2026 Edexcel IAL Physics (YPH11) Practice Paper with Answers

Initial vertical velocity is: \(u_y = u \sin\theta = 20 \sin(30^\circ) = 10\text{ m s}^{-1}\) upwards. Using \(v_y = u_y + a t\), where \(a = -9.81\text{ m s}^{-2}\): \(v_y = 10 + (-9.81 \times 1.5) = 10 - 14.715 = -4.715\text{ m s}^{-1}\). Since the value is negative, the direction is downwards.

1 mark for the correct calculation of magnitude and direction of vertical velocity.

The parallel component of weight is: \(F_g = mg \sin\theta = 1200 \times 9.81 \times 0.10 = 1177.2\text{ N}\). The total force opposing the motion is: \(F_{\text{total}} = F_g + F_{\text{resistive}} = 1177.2 + 400 = 1577.2\text{ N}\). Since the car travels at a constant speed, the engine must supply a forward force equal to \(F_{\text{total}}\). The useful power output is: \(P = F_{\text{total}} v = 1577.2 \times 15 = 23658\text{ W} \approx 23.7\text{ kW}\).

1 mark for the correct calculation of the useful power output.

The extension is given by \(\Delta x = \frac{F L}{A E}\), where \(A = \frac{\pi d^2}{4}\). Thus, \(\Delta x \propto \frac{F L}{d^2}\). For the first wire: \(\Delta x \propto \frac{W L}{d^2}\). For the second wire: \(\Delta x' \propto \frac{(2W) (2L)}{(2d)^2} = \frac{4 W L}{4 d^2} = \frac{W L}{d^2}\). Therefore, the extension of the second wire is also \(\Delta x\).

1 mark for correctly determining that the extension is unchanged.

Taking the initial direction of motion as positive: initial momentum \(p_i = mv\), final momentum \(p_f = m(-\frac{v}{2}) = -0.5mv\). The change in momentum is \(\Delta p = p_f - p_i = -1.5mv\). The magnitude of the average force is given by \(F = \frac{|\Delta p|}{t} = \frac{1.5mv}{t} = \frac{3mv}{2t}\).

1 mark for the correct calculation of the average force magnitude.

The spring constant of a spring is inversely proportional to its length. When cut in half, each half has a spring constant of \(2k\). The elastic strain energy stored when a force \(F\) is applied is \(E = \frac{F^2}{2k}\). For the half-spring with spring constant \(2k\), the energy stored by the same force \(F\) is \(E' = \frac{F^2}{2(2k)} = \frac{F^2}{4k} = 0.5E\).

1 mark for correctly determining the new stored energy in terms of E.

The tension \(T\) in both segments of the cord is equal due to symmetry. The vertical component of each tension force is \(T \cos\theta\), because the angle between each segment and the vertical axis is \(\theta\). For vertical equilibrium, the sum of the vertical components of tension must equal the weight of the picture frame: \(2 T \cos\theta = W\). Therefore, \(T = \frac{W}{2\cos\theta}\).

1 mark for the correct resolution of forces to find tension.

The forces on the ball as it falls are weight \(W\), upthrust \(U\), and viscous drag \(D\). By Stokes' law, the viscous drag is proportional to velocity: \(D = kv\), where \(k = 6\pi\eta r\). Applying Newton's second law: \(W - U - D = ma \implies W - U - kv = ma\). Rearranging gives \(a = \frac{W-U}{m} - \frac{k}{m}v\). Since \(W\), \(U\), \(m\), and \(k\) are constants, this represents a linear equation with a negative gradient. Thus, \(a\) decreases linearly as \(v\) increases.

1 mark for identifying that the acceleration decreases linearly with velocity.

At terminal velocity, the weight of the sphere equals the viscous drag force: \(W = D\). The weight is given by \(W = mg = \rho V g = \rho \left(\frac{4}{3}\pi r^3\right) g\), so \(W \propto r^3\). By Stokes' law, the viscous drag is \(D = 6\pi\eta r v\), so \(D \propto r v\). Equating these gives: \(r v \propto r^3 \implies v \propto r^2\).

1 mark for the correct proportionality relation.

At a constant speed, the net force on the box is zero. Balancing the forces parallel to the slope: \( F - f - mg \sin\theta = 0 \), where \( f \) is the frictional force. This gives \( f = F - mg \sin\theta \). The rate of work done against friction (power) is \( P = f \times v = (F - mg \sin\theta)v \).

1 mark for the correct option B. Reject A, C, D.

The elastic strain energy is given by \( E = \frac{1}{2} F \Delta x \). Since \( \Delta x = \frac{FL}{AY} \) where \( Y \) is the Young modulus and \( A = \pi r^2 \), we have \( E = \frac{F^2 L}{2\pi r^2 Y} \). For wire X: \( E_X = \frac{F^2 L}{2\pi r^2 Y} \). For wire Y: \( E_Y = \frac{F^2 (2L)}{2\pi (2r)^2 Y} = \frac{2 F^2 L}{8\pi r^2 Y} = \frac{F^2 L}{4\pi r^2 Y} \). The ratio \( \frac{E_X}{E_Y} = \frac{1/2}{1/4} = 2 \).

1 mark for the correct option C. Reject A, B, D.

(a) The vertical component of the initial velocity \(u_y\) is:
\(u_y = u \sin(\theta) = 22.0 \times \sin(30.0^\circ) = 11.0\text{ m s}^{-1}\).

(b) Taking upwards as positive:
\(s = u_y t + \frac{1}{2} a_y t^2\)
\(-45.0 = 11.0 t - \frac{1}{2}(9.81)t^2\)
\(4.905 t^2 - 11.0 t - 45.0 = 0\)
Using the quadratic formula:
\(t = \frac{-(-11.0) \pm \sqrt{(-11.0)^2 - 4(4.905)(-45.0)}}{2(4.905)}\)
\(t = \frac{11.0 \pm \sqrt{121.0 + 882.9}}{9.81} = \frac{11.0 \pm \sqrt{1003.9}}{9.81}\)
Since time must be positive:
\(t = \frac{11.0 + 31.68}{9.81} = 4.35\text{ s}\).

(c) The horizontal component of the velocity is constant:
\(v_x = u \cos(30.0^\circ) = 22.0 \times \cos(30.0^\circ) = 19.05\text{ m s}^{-1}\).

The vertical component of velocity just before impact is:
\(v_y^2 = u_y^2 + 2 a_y s = (11.0)^2 + 2(-9.81)(-45.0) = 121 + 882.9 = 1003.9\)
\(v_y = -31.68\text{ m s}^{-1}\).

The magnitude of the final velocity is:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{(19.05)^2 + (-31.68)^2} = \sqrt{362.9 + 1003.6} = \sqrt{1366.5} = 37.0\text{ m s}^{-1}\).

(a) [1 Mark]
- Uses vertical component formula: \(22.0 \times \sin(30.0^\circ) = 11.0\text{ m s}^{-1}\).

(b) [3 Marks]
- Selects \(s = ut + \frac{1}{2}at^2\) with correct signs for displacement and acceleration (1 mark).
- Attempts to solve the quadratic equation (or uses alternative correct multi-step method) (1 mark).
- Correctly calculates \(t = 4.35\text{ s}\) (1 mark).

(c) [3.78 Marks]
- Calculates horizontal velocity \(v_x = 19.1\text{ m s}^{-1}\) (1 mark).
- Calculates vertical velocity at ground \(v_y = -31.7\text{ m s}^{-1}\) (using either \(v = u + at\) or \(v^2 = u^2 + 2as\)) (1 mark).
- Applies Pythagoras' theorem \(v = \sqrt{v_x^2 + v_y^2}\) (1 mark).
- Correctly obtains \(v = 37.0\text{ m s}^{-1}\) [range 36.9 - 37.1] (0.78 mark).

(a) The free-body force diagram must include:
- Weight, \(W\) (or \(mg\)), acting vertically downwards.
- Normal reaction, \(R\) (or \(N\)), acting perpendicular to the slope, away from the surface.
- Tension, \(T\), acting up the slope parallel to the surface.
- Friction, \(F\), acting down the slope parallel to the surface.

(b) Resolving forces parallel to the slope:
Weight component down the slope: \(W_{\parallel} = mg \sin(25.0^\circ) = 85.0 \times 9.81 \times \sin(25.0^\circ) = 352.1\text{ N}\).
Applying Newton's Second Law parallel to the slope:
\(T - F - W_{\parallel} = ma\)
\(650 - 180 - 352.1 = 85.0 \times a\)
\(117.9 = 85.0 \times a\)
\(a = 1.39\text{ m s}^{-2}\).

(c) When the cable breaks, the tension \(T\) becomes zero. According to Newton's First Law, there is now a net resultant force acting down the slope (consisting of the weight component and friction). Under Newton's Second Law, this net force causes the crate to decelerate until its velocity is temporarily zero. Since the weight component down the slope (\(352.1\text{ N}\)) is larger than the static frictional force, the crate will then reverse direction and accelerate down the slope.

(a) [3 Marks]
- Weight drawn vertically downwards (1 mark).
- Normal reaction drawn perpendicular to slope (1 mark).
- Tension drawn up the slope and friction drawn down the slope (1 mark).

(b) [3 Marks]
- Calculates weight component parallel to slope \(W_{\parallel} = 352\text{ N}\) (1 mark).
- Formulates complete equation \(F_{net} = ma\) with correct signs: \(650 - 180 - 352 = 85a\) (1 mark).
- Obtains acceleration \(a = 1.39\text{ m s}^{-2}\) [accept 1.38 - 1.40] (1 mark).

(c) [1.78 Marks]
- Explains that the resultant force is downwards, causing deceleration (Newton's 1st/2nd laws) (1 mark).
- States that the crate stops and subsequently accelerates down the slope because the weight component exceeds maximum friction (0.78 mark).

(a) The total linear momentum of a closed system remains constant in the absence of external forces.

(b) During the upward swing, kinetic energy just after the collision is converted to gravitational potential energy at the maximum height:
\(\frac{1}{2} M_{total} v^2 = M_{total} g h\)
\(v = \sqrt{2 g h}\)
\(v = \sqrt{2 \times 9.81 \times 0.180} = 1.879\text{ m s}^{-1} \approx 1.88\text{ m s}^{-1}\).

(c) Let \(u\) be the initial velocity of the pellet. Applying conservation of momentum during the collision:
\(m_{pellet} u = (m_{pellet} + M_{block}) v\)
\(0.0150 \times u = (0.0150 + 0.450) \times 1.879\)
\(0.0150 \times u = 0.465 \times 1.879\)
\(0.0150 \times u = 0.8737\)
\(u = \frac{0.8737}{0.0150} = 58.25\text{ m s}^{-1} \approx 58.3\text{ m s}^{-1}\).

(a) [1 Mark]
- States that momentum remains constant in a closed system (or when no external forces act) (1 mark).

(b) [3 Marks]
- Equates kinetic energy to gravitational potential energy: \(\frac{1}{2}v^2 = gh\) (1 mark).
- Rearranges to solve for velocity: \(v = \sqrt{2gh}\) (1 mark).
- Obtains \(v = 1.88\text{ m s}^{-1}\) [accept 1.87 - 1.89] (1 mark).

(c) [3.78 Marks]
- Identifies the total mass after collision as \(0.465\text{ kg}\) (1 mark).
- Uses conservation of momentum equation: \(m_{pellet} u = M_{total} v\) (1 mark).
- Substitutes values correctly (1 mark).
- Obtains \(u = 58.3\text{ m s}^{-1}\) [accept 58.1 - 58.4] (0.78 mark).

(a) Tensile strain is defined as the change in length (or extension) per unit original length.

(b) The tension force in the wire is equal to the weight of the mass:
\(F = mg = 15.0 \times 9.81 = 147.15\text{ N}\).

Using the Young modulus formula:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L}\)
\(\Delta L = \frac{F L}{A E}\)
\(\Delta L = \frac{147.15 \times 2.40}{1.20 \times 10^{-6} \times 2.00 \times 10^{11}}\)
\(\Delta L = \frac{353.16}{240000} = 1.472 \times 10^{-3}\text{ m}\) (or \(1.47\text{ mm}\)).

(c) The elastic strain energy \(E_{el}\) stored in the wire is:
\(E_{el} = \frac{1}{2} F \Delta L\)
\(E_{el} = 0.5 \times 147.15 \times 1.4715 \times 10^{-3}\)
\(E_{el} = 0.1083\text{ J} \approx 0.108\text{ J}\).

(a) [1 Mark]
- Defines strain as extension divided by original length (1 mark).

(b) [3.78 Marks]
- Calculates tension: \(F = 15.0 \times 9.81 = 147.15\text{ N}\) (1 mark).
- Rearranges Young modulus equation for extension: \(\Delta L = \frac{FL}{AE}\) (1 mark).
- Correctly substitutes values into the equation (1 mark).
- Calculates \(\Delta L = 1.47 \times 10^{-3}\text{ m}\) [accept 1.47 - 1.48] (0.78 mark).

(c) [3 Marks]
- Uses the formula \(E_{el} = \frac{1}{2} F \Delta L\) (1 mark).
- Substitutes values of force and extension correctly (1 mark).
- Calculates \(E_{el} = 0.108\text{ J}\) [accept 0.108 - 0.109] (1 mark).

(a) The component of the weight parallel to the slope \(W_{\parallel}\) is:
\(W_{\parallel} = mg \sin(\theta) = 1200 \times 9.81 \times \sin(6.0^\circ) = 11772 \times 0.1045 = 1230\text{ N}\), which is approximately \(1200\text{ N}\).

(b) At constant speed, the net force is zero. Thus, the forward force \(F\) exerted by the engine must balance the resistive force plus the weight component acting down the slope:
\(F = F_{resistive} + W_{\parallel} = 450 + 1230 = 1680\text{ N}\).

