^\circ\) (accept range 64 to 66) (0.5)

(b)(iii)
- Stating that the angle of incidence is greater than the critical angle (1)
- Describing that the ray undergoes total internal reflection (TIR) (1)

(a) The principle of conservation of momentum states that the total momentum of a closed system remains constant before and after a collision, provided no external forces act on the system.

(b)(i) Taking motion to the right as positive:
- Momentum of Car A (moving to the right):
$$p_A = m_A \times u_A = 0.45\text{ kg} \times (+2.0\text{ m/s}) = +0.90\text{ kg m/s}$$
- Momentum of Car B (moving to the left):
$$p_B = m_B \times u_B = 0.30\text{ kg} \times (-1.5\text{ m/s}) = -0.45\text{ kg m/s}$$
- Total initial momentum:
$$p_{\text{total}} = p_A + p_B = 0.90 - 0.45 = +0.45\text{ kg m/s}$$

(b)(ii) After the collision, the cars stick together and move as a single combined mass:
$$M = m_A + m_B = 0.45 + 0.30 = 0.75\text{ kg}$$
By the conservation of momentum, the total momentum after the collision must equal the total momentum before the collision:
$$p_{\text{after}} = M \times v = 0.45\text{ kg m/s}$$
$$0.75 \times v = 0.45$$
$$v = \frac{0.45}{0.75} = +0.60\text{ m/s}$$
Since the velocity is positive, the combined cars move to the right at a speed of \(0.60\text{ m/s}\).

(c) The relationship is:
$$\text{force} = \frac{\text{change in momentum}}{\text{time taken}}$$
$$F = \frac{mv - mu}{t}$$

(a)
- Total momentum remains constant / total momentum before = total momentum after (1)
- Provided no external forces act / in a closed system (1)

(b)(i)
- Correct calculation of Car A's momentum: \(0.90\text{ kg m/s}\) (1)
- Correct calculation of Car B's momentum with negative sign: \(-0.45\text{ kg m/s}\) (1)
- Correct sum of total momentum: \(0.45\text{ kg m/s}\) (1)

(b)(ii)
- Combined mass of the cars: \(0.75\text{ kg}\) (1)
- Correct division of momentum by combined mass: \(v = 0.45 / 0.75\) (1)
- Correct magnitude \(0.60\text{ m/s}\) and direction (to the right) (1)

(c)
- Correct statement of relationship: \(\text{force} = \frac{\text{change in momentum}}{\text{time taken}}\) or equivalent symbols (1.5)