Using the relation for power:
\(P = F v\)
\(P = 1680 \times 15 = 25200\text{ W}\) (or \(25.2\text{ kW}\)).
*(Note: If using the rounded value of 1200 N from part (a), \(F = 1650\text{ N}\), leading to \(P = 24750\text{ W}\) or \(2.48 \times 10^4\text{ W}\)).*

(c) Chemical energy (stored in the fuel) is converted to gravitational potential energy (due to gaining height) and thermal energy (dissipated as heat to the surroundings due to resistive forces/friction).

(a) [2 Marks]
- Uses the component formula \(W_{\parallel} = mg \sin\theta\) (1 mark).
- Correctly evaluates \(1200 \times 9.81 \times \sin(6.0^\circ) = 1230\text{ N}\) (1 mark).

(b) [3.78 Marks]
- Recognizes that forward force equals the sum of resistance and weight component (1 mark).
- Calculates total force \(F = 1680\text{ N}\) (or \(1650\text{ N}\)) (1 mark).
- Applies the formula \(P = Fv\) (1 mark).
- Obtains a correct power of \(2.5 \times 10^4\text{ W}\) [accept range 2.45 - 2.55] (0.78 mark).

(c) [2 Marks]
- Mentions initial energy is chemical energy (1 mark).
- Identifies correct final energy types: gravitational potential energy and thermal/heat energy (1 mark).

(a) In elastic deformation, the material returns to its original shape and dimensions when the deforming force/stress is removed. In plastic deformation, the material remains permanently deformed and does not return to its original shape when the force is removed.

(b) Under high stresses, planes of atoms in the metallic crystal lattice slide past one another (often facilitated by the movement of dislocations). Once the stress is removed, the atoms do not return to their original positions, but remain in their new positions, resulting in permanent deformation.

(c) A ductile material can undergo significant plastic deformation under tensile stress before fracturing (and can be drawn into a wire), e.g., copper. A brittle material undergoes little or no plastic deformation before fracturing; it cracks or snaps suddenly when its elastic limit is exceeded, e.g., glass.

(a) [2 Marks]
- States that elastic deformation is reversible / returns to original shape (1 mark).
- States that plastic deformation is permanent / irreversible (1 mark).

(b) [3 Marks]
- Mentions planes/layers of atoms sliding past one another (1 mark).
- Refers to the movement of dislocations in the lattice (1 mark).
- Explains that atoms do not return to their original positions once the force is removed (1 mark).

(c) [2.78 Marks]
- Defines ductile (large plastic region/drawn into wires) and gives example (e.g. copper, gold) (1 mark).
- Defines brittle (breaks with little/no plastic deformation) and gives example (e.g. glass, concrete) (1 mark).
- Clear and precise comparative language used throughout (0.78 mark).

(a) Centre of gravity is the point at which the entire weight of an object can be considered to act.

(b) Since the shelf is uniform, its weight of \(35.0\text{ N}\) acts at its midpoint, \(0.60\text{ m}\) from the hinge.
Taking moments about the hinge \(P\) for rotational equilibrium:
\(\sum \text{Clockwise moments} = \sum \text{Counter-clockwise moments}\)
\(W \times \frac{L}{2} = T \sin(\theta) \times L\)
\(35.0 \times 0.60 = T \sin(40.0^\circ) \times 1.20\)
\(21.0 = T \times 0.6428 \times 1.20\)
\(21.0 = T \times 0.7714\)
\(T = \frac{21.0}{0.7714} = 27.22\text{ N} \approx 27.2\text{ N}\).

(c) For vertical translational equilibrium:
\(\sum F_y = 0\)
\(V_{hinge} + T \sin(40.0^\circ) - W = 0\)
\(V_{hinge} = 35.0 - 27.22 \sin(40.0^\circ)\)
\(V_{hinge} = 35.0 - 17.5 = 17.5\text{ N}\) (acting upwards).

(a) [1.78 Marks]
- Mentions that it is the point where the weight of the body (1 mark)
- is considered/appears to act (0.78 mark).

(b) [4 Marks]
- Identifies that the weight acts at \(0.60\text{ m}\) (1 mark).
- Correctly sets up the principle of moments: \(35.0 \times 0.60 = T \sin(40^\circ) \times 1.20\) (2 marks).
- Obtains a correct value of \(T = 27.2\text{ N}\) [accept 27.0 - 27.5] (1 mark).

(c) [2 Marks]
- Applies vertical equilibrium equation: \(V + T \sin(\theta) = W\) (1 mark).
- Calculates the vertical component of the hinge force as \(17.5\text{ N}\) (1 mark).

(a) The three forces acting on the falling sphere are:
1. Weight (acting downwards)
2. Upthrust (acting upwards)
3. Viscous drag (acting upwards)

(b) Upthrust \(U = V \rho_l g\)

(c) Volume of the sphere:
\(V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (2.50 \times 10^{-3})^3 = 6.545 \times 10^{-8}\text{ m}^3 \approx 6.5 \times 10^{-8}\text{ m}^3\).

At terminal velocity, the upward forces equal the downward forces:
\(W = U + F_D\)
\(F_D = W - U = V \rho_s g - V \rho_l g = V g (\rho_s - \rho_l)\)
\(F_D = 6.545 \times 10^{-8} \times 9.81 \times (7800 - 1260)\)
\(F_D = 6.421 \times 10^{-7} \times 6540 = 4.199 \times 10^{-3}\text{ N}\).

According to Stokes' Law:
\(F_D = 6 \pi \eta r v\)
\(4.199 \times 10^{-3} = 6 \pi \eta (2.50 \times 10^{-3})(0.180)\)
\(4.199 \times 10^{-3} = \eta \times 8.482 \times 10^{-3}\)
\[\eta = \frac{4.199 \times 10^{-3}}{8.482 \times 10^{-3}} = 0.495\text{ Pa s}\.\]

(a) [2 Marks]
- Identifies weight, upthrust, and drag/viscous force (2 marks for all three, 1 mark for any two).

(b) [1 Mark]
- States \(U = V \rho_l g\) (1 mark).
- Note: Accept symbols if defined.

(c) [4.78 Marks]
- Shows volume calculation: \(V = \frac{4}{3} \pi (2.50 \times 10^{-3})^3 = 6.5 \times 10^{-8}\text{ m}^3\) (1 mark).
- Expresses force balance: \(W = U + F_D\) or \(F_D = V g (\rho_s - \rho_l)\) (1 mark).
- Calculates the net downward force \(W - U = 4.2 \times 10^{-3}\text{ N}\) (1 mark).
- Correctly substitutes values into Stokes' Law \(F_D = 6 \pi \eta r v\) (1 mark).
- Calculates \(\eta = 0.495\text{ Pa s}\) [accept range 0.49 - 0.50] (0.78 mark).

(a)
Using the component of weight parallel to the slope:
\( W_{\parallel} = m g \sin\theta \)
\( W_{\parallel} = 45.0 \text{ kg} \times 9.81 \text{ m s}^{-2} \times \sin(28.0^\circ) \)
\( W_{\parallel} = 441.45 \text{ N} \times 0.46947 \approx 207.2 \text{ N} \)
Which is approximately \( 210 \text{ N} \).

(b)
Since the block moves up the slope at a constant speed, the acceleration is zero and the forces parallel to the ramp are in equilibrium:
\( T = W_{\parallel} + F_{\text{friction}} \)
\( 320 \text{ N} = 207.2 \text{ N} + F_{\text{friction}} \)
\( F_{\text{friction}} = 320 \text{ N} - 207.2 \text{ N} = 112.8 \text{ N} \)
Using 3 significant figures, this is \( 113 \text{ N} \) (or \( 110 \text{ N} \) if using the given value of \( 210 \text{ N} \)).

(c)
First, calculate the useful mechanical power output of the motor:
\( P_{\text{out}} = T \times v \)
\( P_{\text{out}} = 320 \text{ N} \times 1.50 \text{ m s}^{-1} = 480 \text{ W} \)

Then, use the efficiency formula to find the electrical power input:
\( \text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \)
\( 0.750 = \frac{480 \text{ W}}{P_{\text{in}}} \)
\( P_{\text{in}} = \frac{480}{0.750} = 640 \text{ W} \)

**Part (a)**
* **(1)** Use of \( W_{\parallel} = m g \sin\theta \)
* **(1)** Correct calculation showing at least 3 significant figures \( (207 \text{ N}) \) to justify the rounded show-that value of \( 210 \text{ N} \)

**Part (b)**
* **(1)** Realisation that constant speed implies zero net force, i.e., \( T = W_{\parallel} + F_{\text{friction}} \)
* **(1)** Subtraction of weight component from tension
* **(1)** Correct answer of \( 113 \text{ N} \) (accept \( 112.8 \text{ N} \); accept \( 110 \text{ N} \) if \( 210 \text{ N} \) is used)

**Part (c)**
* **(1)** Use of \( P = F v \) with the tension force to calculate useful power output \( (480 \text{ W}) \)
* **(1)** Use of \( \text{efficiency} = \frac{\text{power output}}{\text{power input}} \)
* **(1)** Correct answer of \( 640 \text{ W} \)

The critical angle \(\theta_c\) is given by the relation \(\sin\theta_c = \frac{n_2}{n_1}\) where \(n_1 > n_2\). Rearranging this formula for \(n_2\) gives: \(n_2 = n_1 \sin\theta_c = 1.55 \times \sin(62^\circ) \approx 1.55 \times 0.8829 = 1.37\).

A: 1 mark for the correct answer.

The current in a conductor is given by the formula \(I = nAvq\), where \(n\) is the number density of conduction electrons, \(A\) is the cross-sectional area, \(v\) is the drift velocity, and \(q\) is the elementary charge. Rearranging this gives \(v = \frac{I}{nAq}\). For the second wire, the drift velocity is \(v' = \frac{2I}{n(3A)q} = \frac{2}{3} \left(\frac{I}{nAq}\right) = \frac{2}{3}v\).

A: 1 mark for the correct answer.

The diffraction grating equation is \(d \sin\theta = n \lambda\). For the first case, \(d \sin\theta = 3\lambda\). For the second case with wavelength \(1.5\lambda\) and second-order (\(n = 2\)), the equation is \(d \sin\theta' = 2(1.5\lambda) = 3\lambda\). Since both equations equal \(3\lambda\), it follows that \(d \sin\theta' = d \sin\theta\), meaning \(\theta' = \theta\).

B: 1 mark for the correct answer.

The terminal potential difference of a battery is given by the equation \(V = V_0 - Ir\). This can be written in the straight-line form \(y = mx + c\) as \(V = (-r)I + V_0\). Thus, a graph of \(V\) against \(I\) is a straight line with a negative gradient of magnitude \(r\) and a vertical intercept of \(V_0\).

A: 1 mark for the correct answer.

According to Einstein's photoelectric equation, \(E_k = hf - \Phi\). Since both the frequency \(f\) and work function \(\Phi\) remain constant, the maximum kinetic energy of the emitted photoelectrons remains unchanged at \(E_k\). Doubling the intensity of the light doubles the number of photons incident on the metal per second, which doubles the rate of emission of photoelectrons.

A: 1 mark for the correct answer.

The electrical resistance is given by \(R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}\). For the second resistor, the resistance is \(R' = \rho \frac{2L}{\pi (2r)^2} = \rho \frac{2L}{4\pi r^2} = \frac{1}{2} \left(\rho \frac{L}{\pi r^2}\right) = \frac{1}{2}R\).

C: 1 mark for the correct answer.

For a string of length \(L\) fixed at both ends vibrating in its third harmonic, the length of the string contains three half-wavelengths: \(L = 3\left(\frac{\lambda}{2}\right) = 1.5\lambda\). Therefore, the wavelength is \(\lambda = \frac{2}{3}L = \frac{2}{3} \times 1.2\text{ m} = 0.8\text{ m}\). The wave speed is \(v = f\lambda = 240\text{ Hz} \times 0.8\text{ m} = 192\text{ m s}^{-1}\).

C: 1 mark for the correct answer.

An increase in light intensity on an LDR causes more charge carriers to be released, which decreases its electrical resistance. In a potential divider circuit, as the resistance of the LDR decreases relative to the fixed resistor, it takes a smaller fraction of the total supply voltage, causing the output voltage \(V_{\text{out}}\) across it to decrease.

A: 1 mark for the correct answer.

According to the photoelectric equation, the initial maximum kinetic energy is given by: \(E_k = hf - \Phi\). Rearranging this gives \(hf = E_k + \Phi\). When the frequency of the incident light is doubled, the new maximum kinetic energy \(E_k'\) is: \(E_k' = h(2f) - \Phi = 2hf - \Phi\). Substituting the expression for \(hf\) into this equation: \(E_k' = 2(E_k + \Phi) - \Phi = 2E_k + 2\Phi - \Phi = 2E_k + \Phi\). This matches option C.

1 mark for the correct option C. No partial marks are awarded.

The relationship between current \(I\) and drift velocity \(v\) is given by \(I = nAve\), where \(n\) is the number density of charge carriers, \(A\) is the cross-sectional area, and \(e\) is the elementary charge. Since the wires are connected in series, the current \(I\) through both wires is identical. Since they are made of the same material, the number density \(n\) is also identical. Therefore, \(A_X v_X = A_Y v_Y\), which simplifies to \(\frac{v_X}{v_Y} = \frac{A_Y}{A_X}\). The cross-sectional area is proportional to the square of the diameter, so \(A \propto d^2\). Thus, \(\frac{v_X}{v_Y} = \left(\frac{d_Y}{d_X}\right)^2 = \left(\frac{2d}{d}\right)^2 = 4\). This matches option A.

1 mark for the correct option A. No partial marks are awarded.

For (a): The slit spacing \(d\) is given by \(d = 1 / N\), where \(N\) is the number of lines per metre. \(N = 600\\ \text{ lines mm}^{-1} = 6.00 \times 10^5\\ \text{ lines m}^{-1}\). Therefore, \(d = 1 / (6.00 \times 10^5) = 1.67 \times 10^{-6}\\ \text{ m}\).

For (b): Use the formula \(d \sin\theta = n\lambda\). First, calculate the angle \(\theta\) of the second-order maximum (\(n = 2\)). \(\tan\theta = \frac{1.12\text{ m}}{1.50\text{ m}} = 0.7467\), which gives \(\theta = 36.75^\circ\). Alternatively, \(\sin\theta = \frac{1.12}{\sqrt{1.12^2 + 1.50^2}} = 0.5983\). Then, \(\lambda = \frac{d \sin\theta}{n} = \frac{1.67 \times 10^{-6} \times 0.5983}{2} = 5.00 \times 10^{-7}\\ \text{ m}\).

For (c): A shorter wavelength means \(\lambda\) decreases. Since \(d \sin\theta = n\lambda\), \(\sin\theta\) must also decrease. Consequently, the maxima will be closer together (the pattern will be less spread out).

Part (a):
- Use of \(d = 1/N\) (1 mark)
- Correct conversion to metres to show \(d = 1.67 \times 10^{-6}\text{ m}\). (1 mark)

Part (b):
- Correct calculation of angle \(\theta\) or \(\sin\theta\) (1 mark)
- Use of \(d \sin\theta = n\lambda\) with \(n = 2\) (1 mark)
- Calculation of wavelength \(\lambda = 5.00 \times 10^{-7}\text{ m}\). (1 mark)
- Correct unit and 3 significant figures (1 mark)

Part (c):
- Maxima are closer together / pattern is less spread out (1 mark)
- Because angle \(\theta\) is smaller for a smaller wavelength / \(\sin\theta \propto \lambda\) (1 mark)

For (a): Use a micrometer screw gauge to measure the diameter of the wire. Take measurements at several different positions along the wire and at different orientations (perpendicular directions) at each position, then calculate an average diameter.

For (b): The area of the wire is \(A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\). Since \(R = \frac{\rho L}{A}\), the gradient of the R-L graph is \(\frac{\rho}{A}\). Therefore, resistivity \(\rho = \text{gradient} \times A = 4.25 \times 1.134 \times 10^{-7} = 4.82 \times 10^{-7}\\ \Omega\text{ m}\).

For (c): Keeping the current low prevents the wire from heating up. If the temperature of the wire increases, its resistance increases, which would introduce systematic error as resistivity is temperature-dependent.

Part (a):
- Use a micrometer screw gauge / digital calipers (1 mark)
- Take measurements at multiple positions and orientations (1 mark)
- Calculate an average/mean value (1 mark)

Part (b):
- Calculate cross-sectional area \(A = 1.13 \times 10^{-7}\text{ m}^2\) (1 mark)
- Relate gradient to \(\rho / A\) (1 mark)
- Calculate \(\rho = 4.82 \times 10^{-7}\\ \Omega\text{ m}\) (1 mark)

Part (c):
- Prevent heating of the wire / temperature increase (1 mark)
- As resistance/resistivity varies with temperature (1 mark)

For (a): Polarised light is light in which the oscillations of the electric field vector are restricted to a single plane, which contains the direction of wave propagation.

For (b): Sound waves are longitudinal waves. In longitudinal waves, oscillations are parallel to the direction of wave propagation, so there is no perpendicular component to restrict to a single plane.

For (c): When unpolarised light passes through the first filter, its intensity is halved, so \(I_1 = 0.5 I_0\). When this polarised light passes through the second filter, Malus's Law applies: \(I_2 = I_1 \cos^2\theta\). Here, \(\theta = 30^\circ\). Thus, \(I_2 = 0.5 I_0 \cos^2(30^\circ) = 0.5 I_0 \times (\frac{\sqrt{3}}{2})^2 = 0.5 I_0 \times 0.75 = 0.375 I_0\).

Part (a):
- Oscillations are restricted to a single plane (1 mark)
- Plane includes the direction of energy transfer / propagation (1 mark)

Part (b):
- Sound waves are longitudinal (1 mark)
- Oscillations are already only parallel to the direction of wave travel / no transverse component to restrict (1 mark)

Part (c):
- Recall that intensity after first filter is \(0.5 I_0\) (1 mark)
- State Malus's law \(I = I_{\max} \cos^2\theta\) (1 mark)
- Substitute \(\theta = 30^\circ\) and \(I_{\max} = 0.5 I_0\) (1 mark)
- Show calculation leading to \(0.375 I_0\) (1 mark)

For (a): The circuit diagram must show a cell (with a dashed box indicating internal resistance, optional but good), connected in series with a variable resistor and an ammeter, with a voltmeter connected in parallel across the cell.

For (b): The equation for terminal potential difference is \(V = \mathcal{E} - Ir\), which can be rearranged to \(V = -rI + \mathcal{E}\). This is in the form \(y = mx + c\). A graph of \(V\) on the y-axis against \(I\) on the x-axis will yield a straight line where: the y-intercept is equal to the emf \(\mathcal{E}\), and the gradient is equal to \(-r\) (the negative of the internal resistance).

For (c): Using \(\mathcal{E} = I(R + r)\):
\(\mathcal{E} = 0.16(8.5 + r)\)
\(\mathcal{E} = 0.10(14.0 + r)\)
Equating the two expressions for \(\mathcal{E}\):
\(0.16(8.5 + r) = 0.10(14.0 + r)\)
\(1.36 + 0.16r = 1.40 + 0.10r\)
\(0.06r = 0.04\)
\(r = 0.67\\ \Omega\).

Part (a):
- Variable resistor, ammeter, and cell in series (1 mark)
- Voltmeter connected in parallel across the cell (1 mark)

Part (b):
- Use \(V = \mathcal{E} - Ir\) (1 mark)
- State that y-intercept represents the emf \(\mathcal{E}\) (1 mark)
- State that the gradient represents \(-r\) (internal resistance is the negative of the gradient) (1 mark)

Part (c):
- Set up simultaneous equations: \(0.16(8.5 + r) = 0.10(14.0 + r)\) (1 mark)
- Correct expansion and rearrangement to \(0.06r = 0.04\) (1 mark)
- Calculate \(r = 0.67\\ \Omega\) (1 mark)

For (a): The work function is the minimum energy required to release an electron from the surface of a metal.

For (b): The energy of an incident photon is \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{2.40 \times 10^{-7}\text{ m}} = 8.2875 \times 10^{-19}\text{ J}\).
The work function in joules is \(\Phi = 2.36\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.776 \times 10^{-19}\text{ J}\).
Using Einstein's photoelectric equation: \(E_{k,\max} = E - \Phi = 8.2875 \times 10^{-19}\text{ J} - 3.776 \times 10^{-19}\text{ J} = 4.5115 \times 10^{-19}\text{ J} \approx 4.51 \times 10^{-19}\text{ J}\).

For (c): There is no change to the maximum kinetic energy. The intensity only determines the number of photons incident per second, not the energy of individual photons. Since each electron absorbs only one photon, the maximum kinetic energy of the emitted photoelectrons remains unchanged.

Part (a):
- Minimum energy (1 mark)
- Required to liberate/remove an electron from the surface of the metal (1 mark)

Part (b):
- Calculate photon energy \(E = 8.29 \times 10^{-19}\text{ J}\) (1 mark)
- Convert work function to joules: \(3.78 \times 10^{-19}\text{ J}\) (1 mark)
- Use of \(E_{k,\max} = hf - \Phi\) (1 mark)
- Correct value of \(4.51 \times 10^{-19}\text{ J}\) (1 mark)

Part (c):
- No change in maximum kinetic energy (1 mark)
- Because intensity increases number of photons but photon energy is unchanged / one-to-one interaction (1 mark)

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

For (b): In daylight, \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}} = 9.0\text{ V} \times \frac{350\\ \Omega}{1200\\ \Omega + 350\\ \Omega} = 9.0 \times \frac{350}{1550} = 2.03\text{ V}\).

For (c): At night, \(V_{\text{out}} = 7.2\text{ V}\).
\(7.2 = 9.0 \times \frac{R_{\text{LDR}}}{1200 + R_{\text{LDR}}}\)
\(\frac{7.2}{9.0} = 0.8 = \frac{R_{\text{LDR}}}{1200 + R_{\text{LDR}}}\)
\(0.8(1200 + R_{\text{LDR}}) = R_{\text{LDR}}\)
\(960 + 0.8 R_{\text{LDR}} = R_{\text{LDR}}\)
\(0.2 R_{\text{LDR}} = 960\)
\(R_{\text{LDR}} = 4800\\ \Omega\).

Part (a):
- LDR resistance decreases as intensity increases (1 mark)

Part (b):
- Use of potential divider equation (1 mark)
- Total resistance = 1550 \(\Omega\) (1 mark)
- Calculate \(V_{\text{out}} = 2.03\text{ V}\) (1 mark)

Part (c):
- Substitute values: \(7.2 = 9.0 \times R_{\text{LDR}} / (1200 + R_{\text{LDR}})\) (1 mark)
- Rearrange to show \(0.8 = R_{\text{LDR}} / (1200 + R_{\text{LDR}})\) (1 mark)
- Solve for \(R_{\text{LDR}}\) to get \(0.2 R_{\text{LDR}} = 960\) (1 mark)
- Correct resistance of \(4800\\ \Omega\) (1 mark)

For (a): Waves travel from the vibration generator along the string and reflect at the fixed end. The incident and reflected waves, which have the same frequency, wavelength, and amplitude, travel in opposite directions and superpose/interfere. Where they interfere constructively, antinodes (maximum amplitude) are formed; where they interfere destructively, nodes (zero amplitude) are formed.

For (b): In the third harmonic, there are three half-wavelength loops along the length of the string \(L = 1.20\text{ m}\). Therefore, \(L = 3 \times \frac{\lambda}{2} = 1.50\lambda\). This gives \(\lambda = \frac{1.20\text{ m}}{1.50} = 0.80\text{ m}\).
Use the wave equation: \(v = f\lambda = 120\text{ Hz} \times 0.80\text{ m} = 96.0\text{ m s}^{-1}\).

For (c): The frequency of the nth harmonic is \(f_n = n \times f_1\). For the third harmonic, \(f_3 = 3 f_1\). So, the first harmonic frequency is \(f_1 = \frac{f_3}{3} = \frac{120\text{ Hz}}{3} = 40.0\text{ Hz}\).

Part (a):
- Wave reflects at the boundary/fixed end (1 mark)
- Two waves of same frequency/wavelength traveling in opposite directions superpose/interfere (1 mark)
- Nodes are formed at points of destructive interference and antinodes at points of constructive interference (1 mark)

Part (b):
- Use \(L = 1.5\lambda\) to find \(\lambda = 0.80\text{ m}\) (1 mark)
- Use of \(v = f\lambda\) (1 mark)
- Calculate speed \(v = 96.0\text{ m s}^{-1}\) (1 mark)

Part (c):
- Realise \(f_1 = f_3 / 3\) (1 mark)
- Calculate \(f_1 = 40.0\text{ Hz}\) (1 mark)

For (a): Drift velocity is the average displacement of charge carriers per unit time along the wire when an electric field is applied (causing a current).

For (b): Use the transport equation: \(I = nAvq\), where \(q = e = 1.60 \times 10^{-19}\text{ C}\).
Rearranging for drift velocity \(v\):
\(v = \frac{I}{nAe} = \frac{4.50}{8.50 \times 10^{28} \times 1.50 \times 10^{-6} \times 1.60 \times 10^{-19}}\)
\(v = \frac{4.50}{20400} = 2.21 \times 10^{-4}\text{ m s}^{-1}\).

For (c): Doubling the diameter \(d\) increases the cross-sectional area \(A\) by a factor of 4, since \(A \propto d^2\). Since the current \(I\), charge carrier density \(n\), and electron charge \(e\) remain constant, from \(I = nAve\), the drift velocity \(v\) must be inversely proportional to \(A\). Thus, the drift velocity decreases by a factor of 4 (becomes 1/4 of the original value, i.e., \(5.51 \times 10^{-5}\text{ m s}^{-1}\).

Part (a):
- Average velocity of charge carriers along the conductor (1 mark)

Part (b):
- Use of \(I = nAvq\) (1 mark)
- Correct substitution of \(e = 1.60 \times 10^{-19}\text{ C}\) (1 mark)
- Calculation of denominator as \(2.04 \times 10^4\text{ C m}^{-1}\) (1 mark)
- Calculate \(v = 2.21 \times 10^{-4}\text{ m s}^{-1}\) (1 mark)

Part (c):
- Area increases by a factor of 4 (1 mark)
- Since \(I\) and \(n\) are constant, \(v\) is inversely proportional to Area (1 mark)
- Drift velocity is quartered / decreases by a factor of 4 (1 mark)

(a) As the temperature of a semiconductor (such as an NTC thermistor) increases, thermal energy is supplied to the lattice. This energy liberates more charge carriers (electrons) by promoting them from the valence band to the conduction band. Consequently, the number density of charge carriers \(n\) increases significantly. While the increased temperature also causes more frequent collisions between charge carriers and lattice ions, the large increase in \(n\) dominates, resulting in an overall decrease in resistance. (b) The circuit forms a potential divider. The total resistance of the series circuit is \(R_{\text{total}} = R_{\text{thermistor}} + R_{\text{fixed}} = 3.8\text{ k}\Omega + 1.5\text{ k}\Omega = 5.3\text{ k}\Omega\). Using the potential divider formula, the potential difference across the fixed resistor is \(V_{\text{fixed}} = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{total}}} = 9.0\text{ V} \times \frac{1.5\text{ k}\Omega}{5.3\text{ k}\Omega} = 2.55\text{ V}\). (c) As temperature increases, the resistance of the thermistor decreases. Since the fixed resistor now represents a larger fraction of the total circuit resistance, the potential difference across it must increase (or, the decrease in total resistance causes circuit current to increase, and since \(V = IR\), the potential difference across the fixed resistor increases).

(a) MP1: Increased temperature provides thermal energy which liberates charge carriers / free electrons. MP2: The number density of charge carriers \(n\) increases. MP3: The effect of the increase in \(n\) outweighs the effect of increased collisions with lattice ions. (b) MP1: Calculates total resistance of the circuit: \(R_{\text{total}} = 5.3\text{ k}\Omega\) (or \(5300\text{ }\Omega\)). MP2: Uses a valid potential divider ratio or calculates circuit current: \(I = 9.0\text{ V} / 5300\text{ }\Omega = 1.70\times 10^{-3}\text{ A}\). MP3: Final value of potential difference in range \(2.5\text{ V}\) to \(2.6\text{ V}\). (c) MP1: States that resistance of the thermistor decreases. MP2: Explains that this leads to an increase in circuit current (or the fixed resistor takes a larger share of the total voltage), so the potential difference across the fixed resistor increases.

Detailed worked solution:

(a) Measure the diameter of the wire using a micrometer screw gauge. Take measurements at several different positions along the wire and at perpendicular orientations at each position to check for non-circularity. Calculate the mean diameter \(d\) and use the formula \(A = \frac{\pi d^2}{4}\) to find the cross-sectional area.

(b) Attach a reference marker (such as a piece of tape or a flag) to the wire near the scale. Read its position against a vertically mounted meter rule. A vernier scale or travel microscope can be used to measure the small extensions with high resolution. To ensure the elastic limit is not exceeded, the student should plot a force-extension graph as they add loads and stop immediately if the relationship ceases to be linear, or remove the loads periodically to verify that the wire returns to its original length.

(c) Use a meter rule to measure the original length \(L\) of the wire. The gradient of the linear section of the load-extension graph corresponds to \(\frac{F}{\Delta x}\). Since the Young modulus is defined as \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta x}\), it can be calculated as \(E = \text{gradient} \times \frac{L}{A}\).

(a) [4 marks total]:
- Use micrometer screw gauge [1 mark]
- Measure diameter at multiple positions and perpendicular orientations [1 mark]
- Check and correct for zero error [1 mark]
- Use formula \(A = \frac{\pi d^2}{4}\) or equivalent [1 mark]

(b) [4 marks total]:
- Use reference marker on wire and vertical meter rule / vernier scale / travel microscope [1 mark]
- Measure extension as difference in position under load [1 mark]
- Plot force vs extension during experiment and stop if line curves [1 mark]
- Alternatively, remove load to check for zero permanent set/deformation [1 mark]

(c) [4 marks total]:
- Measure original length \(L\) using meter rule [1 mark]
- Identify gradient of linear region as \(\frac{F}{\Delta x}\) [1 mark]
- Recall formula \(E = \frac{F L}{A \Delta x}\) [1 mark]
- Calculate \(E = \text{gradient} \times \frac{L}{A}\) [1 mark]

Detailed worked solution:

(a) Draw a circuit showing a DC power supply in series with an ammeter, a variable resistor, and the test wire. Connect a voltmeter in parallel across a portion of the test wire using movable contacts (such as crocodile clips or flying leads).

(b) Use a micrometer screw gauge to measure the diameter of the wire. Take measurements at various points along the length and at perpendicular angles at each point. Find the average diameter \(d\) and calculate the cross-sectional area using \(A = \frac{\pi d^2}{4}\).

(c) Measure the potential difference \(V\) and current \(I\) for at least five different lengths \(L\) of the wire. For each length, calculate resistance \(R = \frac{V}{I}\). Plot a graph of \(R\) (y-axis) against \(L\) (x-axis). The relationship is \(R = \frac{\rho L}{A}\), so the gradient of this linear graph is \(\frac{\rho}{A}\) . The resistivity is then calculated as \(\rho = \text{gradient} \times A\).

(d) A major systematic error is contact resistance between the voltmeter leads/clips and the wire, which adds a constant resistance. This is minimized by using the gradient of the graph rather than direct calculation of individual points, as the contact resistance only affects the y-intercept. Alternatively, current can heat the wire and increase resistance; this is minimized by using a variable resistor to keep the current low and switching off the power supply between readings.

(a) [3 marks total]:
- Power supply and test wire in series with ammeter [1 mark]
- Voltmeter connected in parallel across the length of the test wire [1 mark]
- Switch or variable resistor included in the circuit [1 mark]

(b) [3 marks total]:
- Micrometer screw gauge [1 mark]
- Measure diameter at multiple positions and perpendicular directions to find average [1 mark]
- Area formula \(A = \frac{\pi d^2}{4}\) [1 mark]

(c) [5 marks total]:
- Measure voltage and current for at least five different wire lengths [1 mark]
- Calculate resistance \(R = \frac{V}{I}\) for each length [1 mark]
- Plot a graph of \(R\) against \(L\) [1 mark]
- State that the gradient equals \(\frac{\rho}{A}\) [1 mark]
- Calculate resistivity \(\rho = \text{gradient} \times A\) [1 mark]

(d) [2 marks total]:
- Identify source of error (e.g., contact resistance / heating effect) [1 mark]
- Describe how to minimize it (e.g., use graphical method to eliminate constant contact resistance / keep current low and turn off between readings) [1 mark]

Detailed worked solution:

(a) The schematic diagram should show a heavy cylinder falling vertically through two light gates mounted on a stand. The vertical distance between the two gates should be labelled \(h\), and the light gates should be connected to a computer or data logger.

(b) As the cylinder falls, the attached card blocks the light beam of each gate. The data logger records the duration of the interruption, \(\Delta t_1\) at gate 1 and \(\Delta t_2\) at gate 2. The average velocity at each gate is calculated using \(v_1 = \frac{d}{\Delta t_1}\) and \(v_2 = \frac{d}{\Delta t_2}\). Since \(d\) is very small, these are excellent approximations of the instantaneous velocities.

(c) Applying the equation of motion \(v^2 = u^2 + 2as\), we define \(u = v_1\) (velocity at first gate), \(v = v_2\) (velocity at second gate), \(a = g\) (acceleration), and \(s = h\) (distance fallen). Substituting these gives \(v_2^2 = v_1^2 + 2gh\), which rearranges to \(v_2^2 - v_1^2 = 2gh\). Comparing this to the equation of a straight line through the origin \(y = mx\), where \(y = v_2^2 - v_1^2\) and \(x = h\), the gradient is \(m = 2g\). Therefore, \(g\) can be determined as \(g = \frac{\text{gradient}}{2}\).

(d) Advantages: 1) It eliminates human reaction time, which can introduce significant systematic and random errors in manual stopwatch timing. 2) It provides a much higher level of precision and resolution (typically down to milliseconds). Air resistance acts as a drag force in the upward direction, reducing the net downward acceleration. This results in measured velocities that are lower than they would be in a vacuum, causing the calculated value of \(g\) to be systematically lower than the true value.

(a) [3 marks total]:
- Cylinder shown falling vertically through light gates [1 mark]
- Light gates mounted with vertical separation \(h\) clearly labelled [1 mark]
- Computer/data logger shown connected to both gates [1 mark]

(b) [2 marks total]:
- Data logger measures the time intervals \(\Delta t_1\) and \(\Delta t_2\) during which the light beam is interrupted [1 mark]
- Calculates velocity as \(v = \frac{d}{\Delta t}\) [1 mark]

(c) [3 marks total]:
- Substitute into \(v^2 = u^2 + 2as\) to derive \(v_2^2 - v_1^2 = 2gh\) [1 mark]
- Identify \(v_2^2 - v_1^2\) as y-axis and \(h\) as x-axis [1 mark]
- Show that \(g = \frac{\text{gradient}}{2}\) [1 mark]

(d) [4 marks total]:
- Advantage 1: Eliminates human reaction time [1 mark]
- Advantage 2: Higher resolution/precision in time measurement [1 mark]
- Explain that air resistance reduces the acceleration below the true value of \(g\) [1 mark]
- Conclude that the experimental value of \(g\) will be systematically lower than the actual value [1 mark]

Detailed worked solution:

(a) Place the rectangular glass block on a piece of paper and trace its outline with a sharp pencil. Direct a light ray from a ray box at an angle to the normal of one face. Mark the path of the incident and emergent rays with at least two fine pencil dots each. Remove the glass block, draw straight lines through the dots to represent the rays, and connect the entry and exit points to show the path inside the block. Draw a normal (perpendicular line) at the point of incidence to define the angle of incidence \(i\) and angle of refraction \(r\).

(b) The table layout should have the following column headings:
| Angle of incidence \(i\) / \(^\circ\) | Angle of refraction \(r\) / \(^\circ\) | \(\sin i\) | \(\sin r\) |
According to Snell's Law, \(n = \frac{\sin i}{\sin r}\). Plot a graph of \(\sin i\) on the y-axis against \(\sin r\) on the x-axis. The data points will form a straight line passing through the origin, and the refractive index \(n\) of the glass is determined as the gradient of this line.

(c) To minimize the wide ray issue, insert a narrow single-slit plate in the ray box collimator. To ensure high precision, use a very sharp pencil to draw the lines through the exact center of the ray, and draw the normal using a set square to ensure accurate perpendicularity. Measuring over a wide range of angles (e.g., up to \(70^\circ\)) also minimizes percentage uncertainties in the angles.

(d) If they measured the angle to the face \(\theta_i\) and \(\theta_r\) instead of the normal, they would effectively be calculating the cosines of the real angles, since \(\theta_i = 90^\circ - i\) and \(\theta_r = 90^\circ - r\). Thus, they would be plotting \(\cos i\) against \(\cos r\). This would yield incorrect and variable ratios rather than a constant refractive index, and the value of \(n\) would systematically deviate as the angle changes.

(a) [4 marks total]:
- Trace block outline using a sharp pencil [1 mark]
- Use dots/crosses to mark incident and emergent rays [1 mark]
- Join the entry and exit points after removing the block [1 mark]
- Draw the normal perpendicular to the boundary at the point of incidence [1 mark]

(b) [4 marks total]:
- Correct table headers: \(i\) / \(^\circ\), \(r\) / \(^\circ\), \(\sin i\), \(\sin r\) [1 mark]
- Plot graph of \(\sin i\) against \(\sin r\) [1 mark]
- State Snell's law relation: \(n = \frac{\sin i}{\sin r}\) [1 mark]
- Conclude that \(n\) is the gradient of the graph [1 mark]

(c) [3 marks total]:
- Use a narrow slit plate in the ray box [1 mark]
- Mark the center of the ray with a sharp pencil [1 mark]
- Use a set square to draw the normal / take readings over a broad range of angles to minimize percentage error [1 mark]

(d) [2 marks total]:
- Recognize that they are calculating the ratio of cosines \(\frac{\cos i}{\cos r}\) instead of sines [1 mark]
- Conclude that this results in an incorrect and inconsistent value for the refractive index that varies with angle [1 mark]

At the highest point of the hump-backed bridge, both the weight \(mg\) (acting downwards, towards the center of the circular arc) and the normal contact force \(R\) (acting upwards, away from the center of the circular arc) act on the car.

The net force acting towards the center of the circular path provides the necessary centripetal force:
\(F_c = mg - R\)

Since \(F_c = \frac{mv^2}{r}\):
\(mg - R = \frac{mv^2}{r}\)

Rearranging this equation for \(R\) gives:
\(R = mg - \frac{mv^2}{r}\)

1 mark: Correct option chosen (A).

By conservation of momentum:
\(m u = (m + 2m) v\)
\(v = \frac{u}{3}\)

The initial kinetic energy is:
\(E_{ki} = \frac{1}{2} m u^2\)

The final kinetic energy is:
\(E_{kf} = \frac{1}{2} (3m) v^2 = \frac{3}{2} m \left(\frac{u}{3}\right)^2 = \frac{1}{6} m u^2\)

Therefore, the ratio of final to initial kinetic energy is:
\(\frac{E_{kf}}{E_{ki}} = \frac{\frac{1}{6} m u^2}{\frac{1}{2} m u^2} = \frac{1}{3}\)

1 mark: Correct option chosen (B).

The electric potential \(V\) due to a point charge is given by \(V = \frac{kQ}{r}\).
Let \(x\) be the distance from the charge \(+Q\) where the potential is zero, where \(0 < x < d\). The distance from this point to the charge \(-4Q\) is therefore \(d - x\).

The total electric potential is the scalar sum of the potentials from both charges:
\(V = \frac{kQ}{x} + \frac{k(-4Q)}{d - x} = 0\)

This simplifies to:
\(\frac{1}{x} = \frac{4}{d - x}\)

Cross-multiplying:
\(d - x = 4x\)
\(5x = d\)
\(x = \frac{d}{5}\)

1 mark: Correct option chosen (C).

The potential difference across the discharging capacitor at time \(t\) is given by:
\(V = V_0 e^{-\frac{t}{RC}}\)

At time \(t = RC\), the potential difference is:
\(V = V_0 e^{-1}\)

The energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\).

The initial energy is:
\(E_0 = \frac{1}{2} C V_0^2\)

The energy remaining at \(t = RC\) is:
\(E = \frac{1}{2} C \left(V_0 e^{-1}\right)^2 = \frac{1}{2} C V_0^2 e^{-2} = E_0 e^{-2}\)

The fraction of energy remaining is:
\(\frac{E}{E_0} = e^{-2}\)

1 mark: Correct option chosen (D).

For a charged particle of mass \(m\), charge \(q\), and speed \(v\) moving in a circle of radius \(r\) perpendicular to a magnetic field \(B\), the magnetic force provides the centripetal force:
\(Bqv = \frac{mv^2}{r}\)

Rearranging this gives the momentum \(p = mv\):
\(p = Bqr\)

Since the magnetic field \(B\) and the radius \(r\) are the same for both the proton and the alpha particle, the momentum is directly proportional to the charge:
\(p \propto q\)

Therefore, the ratio of the momentum of the proton to the momentum of the alpha particle is equal to the ratio of their charges:
\(\frac{p_p}{p_\alpha} = \frac{e}{2e} = \frac{1}{2}\)

So the ratio is \(1 : 2\).

1 mark: Correct option chosen (B).

In the decay of a neutron: \(n \rightarrow p + e^{-} + \bar{
u}_e\):
- The total baryon number is conserved: \(1 \rightarrow 1 + 0 + 0\).
- The total electric charge is conserved: \(0 \rightarrow (+1) + (-1) + 0\).
- To conserve the total lepton number, we look at the values before and after. Before the decay, there are no leptons, so the lepton number \(L = 0\). After the decay, the electron has a lepton number of \(L = +1\). To ensure the total lepton number remains zero, a lepton with \(L = -1\) must be emitted. The electron antineutrino has \(L = -1\), conserving the lepton number.

1 mark: Correct option chosen (A).

For a linac to operate correctly, the particles must reach the gap between successive drift tubes at the precise moment when the alternating voltage switches polarity to accelerate them. This means the time \(t\) spent inside each drift tube must be constant and equal to half the period of the alternating voltage: \(t = \frac{T}{2}\).

As the particles are accelerated, their velocity \(v\) increases. Since speed is defined as \(v = \frac{L}{t}\) (where \(L\) is the length of the drift tube), and \(t\) must remain constant, the length of each successive tube must increase in proportion to the speed of the particles: \(L = v \cdot t\).

1 mark: Correct option chosen (C).

According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. is equal to the rate of change of magnetic flux linkage:
\(E = \left| \frac{\Delta \Phi}{\Delta t} \right|\)

Initial flux linkage, when the plane of the coil is perpendicular to the field (angle between normal to coil and field is \(0^\circ\)):
\(\Phi_i = N B A\)

Final flux linkage, when the plane of the coil is parallel to the field (angle between normal to coil and field is \(90^\circ\)):
\(\Phi_f = 0\)

Therefore, the magnitude of the change in flux linkage is:
\(|\Delta \Phi| = |0 - NBA| = NBA\)

The average induced e.m.f. is:
\(E = \frac{NBA}{\Delta t}\)

1 mark: Correct option chosen (D).

The kinetic energy \(E_k\) gained by an ion of charge \(q\) accelerated through potential difference \(V\) is \(E_k = qV\). Since \(E_k = \frac{1}{2}mv^2\), the velocity of the ion is \(v = \sqrt{\frac{2qV}{m}}\). In a uniform magnetic field \(B\), the magnetic force provides the centripetal acceleration: \(Bqv = \frac{mv^2}{r}\), which gives the radius of curvature as \(r = \frac{mv}{Bq}\). Substituting the expression for \(v\) yields: \(r = \frac{m}{Bq}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}}\). Therefore, the radius is proportional to \(\sqrt{\frac{m}{q}}\). For ion \(X^{2+}\), the ratio of mass to charge is \(\frac{2m}{2e} = \frac{m}{e}\). For ion \(Y^+\), the ratio of mass to charge is \(\frac{m}{e}\). Since their mass-to-charge ratios are equal, the ratio of their path radii is \(1 : 1\).

1 mark for the correct option B. (Award 1 mark for identifying that \(r \propto \sqrt{m/q}\) and correctly determining that the ratio is \(1:1\)).

Let us check the conservation of each quantum number. Baryon number: \(\Sigma^+\) has baryon number \(+1\); proton has \(+1\) and pion has \(0\), so baryon number is conserved (\(1 \rightarrow 1 + 0\)). Charge: \(\Sigma^+\) has charge \(+1\); proton has \(+1\) and pion has \(0\), so charge is conserved (\(+1 \rightarrow +1 + 0\)). Lepton number is \(0\) on both sides, so it is conserved. Strangeness: \(\Sigma^+\) has strangeness \(-1\) (due to the strange quark), while the proton and pion have strangeness \(0\). Therefore, strangeness is not conserved (it changes by \(+1\)), which is allowed in weak interactions.

1 mark for the correct option D. (Award 1 mark for identifying that strangeness changes from \(-1\) to \(0\) and is therefore not conserved).

For part (a):
To maintain contact with the track, the centripetal force must be equal to or greater than the gravitational force at the top of the loop:
\( \frac{m v^2}{r} \ge mg \)
which simplifies to \( v_{min} = \sqrt{gr} \).
Calculate the minimum speed:
\( v_{min} = \sqrt{9.81\text{ m s}^{-2} \times 12\text{ m}} = 10.85\text{ m s}^{-1} \).
Since the actual speed is 11 m s\(^{-1}\), which is greater than 10.85 m s\(^{-1}\), the centripetal force required is greater than gravity, meaning the track must exert a downward normal contact force, and thus contact is not lost.

For part (b):
Apply the conservation of mechanical energy between the top and the bottom of the loop:
\( E_{\text{top}} = E_{\text{bottom}} \)
\( \frac{1}{2} m v_t^2 + m g h = \frac{1}{2} m v_b^2 \)
Multiply by \( 2/m \):
\( v_b^2 = v_t^2 + 2 g h \)
\( v_b^2 = (11)^2 + 2(9.81)(24) = 121 + 470.88 = 591.88\text{ m}^2\text{ s}^{-2} \)

At the bottom of the loop, the forces acting on the car are the normal contact force \( R \) acting upwards and gravity \( mg \) acting downwards. The resultant force is the centripetal force:
\( R - mg = \frac{m v_b^2}{r} \)
\( R = m \left( g + \frac{v_b^2}{r} \right) \)
\( R = 450 \left( 9.81 + \frac{591.88}{12} \right) \)
\( R = 450 (9.81 + 49.32) = 450 (59.13) \approx 26610\text{ N} \approx 2.7 \times 10^4\text{ N} \).

(a)
- Identifies the condition for not losing contact: \( v \ge \sqrt{gr} \) or \( R \ge 0 \) (1)
- Calculates minimum speed \( v_{min} = 10.8\text{ m s}^{-1} \) (or 10.9) (1)
- Makes a valid comparison and concludes that the car does not lose contact because \( 11\text{ m s}^{-1} > 10.8\text{ m s}^{-1} \) (1)

(b)
- Uses conservation of energy to write: \( v_b^2 = v_t^2 + 2gh \) (1)
- Calculates \( v_b^2 = 592\text{ m}^2\text{ s}^{-2} \) (or speed \( v_b = 24.3\text{ m s}^{-1} \)) (1)
- Identifies the centripetal force equation at the bottom: \( R - mg = \frac{m v^2}{r} \) (1)
- Substitutes values into the force equation (1)
- Correct final calculation of normal force \( 2.7 \times 10^4\text{ N} \) (accept 2.66 × 10^4 N) (1)

For part (a):
\( E_k = qV \)
\( E_k = (2 \times 1.60 \times 10^{-19}\text{ C}) \times (1.5 \times 10^6\text{ V}) = 4.80 \times 10^{-13}\text{ J} \)
This matches the shown value.

For part (b):
1. Find the horizontal velocity \( v_x \):
\( E_k = \frac{1}{2} m v_x^2 \Rightarrow v_x = \sqrt{\frac{2 E_k}{m}} \)
\( v_x = \sqrt{\frac{2 \times 4.80 \times 10^{-13}\text{ J}}{6.64 \times 10^{-27}\text{ kg}}} \approx 1.202 \times 10^7\text{ m s}^{-1} \)

2. Find the travel time \( t \) through the plates:
\( t = \frac{L}{v_x} = \frac{0.15\text{ m}}{1.202 \times 10^7\text{ m s}^{-1}} \approx 1.248 \times 10^{-8}\text{ s} \)

3. Calculate the vertical electric force \( F_y \) and acceleration \( a_y \):
\( F_y = qE = 2 \times (1.60 \times 10^{-19}\text{ C}) \times (8.0 \times 10^5\text{ V m}^{-1}) = 2.56 \times 10^{-13}\text{ N} \)
\( a_y = \frac{F_y}{m} = \frac{2.56 \times 10^{-13}\text{ N}}{6.64 \times 10^{-27}\text{ kg}} \approx 3.855 \times 10^{13}\text{ m s}^{-2} \)

4. Calculate the vertical deflection \( s_y \):
\( s_y = \frac{1}{2} a_y t^2 \)
\( s_y = 0.5 \times (3.855 \times 10^{13}\text{ m s}^{-2}) \times (1.248 \times 10^{-8}\text{ s})^2 \)
\( s_y = 0.5 \times 3.855 \times 10^{13} \times 1.558 \times 10^{-16} \approx 3.00 \times 10^{-3}\text{ m} = 3.0\text{ mm} \).

(a)
- Uses \( E_k = qV \) with correct charge of \( 2 \times 1.60 \times 10^{-19}\text{ C} \) (1)
- Obtains value \( 4.8 \times 10^{-13}\text{ J} \) (1)

(b)
- Relates kinetic energy to horizontal speed to find \( v_x \approx 1.2 \times 10^7\text{ m s}^{-1} \) (1)
- Uses \( t = L / v_x \) to find transition time (1)
- Uses \( F = qE \) to find the force on the alpha particle (1)
- Uses Newton's second law to find vertical acceleration \( a_y \approx 3.9 \times 10^{13}\text{ m s}^{-2} \) (1)
- Uses \( s = \frac{1}{2} a t^2 \) to determine deflection (1)
- Calculates correct vertical deflection \( 3.0\text{ mm} \) (or \( 3.0 \times 10^{-3}\text{ m} \)) (1)

For part (a):
\( C_0 = \frac{\varepsilon_0 A}{d_0} = \frac{8.85 \times 10^{-12}\text{ F m}^{-1} \times 1.8 \times 10^{-3}\text{ m}^2}{1.2 \times 10^{-3}\text{ m}} = 1.3275 \times 10^{-11}\text{ F} \)
\( Q = C_0 V = 1.3275 \times 10^{-11}\text{ F} \times 9.0\text{ V} = 1.195 \times 10^{-10}\text{ C} \approx 1.2 \times 10^{-10}\text{ C} \).

For part (b)(i):
1. Capacitance is given by \( C = \frac{\varepsilon_0 A}{d} \), so decreasing separation \( d \) increases capacitance \( C \).
2. Since the capacitor is disconnected from the supply, the stored charge \( Q \) remains constant.
3. Stored energy is \( E_c = \frac{1}{2} \frac{Q^2}{C} \). As \( C \) increases with \( Q \) constant, the energy stored decreases (the attractive force between the opposite charges on the plates does work as they move closer).

For part (b)(ii):
Since \( Q \) is constant, and \( Q = C V \):
\( C_0 V_0 = C_1 V_1 \)
\( V_1 = V_0 \frac{C_0}{C_1} = V_0 \frac{d_1}{d_0} \) (since capacitance is inversely proportional to plate distance)
\( V_1 = 9.0\text{ V} \times \frac{0.80\text{ mm}}{1.2\text{ mm}} = 6.0\text{ V} \)
Change in potential difference is:
\( \Delta V = V_0 - V_1 = 9.0\text{ V} - 6.0\text{ V} = 3.0\text{ V} \) (a decrease of 3.0 V).

(a)
- Calculates capacitance \( C_0 \approx 1.33 \times 10^{-11}\text{ F} \) (1)
- Uses \( Q = CV \) to find charge \( 1.2 \times 10^{-10}\text{ C} \) (1)

(b)(i)
- Explains that capacitance increases because plate separation decreases (1)
- States that charge \( Q \) is constant because the capacitor is disconnected (1)
- References \( E_c = \frac{1}{2} \frac{Q^2}{C} \) or explains that energy decreases because work is done by attractive forces (1)

(b)(ii)
- Uses constant charge to relate potential differences: \( V \propto d \) or calculates new capacitance \( C_1 \approx 1.99 \times 10^{-11}\text{ F} \) (1)
- Calculates new voltage \( V_1 = 6.0\text{ V} \) (1)
- States change in potential difference is \( 3.0\text{ V} \) (1)

For part (a):
The K-minus meson has a strangeness of \(-1\) and a charge of \(-1\). Its quark composition is \(\bar{u}s\) (since strange quark \(s\) has charge \(-1/3\) and anti-up quark \(\bar{u}\) has charge \(-2/3\)).
Charge on left hand side: \(-1\) (for \(\text{K}^-\)).
Charge on right hand side: \(-1\) (for \(\pi^-\)) + \(0\) (for \(\pi^0\)) = \(-1\).
Since both sides have a net charge of \(-1\), charge is conserved.

For part (b):
Calculate the rest mass difference (mass defect):
\( \Delta m = m_{K^-} - (m_{\pi^-} + m_{\pi^0}) = 494\text{ MeV/c}^2 - (140\text{ MeV/c}^2 + 135\text{ MeV/c}^2) = 219\text{ MeV/c}^2 \).

This mass difference is converted into the total kinetic energy \( E_k \) of the decay products:
\( E_k = 219\text{ MeV} \)

Convert this energy to Joules:
\( 219\text{ MeV} = 219 \times 10^6\text{ eV} \)
\( E_k = 219 \times 10^6 \times 1.60 \times 10^{-19}\text{ J/eV} = 3.504 \times 10^{-11}\text{ J} \approx 3.5 \times 10^{-11}\text{ J} \).

(a)
- States quark composition is \(\bar{u}s\) (1)
- Identifies charges: \(\text{K}^-\): \(-1\), \(\pi^-\): \(-1\), \(\pi^0\): \(0\) (1)
- Concludes that net charge before is equal to net charge after (-1 = -1), so charge is conserved (1)

(b)
- Calculates mass defect in \(\text{MeV/c}^2\) as \( 219\text{ MeV/c}^2 \) (1)
- States that the kinetic energy released is equivalent to this mass energy difference (1)
- Multiplies by \( 10^6 \) to convert to \(\text{eV}\) (1)
- Multiplies by \( 1.60 \times 10^{-19}\text{ J/eV}\) (1)
- Obtains final value of \( 3.5 \times 10^{-11}\text{ J} \) (accept range \( 3.50 \times 10^{-11}\text{ J} \) to \( 3.51 \times 10^{-11}\text{ J} \)) (1)

For part (a):
According to Faraday's law of electromagnetic induction:
\( \text{e.m.f.} = -N \frac{\Delta \Phi}{\Delta t} = -N A \frac{\Delta B}{\Delta t} \)
where \( N = 400 \), \( A = 2.5 \times 10^{-4}\text{ m}^2 \), \( \Delta B = 0 - 0.18\text{ T} = -0.18\text{ T} \), and \( \Delta t = 0.060\text{ s} \).
Magnitude of average induced e.m.f.:
\( \text{e.m.f.} = 400 \times 2.5 \times 10^{-4}\text{ m}^2 \times \frac{0.18\text{ T}}{0.060\text{ s}} \)
\( \text{e.m.f.} = 0.10 \times 3.0 = 0.30\text{ V} \).

For part (b)(i):
Using Ohm's law:
\( I = \frac{V}{R} = \frac{0.30\text{ V}}{50\text{ \Omega}} = 6.0 \times 10^{-3}\text{ A} = 6.0\text{ mA} \).

For part (b)(ii):
1. Lenz's law states that the direction of the induced current is such that it opposes the change in magnetic flux that produces it.
2. In this situation, the magnetic flux through the coil is decreasing as the external magnetic field is reduced to zero.
3. Therefore, the induced current flows in a direction that creates its own magnetic field pointing in the same direction as the original external magnetic field, in order to oppose the decrease in flux.

(a)
- Recalls Faraday's law \( \text{e.m.f.} = N \frac{\Delta \Phi}{\Delta t} \) (1)
- Substitutes area, turns, flux change, and time correctly (1)
- Obtains \( 0.30\text{ V} \) (1)

(b)(i)
- Uses \( I = V/R \) (1)
- Obtains \( 6.0 \times 10^{-3}\text{ A} \) (or \( 6.0\text{ mA} \)) (1)

(b)(ii)
- States that Lenz's law means the induction opposes the change in flux (1)
- Identifies that the magnetic flux is decreasing (1)
- Explains that the induced current's field must be in the same direction as the original field to oppose this decrease (1)

For part (a):
1. A charged particle moving perpendicular to a magnetic field experiences a magnetic force given by \( F = Bqv \).
2. By Fleming's Left-Hand Rule, this force is always perpendicular to both the velocity vector and the magnetic field lines.
3. Since the force is constantly perpendicular to the velocity, it does no work on the particle (its speed remains constant) and acts as a centripetal force, directing the particle into a circular path.

For part (b):
Equating centripetal force to the magnetic force:
\( Bqv = \frac{m v^2}{r} \Rightarrow r = \frac{m v}{B q} \)
For a proton: \( m = 1.67 \times 10^{-27}\text{ kg} \), \( q = 1.60 \times 10^{-19}\text{ C} \).
\( r = \frac{1.67 \times 10^{-27}\text{ kg} \times 3.2 \times 10^6\text{ m s}^{-1}}{0.45\text{ T} \times 1.60 \times 10^{-19}\text{ C}} \)
\( r = \frac{5.344 \times 10^{-21}}{7.20 \times 10^{-20}} \approx 0.0742\text{ m} = 7.4\text{ cm} \).

For part (c):
Since kinetic energy is \( E_k = \frac{p^2}{2m} \), the momentum is \( p = \sqrt{2m E_k} \).
Therefore, the radius is:
\( r = \frac{p}{Bq} = \frac{\sqrt{2m E_k}}{Bq} \).
Since both the proton and deuteron have the same charge \( q \), are in the same magnetic field \( B \), and have the same kinetic energy \( E_k \), the radius is proportional to \( \sqrt{m} \).
Because the mass of a deuteron is approximately twice the mass of a proton, the radius will increase by a factor of \( \sqrt{2} \approx 1.41 \) times.

(a)
- States that the magnetic force is perpendicular to velocity (1)
- Mentions that Fleming's left-hand rule determines this direction (1)
- Explains that a perpendicular force acts as a centripetal force causing a circular path (with constant speed) (1)

(b)
- Equates magnetic force to centripetal force to get \( r = mv / Bq \) (1)
- Substitutes correct values for mass and charge of a proton (1)
- Obtains \( 0.074\text{ m} \) or \( 7.4\text{ cm} \) (1)

(c)
- States that momentum \( p = \sqrt{2mE_k} \) or radius \( r \propto \sqrt{m} \) (for constant energy and charge) (1)
- Concludes that because the deuteron has twice the mass, the radius increases by a factor of \( \sqrt{2} \) (or is \( 1.41 \) times larger / \( 10.5\text{ cm} \)) (1)

For part (a):
1. Protons are injected into the center of the cyclotron between two hollow, D-shaped metal electrodes called "Dees".
2. A uniform magnetic field acts perpendicular to the plane of the Dees. This magnetic field exerts a centripetal force on the protons, causing them to move in a circular orbit of radius \( r = mv / Bq \) inside the Dees, but does not change their speed.
3. An alternating electric field is established across the gap between the Dees.
4. When protons cross the gap, the electric field does work on them, increasing their speed and kinetic energy.
5. As the speed increases, the radius of their circular path increases, causing them to spiral outwards. The alternating potential difference changes polarity at the same frequency as the circular motion (which is constant since \( T = 2\pi m / Bq \) is independent of speed), ensuring the electric field is always in the direction to accelerate the protons whenever they cross the gap.

For part (b):
The frequency of the alternating voltage must match the cyclotron frequency:
\( f = \frac{1}{T} = \frac{Bq}{2\pi m} \)
Rearranging for \( B \):
\( B = \frac{2\pi m f}{q} \)
For a proton: \( m = 1.67 \times 10^{-27}\text{ kg} \), \( q = 1.60 \times 10^{-19}\text{ C} \).
\( B = \frac{2\pi \times 1.67 \times 10^{-27}\text{ kg} \times 12 \times 10^6\text{ s}^{-1}}{1.60 \times 10^{-19}\text{ C}} \)
\( B = \frac{1.2592 \times 10^{-19}}{1.60 \times 10^{-19}} \approx 0.787\text{ T} \approx 0.79\text{ T} \).

(a)
- States that the magnetic field is perpendicular to the plane of the Dees (1)
- Explains that the magnetic field causes the protons to travel in circular paths (providing centripetal force) without changing speed (1)
- States that the electric field exists in the gap between the Dees and accelerates the protons (1)
- Explains that the potential difference must alternate so that the electric field direction reverses each time the protons cross the gap (1)
- Mentions that as the protons speed up, the radius of their path increases, resulting in a spiral path (while the time per half-turn remains constant) (1)

(b)
- Recalls or derives \( f = \frac{Bq}{2\pi m} \) or equivalent (1)
- Substitutes correct frequency and values for proton mass and charge (1)
- Calculates \( B \approx 0.79\text{ T} \) (accept 0.78 - 0.79 T) (1)

For part (a):
Define the direction to the right as positive.
Initial momentum:
\( p_i = m_A v_{A,i} + m_B v_{B,i} \)
\( p_i = (0.35\text{ kg} \times 1.8\text{ m s}^{-1}) + (0.15\text{ kg} \times (-1.2\text{ m s}^{-1})) = 0.63 - 0.18 = 0.45\text{ kg m s}^{-1} \)

According to the principle of conservation of momentum, total momentum before collision equals total momentum after collision:
\( p_f = p_i \)
\( m_A v_{A,f} + m_B v_{B,f} = 0.45\text{ kg m s}^{-1} \)
\( (0.35\text{ kg} \times v_{A,f}) + (0.15\text{ kg} \times 2.2\text{ m s}^{-1}) = 0.45 \)
\( 0.35 v_{A,f} + 0.33 = 0.45 \)
\( 0.35 v_{A,f} = 0.12 \)
\( v_{A,f} = \frac{0.12}{0.35} \approx +0.343\text{ m s}^{-1} \)
Since the result is positive, the direction is to the right.
Velocity of Glider A is 0.34 m s\(^{-1}\) to the right.

For part (b):
An elastic collision is one in which kinetic energy is conserved.
Calculate initial kinetic energy:
\( E_{k,i} = \frac{1}{2} m_A v_{A,i}^2 + \frac{1}{2} m_B v_{B,i}^2 \)
\( E_{k,i} = 0.5 \times 0.35 \times (1.8)^2 + 0.5 \times 0.15 \times (-1.2)^2 \)
\( E_{k,i} = 0.567\text{ J} + 0.108\text{ J} = 0.675\text{ J} \)

Calculate final kinetic energy:
\( E_{k,f} = \frac{1}{2} m_A v_{A,f}^2 + \frac{1}{2} m_B v_{B,f}^2 \)
\( E_{k,f} = 0.5 \times 0.35 \times (0.343)^2 + 0.5 \times 0.15 \times (2.2)^2 \)
\( E_{k,f} = 0.0206\text{ J} + 0.363\text{ J} \approx 0.384\text{ J} \)

Since the final kinetic energy (0.38 J) is less than the initial kinetic energy (0.68 J), kinetic energy is not conserved. Thus, the collision is inelastic.

(a)
- Applies conservation of momentum equation: \( m_A v_{A,i} + m_B v_{B,i} = m_A v_{A,f} + m_B v_{B,f} \) (1)
- Uses signs correctly to represent opposing initial velocities (e.g. \( -1.2\text{ m s}^{-1} \)) (1)
- Calculates magnitude of velocity as \( 0.34\text{ m s}^{-1} \) (or \( 0.343\text{ m s}^{-1} \)) (1)
- States direction is to the right (1)

(b)
- States that elastic collisions conserve kinetic energy (1)
- Correctly calculates total initial kinetic energy as \( 0.68\text{ J} \) (or \( 0.675\text{ J} \)) (1)
- Correctly calculates total final kinetic energy as \( 0.38\text{ J} \) (or \( 0.384\text{ J} \)) (1)
- Concludes that the collision is inelastic because kinetic energy is lost/not conserved (1)

(a) Centripetal force is the resultant force acting on an object moving in a circle, directed towards the centre of the circular path.

(b) At the top of the circle, the centripetal force is provided by the tension \(T\) and the weight of the sphere \(mg\):
\(T + mg = \frac{mv^2}{L}\)
For the string to remain taut, the tension \(T \ge 0\). The minimum speed occurs when \(T = 0\):
\(mg = \frac{mv_{\min}^2}{L} \implies v_{\min} = \sqrt{g L}\)
\(v_{\min} = \sqrt{9.81 \times 0.80} = 2.80\text{ m s}^{-1}\)

(c) At the lowest point, the tension acts upwards and weight acts downwards:
\(T - mg = \frac{mv^2}{L} \implies T = m \left( g + \frac{v^2}{L} \right)\)
\(T = 0.15 \times \left( 9.81 + \frac{6.5^2}{0.80} \right)\)
\(T = 0.15 \times (9.81 + 52.8125) = 0.15 \times 62.6225 = 9.39\text{ N}\) (or \(9.4\text{ N}\))

(d) At the lowest point, the tension must support the weight of the sphere and provide the centripetal acceleration: \(T = \frac{mv^2}{L} + mg\). At the highest point, gravity acts towards the centre, reducing the tension needed: \(T = \frac{mv^2}{L} - mg\). Additionally, as the sphere moves upwards, kinetic energy is converted to gravitational potential energy, so its speed decreases, reducing the required centripetal force at the top.

(a)
* Resultant force directed towards the centre of the circle (1) [Do not accept 'force that keeps it moving in a circle' without direction].

(b)
* Setting \(T = 0\) and equating centripetal force to weight: \(mg = \frac{mv^2}{L}\) (1)
* Correct calculation showing \(v = 2.80\text{ m s}^{-1}\) (1)

(c)
* Expresses correct equation for bottom of path: \(T - mg = \frac{mv^2}{L}\) (1)
* Correct substitution of values (1)
* Correct final answer of \(9.4\text{ N}\) (accept \(9.39\text{ N}\)) (1)

(d)
* Explains that at the bottom \(T = F_c + mg\) whereas at the top \(T = F_c - mg\) (or equivalent word description of gravity helping/opposing tension) (1)
* Explains that speed is greater at the bottom than the top because gravitational potential energy is converted to kinetic energy, which increases the required centripetal force at the bottom (1)

(a) The electric field is situated in the gap between the two D-shaped electrodes (dees). It accelerates the protons as they cross the gap. The magnetic field is uniform and directed perpendicularly to the plane of the dees. It exerts a magnetic force that acts perpendicular to the velocity of the protons, causing them to move in semicircular paths of increasing radius inside the dees. The high-frequency alternating potential difference ensures that the electric field reverses direction each time the protons reach the gap, continuously accelerating them.

(b) The magnetic force on a proton provides the necessary centripetal force:
\(B q v = \frac{m v^2}{r} \implies v = \frac{B q r}{m}\)
The time \(t\) for a proton to complete one half-orbit inside a dee is:
\(t = \frac{\pi r}{v}\)
Substituting \(v\):
\(t = \frac{\pi r}{\frac{B q r}{m}} = \frac{\pi m}{B q}\)
The period \(T\) of the full cycle is \(2t\):
\(T = \frac{2 \pi m}{B q}\)
The frequency \(f\) is the reciprocal of the period:
\(f = \frac{1}{T} = \frac{B q}{2 \pi m}\)
Since \(B\), \(q\), and \(m\) are constant, the frequency is independent of the radius \(r\).

(c) Using the derived formula:
\(f = \frac{B q}{2 \pi m}\)
\(f = \frac{1.2 \times 1.60 \times 10^{-19}}{2 \pi \times 1.67 \times 10^{-27}}\)
\(f = \frac{1.92 \times 10^{-19}}{1.0493 \times 10^{-26}} = 1.83 \times 10^7\text{ Hz}\) (or \(18.3\text{ MHz}\))

(a)
* Electric field accelerates protons across the gap (1)
* Magnetic field deflects protons into circular/semicircular paths inside the dees (1)
* Alternating potential difference switches polarity periodically to ensure acceleration occurs at every gap crossing (1)

(b)
* Equates magnetic force to centripetal force: \(Bqv = \frac{mv^2}{r}\) (1)
* Relates speed to time period or frequency: \(v = \frac{2\pi r}{T}\) or \(v = 2\pi r f\) (1)
* Rearranges to show \(f = \frac{Bq}{2\pi m}\) and explicitly states that since all terms on the RHS are constant, \(f\) is independent of \(r\) (1)

(c)
* Correct substitution of values into \(f = \frac{Bq}{2\pi m}\) (1)
* Correct calculation of frequency to 2 or 3 significant figures: \(1.8 \times 10^7\text{ Hz}\) or \(1.83 \times 10^7\text{ Hz}\) (accept \(1.8\text{ MHz}\) as a power-of-ten slip only if the math is shown, but reward correct final answer in Hz or MHz) (1)

The root-mean-square speed of an ideal gas is given by \(v_{\text{rms}} = \sqrt{\frac{3 k T}{m}}\), where \(T\) is the absolute temperature in kelvin. This means that \(v_{\text{rms}} \propto \sqrt{T}\), or \(T \propto v_{\text{rms}}^2\). First, convert the initial temperature to kelvin: \(T_1 = 27 + 273 = 300\text{ K}\). Since the r.m.s. speed is doubled, the absolute temperature must increase by a factor of \(2^2 = 4\): \(T_2 = 4 \times 300\text{ K} = 1200\text{ K}\). Convert the final temperature back to degrees Celsius: \(T_2 = 1200 - 273 = 927\text{ }^\circ\text{C}\).

C : 927 °C (1 mark)

From the formula for thermal energy transfer, \(\Delta Q = m c \Delta T\). Dividing by time \(\Delta t\) gives the power input: \(P = m c \frac{\Delta T}{\Delta t}\). Thus, the rate of temperature increase is: \(R = \frac{\Delta T}{\Delta t} = \frac{P}{m c}\). For the second solid, the rate of temperature increase \(R_2\) is: \(R_2 = \frac{3P}{(2m)(c/2)} = \frac{3P}{m c} = 3 R\).

C : 3.0 R (1 mark)

The activity \(A\) of a radioactive sample is given by: \(A = \lambda N\). The number of undecayed nuclei \(N\) at time \(t\) is: \(N = N_0 e^{-\lambda t}\). Substituting \(t = \frac{2}{\lambda}\): \(N = N_0 e^{-\lambda \left(\frac{2}{\lambda}\right)} = N_0 e^{-2} = \frac{N_0}{e^2}\). Therefore, the activity after this time is: \(A = \lambda N = \lambda \frac{N_0}{e^2} = \frac{\lambda N_0}{e^2}\).

A : \(\frac{\lambda N_0}{e^2}\) (1 mark)

The nuclear reaction is: \(^{3}_{2}\text{He} + \text{n} \rightarrow\ ^{4}_{2}\text{He}\). The total binding energy of helium-3 is: \(3 \times 2.57\text{ MeV} = 7.71\text{ MeV}\). The binding energy of a free neutron is 0. The total binding energy of helium-4 is: \(4 \times 7.07\text{ MeV} = 28.28\text{ MeV}\). The energy released is the increase in binding energy: \(\Delta E = 28.28\text{ MeV} - 7.71\text{ MeV} = 20.57\text{ MeV}\).

C : 20.57 MeV (1 mark)

The luminosity of a star is given by Stefan-Boltzmann's law: \(L = 4\pi R^2 \sigma T^4\). Thus, luminosity is proportional to \(R^2 T^4\). We can find the ratio of luminosities: \(\frac{L_Y}{L_X} = \left(\frac{R_Y}{R_X}\right)^2 \left(\frac{T_Y}{T_X}\right)^4 = \left(\frac{0.5R}{R}\right)^2 \left(\frac{2T}{T}\right)^4 = (0.5)^2 \times (2)^4 = 0.25 \times 16 = 4\).

B : 4 (1 mark)

Redshift is defined as: \(z = \frac{v}{c}\), where \(v\) is the recessional velocity of the galaxy and \(c\) is the speed of light (\(3.0 \times 10^5\text{ km s}^{-1}\)). Calculate the recessional velocity: \(v = z c = 0.050 \times 3.0 \times 10^5\text{ km s}^{-1} = 1.5 \times 10^4\text{ km s}^{-1}\). According to Hubble's law: \(v = H_0 d\). Rearranging for distance \(d\): \(d = \frac{v}{H_0} = \frac{1.5 \times 10^4\text{ km s}^{-1}}{68\text{ km s}^{-1}\text{Mpc}^{-1}} \approx 220.6\text{ Mpc} \approx 220\text{ Mpc}\).

C : 220 Mpc (1 mark)

The total energy in simple harmonic motion is given by: \(E_{\text{total}} = E_k + E_p = \frac{1}{2} k A^2\). The potential energy at displacement \(x\) is: \(E_p = \frac{1}{2} k x^2\). We are given that the kinetic energy is three times the potential energy: \(E_k = 3 E_p\). Substituting this into the total energy equation: \(3 E_p + E_p = \frac{1}{2} k A^2 \Rightarrow 4 E_p = \frac{1}{2} k A^2\). Substitute the formula for \(E_p\): \(4 \left(\frac{1}{2} k x^2\right) = \frac{1}{2} k A^2 \Rightarrow 4 x^2 = A^2 \Rightarrow x^2 = \frac{A^2}{4}\). Taking the square root gives: \(x = \pm 0.50 A\).

B : \(x = \pm 0.50 A\) (1 mark)

As damping in a forced oscillating system increases, the peak amplitude of the oscillations at resonance decreases, the resonance peak becomes wider/broader, and the resonant frequency (the frequency at which the maximum amplitude occurs) shifts slightly to a lower frequency.

A : The peak amplitude decreases and the resonant frequency shifts to a slightly lower frequency. (1 mark)

The root-mean-square speed is related to the absolute temperature by \( c_{\text{rms}} = \sqrt{\frac{3kT}{m}} \), which means \( c_{\text{rms}} \propto \sqrt{T} \).

When the temperature is increased to \( 1.44T \):
\( \frac{\text{new } c_{\text{rms}}}{\text{original } c_{\text{rms}}} = \sqrt{1.44} = 1.2 \)

According to the ideal gas equation, \( pV = NkT \). At constant volume \( V \), the pressure is directly proportional to the absolute temperature (\( p \propto T \)).

When the temperature is increased to \( 1.44T \):
\( \frac{\text{new pressure}}{\text{original pressure}} = 1.44 \)

Therefore, the correct ratios are \( 1.2 \) and \( 1.44 \), which corresponds to option A.

Select Option A.

[1 mark] for selecting the correct option.

The total energy \( E_{\text{total}} \) of a simple harmonic oscillator is given by:
\( E_{\text{total}} = \frac{1}{2}kA^2 \)

The potential energy \( E_p \) at displacement \( x \) is:
\( E_p = \frac{1}{2}kx^2 \)

The kinetic energy \( E_k \) is:
\( E_k = E_{\text{total}} - E_p \)

We are given that the kinetic energy is three times the potential energy:
\( E_k = 3E_p \)

Substitute the energy expressions:
\( E_{\text{total}} - E_p = 3E_p \implies E_{\text{total}} = 4E_p \)

Substitute the algebraic expressions for energy:
\( \frac{1}{2}kA^2 = 4 \left( \frac{1}{2}kx^2 \right) \)

Simplify the equation:
\( A^2 = 4x^2 \implies x = \pm \sqrt{\frac{A^2}{4}} = \pm 0.50A \)

Therefore, the displacement is \( 0.50A \), which corresponds to option B.

Select Option B.

[1 mark] for selecting the correct option.

Part (a): At constant pressure, the work done by an ideal gas during expansion is given by \( W = p \Delta V \). Using the ideal gas equation \( p V = n R T \), the change in volume is related to the change in temperature by \( p \Delta V = n R \Delta T \). Therefore, \( W = n R \Delta T \). Rearranging for the change in temperature: \( \Delta T = \frac{W}{n R} = \frac{195}{0.28 \times 8.31} \approx 83.8\text{ K} \). The final temperature is \( T_2 = T_1 + \Delta T = 290 + 83.8 = 373.8\text{ K} \) which is approximately 374 K. Part (b): According to the First Law of Thermodynamics, the change in internal energy is given by \( \Delta U = Q - W \), where \( Q \) is the heat energy supplied and \( W \) is the work done by the gas. Substituting the values: \( \Delta U = 680\text{ J} - 195\text{ J} = 485\text{ J} \).

Part (a): [M1] Recalls or uses \( W = n R \Delta T \). [M1] Correctly substitutes values to find \( \Delta T = 83.8\text{ K} \). [A1] Obtains \( T_2 = 374\text{ K} \) (accept values in range 373 K to 374 K). Part (b): [M1] Recalls and applies the First Law of Thermodynamics: \( \Delta U = Q - W \). [A1.67] Calculates \( \Delta U = 485\text{ J} \).

Part (a): Using the ideal gas equation: \( p V = N k_B T \). Rearranging for the number of molecules \( N \): \( N = \frac{p V}{k_B T} = \frac{1.80 \times 10^5 \times 0.045}{1.38 \times 10^{-23} \times 293} = \frac{8100}{4.0434 \times 10^{-21}} \approx 2.00 \times 10^{24} \text{ atoms} \). Part (b): The average kinetic energy of a gas molecule is \( \frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k_B T \). The root-mean-square speed is \( c_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} \). First, find the mass \( m \) of a single helium atom: \( m = \frac{\text{Molar mass}}{N_A} = \frac{4.00 \times 10^{-3}\text{ kg mol}^{-1}}{6.02 \times 10^{23}\text{ mol}^{-1}} \approx 6.64 \times 10^{-27}\text{ kg} \). Then, \( c_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 293}{6.64 \times 10^{-27}}} = \sqrt{1.827 \times 10^6} \approx 1350\text{ m s}^{-1} \).

Part (a): [M1] Uses \( p V = N k_B T \) or \( p V = n R T \) with \( N = n N_A \). [M1] Substitutes values correctly. [A1.67] Obtains \( N = 2.00 \times 10^{24} \text{ atoms} \) (accept range 1.99 to 2.01). Part (b): [M1] Recalls expression for mean kinetic energy or root-mean-square speed. [M1] Correctly calculates mass of one helium atom as \( 6.64 \times 10^{-27}\text{ kg} \). [A1] Calculates \( c_{\text{rms}} = 1350\text{ m s}^{-1} \) (accept range 1340 to 1360).

Part (a): The decay constant \( \lambda \) is related to half-life by \( \lambda = \frac{\ln 2}{t_{1/2}} \). First, convert the half-life in years to seconds: \( t_{1/2} = 5730 \times 365.25 \times 24 \times 3600 = 1.808 \times 10^{11}\text{ s} \). Thus, \( \lambda = \frac{0.69315}{1.808 \times 10^{11}} \approx 3.83 \times 10^{-12}\text{ s}^{-1} \), which is approximately \( 3.8 \times 10^{-12}\text{ s}^{-1} \). Part (b): Using the radioactive decay equation \( N = N_0 e^{-\lambda t} \), the ratio of remaining to initial atoms is \( \frac{N}{N_0} = \frac{2.40 \times 10^{11}}{1.25 \times 10^{12}} = 0.192 \). We have \( e^{-\lambda_{\text{yr}} t} = 0.192 \), where \( \lambda_{\text{yr}} = \frac{\ln 2}{5730} \approx 1.21 \times 10^{-4}\text{ yr}^{-1} \). Taking natural logs of both sides: \( -\lambda_{\text{yr}} t = \ln(0.192) = -1.6502 \). Therefore, \( t = \frac{1.6502}{1.21 \times 10^{-4}} \approx 13642\text{ years} \) (accept 13600 years to 3 significant figures).

Part (a): [M1] Converts half-life from years to seconds correctly. [M1] Uses \( \lambda = \frac{\ln 2}{t_{1/2}} \). [A1] Shows intermediate calculation leading to \( 3.83 \times 10^{-12}\text{ s}^{-1} \). Part (b): [M1] Recalls radioactive decay formula or equivalent fraction method. [M1] Calculates the ratio \( \frac{N}{N_0} = 0.192 \) and sets up equation. [A1.67] Obtains \( t = 1.36 \times 10^4 \) years (accept range 13500 to 13700 years).

Part (a): The total initial mass of the reactants is the mass of four protons: \( m_i = 4 \times m_p = 4 \times 1.007276\text{ u} = 4.029104\text{ u} \). The total final mass of the products is the mass of one helium-4 nucleus and two positrons (the mass of neutrinos is negligible): \( m_f = m_{\text{He}} + 2 \times m_e = 4.001506\text{ u} + 2 \times 0.000549\text{ u} = 4.002604\text{ u} \). The mass defect \( \Delta m = m_i - m_f = 4.029104 - 4.002604 = 0.0265\text{ u} \). Part (b): Using the mass-energy conversion factor \( 1\text{ u} = 931.5\text{ MeV} \): \( E = \Delta m \times 931.5\text{ MeV} = 0.0265 \times 931.5 \approx 24.68\text{ MeV} \), which rounds to 24.7 MeV.

Part (a): [M1] Calculates total mass of reactants: \( m_i = 4.029104\text{ u} \). [M1] Calculates total mass of products including 2 positrons: \( m_f = 4.002604\text{ u} \). [A1.67] Obtains mass defect \( \Delta m = 0.0265\text{ u} \). Part (b): [M1] Recalls or uses \( E = \Delta m c^2 \) or the conversion \( 1\text{ u} = 931.5\text{ MeV} \). [A2] Obtains \( E = 24.7\text{ MeV} \) (accept range 24.6 to 24.8 MeV).

Part (a): The angular frequency of the mass-spring system is \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{45}{0.35}} \approx 11.34\text{ rad s}^{-1} \). The maximum velocity of a system undergoing SHM is \( v_{\text{max}} = \omega A \), where \( A \) is the amplitude (maximum displacement). Here, \( A = 0.080\text{ m} \). Thus, \( v_{\text{max}} = 11.34 \times 0.080 \approx 0.907\text{ m s}^{-1} \), which rounds to approximately \( 0.9\text{ m s}^{-1} \). Part (b): Since the motion starts from maximum displacement at \( t = 0 \), the displacement equation is \( x = A \cos(\omega t) \). We need to find \( t \) when \( x = 0.040\text{ m} \): \( 0.040 = 0.080 \cos(11.34 t) \implies \cos(11.34 t) = 0.5 \). Therefore, \( 11.34 t = \frac{\pi}{3} \) radians. Solving for \( t \): \( t = \frac{\pi}{3 \times 11.34} \approx 0.0923\text{ s} \).

Part (a): [M1] Calculates angular frequency \( \omega = 11.34\text{ rad s}^{-1} \). [M1] Uses \( v_{\text{max}} = \omega A \). [A1] Obtains value 0.91 m/s and states it is approximately 0.9 m/s. Part (b): [M1] Recalls displacement-time equation \( x = A \cos(\omega t) \) or equivalent. [M1] Sets up equation to get \( 11.34 t = \frac{\pi}{3} \). [A1.67] Obtains \( t = 0.092\text{ s} \) (accept range 0.092 s to 0.093 s).

Part (a): Resonance is the condition in which an oscillating system vibrates with maximum amplitude. This occurs when the frequency of an applied periodic driving force (the driving frequency) matches the natural frequency of the oscillating system, resulting in highly efficient energy transfer. Part (b): When damping increases: 1. The peak amplitude of the oscillation decreases. 2. The resonance peak becomes wider or broader. 3. The resonant frequency (the frequency of the peak amplitude) shifts slightly to a lower frequency.

Part (a): [M1] Refers to driving frequency matching/equalling natural frequency. [M1] Refers to maximum amplitude of oscillations. [A1] Identifies maximum energy transfer as the cause or consequence. Part (b): [M1] States peak amplitude decreases. [M1] States peak becomes wider / broader. [A1.67] States resonant frequency shifts slightly to a lower value.

Part (a): According to Wien's displacement law, \( \lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K} \). Therefore, \( \lambda_{\text{max}} = \frac{2.898 \times 10^{-3}}{3500} \approx 8.28 \times 10^{-7}\text{ m} \) (or 830 nm). Part (b): According to the Stefan-Boltzmann law, \( L = 4 \pi R^2 \sigma T^4 \). The luminosity of Betelgeuse is \( L = 1.2 \times 10^5 \times 3.85 \times 10^{26}\text{ W} = 4.62 \times 10^{31}\text{ W} \). Rearranging for the radius \( R \): \( R = \sqrt{\frac{L}{4 \pi \sigma T^4}} = \sqrt{\frac{4.62 \times 10^{31}}{4 \pi \times 5.67 \times 10^{-8} \times 3500^4}} = \sqrt{\frac{4.62 \times 10^{31}}{1.0694 \times 10^8}} \approx 6.57 \times 10^{11}\text{ m} \) (or 6.6 \times 10^{11} m).

Part (a): [M1] Recalls Wien's displacement law. [M1] Correctly substitutes values. [A1] Calculates \( \lambda_{\text{max}} = 8.3 \times 10^{-7}\text{ m} \) (accept 8.28 to 8.30). Part (b): [M1] Calculates total luminosity \( L = 4.62 \times 10^{31}\text{ W} \). [M1] Recalls Stefan-Boltzmann law and rearranges for R. [A1.67] Calculates \( R = 6.6 \times 10^{11}\text{ m} \) (accept range 6.5 to 6.6).

Part (a): The redshift parameter \( z \) is given by \( z = \frac{\Delta \lambda}{\lambda} = \frac{682.1 - 656.3}{656.3} = \frac{25.8}{656.3} \approx 0.03931 \). For non-relativistic velocities, the Doppler shift formula is \( z = \frac{v}{c} \). Therefore, the recessional velocity is \( v = z c = 0.03931 \times 3.00 \times 10^8\text{ m s}^{-1} \approx 1.18 \times 10^7\text{ m s}^{-1} \). Part (b): From Hubble's Law, \( v = H_0 d \). First, convert velocity to kilometers per second: \( v = 1.18 \times 10^4\text{ km s}^{-1} \). Then, \( d = \frac{v}{H_0} = \frac{1.18 \times 10^4\text{ km s}^{-1}}{71\text{ km s}^{-1}\text{ Mpc}^{-1}} \approx 166\text{ Mpc} \) (or 167 Mpc using unrounded figures).

Part (a): [M1] Calculates wavelength difference \( \Delta \lambda = 25.8\text{ nm} \). [M1] Recalls Doppler shift equation \( z = \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \). [A1.67] Calculates \( v = 1.18 \times 10^7\text{ m s}^{-1} \) (accept range 1.17 to 1.19). Part (b): [M1] Recalls Hubble's law \( v = H_0 d \). [M1] Converts velocity unit to km/s. [A1] Calculates \( d = 166\text{ Mpc} \) (accept range 165 to 168).

\textbf{(a)} According to the kinetic theory, pressure is given by \( p = \frac{1}{3} \rho \langle c^2 \rangle \). Because the volume is constant, the density \( \rho \) is constant. Therefore, for the pressure to double, the mean square speed \( \langle c^2 \rangle \) of the atoms must double. Since the absolute temperature \( T \) is directly proportional to the mean square speed (average molecular kinetic energy), the temperature must increase as well.

\textbf{(b)} For an ideal monatomic gas, the total internal energy is purely kinetic and is given by:
\( E_k = \frac{3}{2} n R T \)

Since the pressure doubles at constant volume, the temperature must also double according to the ideal gas law (\( pV = nRT \)):
\( T_2 = 2 \times 290 \text{ K} = 580 \text{ K} \)
\( \Delta T = T_2 - T_1 = 580 - 290 = 290 \text{ K} \)

The change in internal energy is:
\( \Delta E_k = \frac{3}{2} n R \Delta T \)
\( \Delta E_k = 1.5 \times 0.25 \text{ mol} \times 8.31 \text{ J K}^{-1} \text{ mol}^{-1} \times 290 \text{ K} = 903.7 \text{ J} \approx 904 \text{ J} \)

\textbf{(c)} The r.m.s. speed is given by:
\( c_{\text{rms}} = \sqrt{\frac{3 R T_2}{M}} \)
where \( T_2 = 580 \text{ K} \) and \( M = 4.0 \times 10^{-3} \text{ kg mol}^{-1} \).
\( c_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \text{ J K}^{-1} \text{ mol}^{-1} \times 580 \text{ K}}{4.0 \times 10^{-3} \text{ kg mol}^{-1}}} = \sqrt{3.615 \times 10^6} = 1.90 \times 10^3 \text{ m s}^{-1} \).

**(a) (Max 2 marks)**
- Relates pressure to the rate of change of momentum of molecular collisions per unit area (1)
- Identifies that constant volume means constant density, so doubling the pressure requires an increase in the mean square speed, which is directly proportional to the absolute temperature (1)

**(b) (Max 2 marks)**
- Calculates final temperature or temperature change as \( 290 \text{ K} \) (1)
- Correct substitution into \( \Delta E_k = \frac{3}{2} n R \Delta T \) giving \( 904 \text{ J} \) (accept range \( 900 \text{ J} \) to \( 905 \text{ J} \)) (1)

**(c) (Max 3 marks)**
- Recalls or uses \( c_{\text{rms}} = \sqrt{\frac{3 R T}{M}} \) (1)
- Substitutes \( T = 580 \text{ K} \) and \( M = 4.0 \times 10^{-3} \text{ kg mol}^{-1} \) (1)
- Obtains \( 1.90 \times 10^3 \text{ m s}^{-1} \) (accept \( 1.9 \times 10^3 \text{ m s}^{-1} \)) (1)

\textbf{(a)} The decay constant \( \lambda \) is related to the half-life \( T_{1/2} \) by:
\( \lambda = \frac{\ln 2}{T_{1/2}} \)
First, convert the half-life into seconds:
\( T_{1/2} = 8.0 \text{ days} \times 24 \text{ h/day} \times 3600 \text{ s/h} = 6.912 \times 10^5 \text{ s} \)
Now, calculate \( \lambda \):
\( \lambda = \frac{0.69315}{6.912 \times 10^5 \text{ s}} = 1.003 \times 10^{-6} \text{ s}^{-1} \approx 1.0 \times 10^{-6} \text{ s}^{-1} \).

\textbf{(b)} The initial activity is given by \( A_0 = \lambda N_0 \), where \( N_0 \) is the initial number of nuclei:
\( N_0 = \frac{A_0}{\lambda} = \frac{3.5 \times 10^7 \text{ Bq}}{1.003 \times 10^{-6} \text{ s}^{-1}} = 3.490 \times 10^{13} \text{ nuclei} \)
The number of moles \( n \) is:
\( n = \frac{N_0}{N_A} = \frac{3.490 \times 10^{13}}{6.02 \times 10^{23} \text{ mol}^{-1}} = 5.797 \times 10^{-11} \text{ mol} \)
The mass \( m \) is:
\( m = n \times M = 5.797 \times 10^{-11} \text{ mol} \times 131 \text{ g mol}^{-1} = 7.59 \times 10^{-9} \text{ g} = 7.59 \times 10^{-12} \text{ kg} \approx 7.6 \times 10^{-12} \text{ kg} \).

\textbf{(c)} The activity \( A \) at \( t = 20 \text{ days} \) is given by:
\( A = A_0 e^{-\lambda t} \)
where \( t = 20 \times 24 \times 3600 = 1.728 \times 10^6 \text{ s} \).
\( A = 3.5 \times 10^7 \text{ Bq} \times e^{-(1.003 \times 10^{-6} \text{ s}^{-1} \times 1.728 \times 10^6 \text{ s})} \)
\( A = 3.5 \times 10^7 \text{ Bq} \times e^{-1.733} = 6.185 \times 10^6 \text{ Bq} \approx 6.2 \times 10^6 \text{ Bq} \).

Alternatively, using the half-life directly:
\( A = A_0 \times (0.5)^{t / T_{1/2}} = 3.5 \times 10^7 \times (0.5)^{20 / 8} = 3.5 \times 10^7 \times (0.5)^{2.5} = 6.19 \times 10^6 \text{ Bq} \).

**(a) (Max 2 marks)**
- Converts 8.0 days into seconds (\( 6.91 \times 10^5 \text{ s} \)) (1)
- Uses \( \lambda = \frac{\ln 2}{T_{1/2}} \) correctly to obtain \( 1.0 \times 10^{-6} \text{ s}^{-1} \) (1)

**(b) (Max 3 marks)**
- Rearranges \( A_0 = \lambda N_0 \) and calculates \( N_0 = 3.49 \times 10^{13} \) atoms (or uses candidate's value of \( \lambda \)) (1)
- Uses Avogadro's constant to find moles or uses the atomic mass unit to find single nucleus mass (1)
- Obtains a mass of \( 7.6 \times 10^{-12} \text{ kg} \) (accept range \( 7.5 \times 10^{-12} \text{ kg} \) to \( 7.7 \times 10^{-12} \text{ kg} \)) (1)

**(c) (Max 2 marks)**
- Uses \( A = A_0 e^{-\lambda t} \) or \( A = A_0 (0.5)^{t/T_{1/2}} \) with appropriate time conversion (1)
- Correctly calculates activity as \( 6.2 \times 10^6 \text{ Bq} \) (accept range \( 6.1 \times 10^6 \text{ Bq} \) to \( 6.3 \times 10^6 \text{ Bq} \)) (1)

\textbf{(a)} From Stefan-Boltzmann Law:
\( L = 4\pi r^2 \sigma T^4 \)

Rearranging for the radius \( r \):
\( r = \sqrt{\frac{L}{4\pi \sigma T^4}} \)
where \( \sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4} \).

\( r = \sqrt{\frac{6.0 \times 10^{24} \text{ W}}{4 \pi \times 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4} \times (3400 \text{ K})^4}} \)
\( r = \sqrt{\frac{6.0 \times 10^{24}}{9.522 \times 10^7}} = \sqrt{6.301 \times 10^{16}} = 2.51 \times 10^8 \text{ m} \approx 2.5 \times 10^8 \text{ m} \).

\textbf{(b)} From Wien's Displacement Law:
\( \lambda_{\text{max}} T = 2.898 \times 10^{-3} \text{ m K} \)
\( \lambda_{\text{max}} = \frac{2.898 \times 10^{-3} \text{ m K}}{3400 \text{ K}} = 8.524 \times 10^{-7} \text{ m} \approx 8.5 \times 10^{-7} \text{ m} \) (or \( 850 \text{ nm} \)).

\textbf{(c)} Using the Doppler redshift equation:
\( z = \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \)

Here, \( \Delta \lambda = 656.8 \text{ nm} - 656.3 \text{ nm} = 0.5 \text{ nm} \).
\( v = c \frac{\Delta \lambda}{\lambda} = 3.00 \times 10^8 \text{ m s}^{-1} \times \frac{0.5 \text{ nm}}{656.3 \text{ nm}} = 2.285 \times 10^5 \text{ m s}^{-1} \approx 2.3 \times 10^5 \text{ m s}^{-1} \).

**(a) (Max 3 marks)**
- Recalls or states Stefan-Boltzmann equation \( L = 4\pi r^2 \sigma T^4 \) (1)
- Rearranges to make \( r \) the subject (1)
- Substitutes values correctly, yielding a calculated value of at least 2 significant figures showing that \( r \approx 2.5 \times 10^8 \text{ m} \) (1)

**(b) (Max 2 marks)**
- Recalls or states Wien's Law \( \lambda_{\text{max}} T = 2.898 \times 10^{-3} \text{ m K} \) (1)
- Calculates wavelength \( 8.5 \times 10^{-7} \text{ m} \) correctly (accept range \( 8.5 \times 10^{-7} \text{ m} \) to \( 8.6 \times 10^{-7} \text{ m} \)) (1)

**(c) (Max 2 marks)**
- Recalls or states \( v = c \frac{\Delta \lambda}{\lambda} \) and identifies \( \Delta \lambda = 0.5 \text{ nm} \) (1)
- Calculates velocity correctly as \( 2.3 \times 10^5 \text{ m s}^{-1} \) (accept range \( 2.28 \times 10^5 \text{ m s}^{-1} \) to \( 2.30 \times 10^5 \text{ m s}^{-1} \)) (1)

\textbf{(a)} At the equilibrium position, the tension in the spring is equal to the weight of the suspended mass:
\( k \Delta x = mg \)
Using \( g = 9.81 \text{ m s}^{-2} \) and \( \Delta x = 0.098 \text{ m} \):
\( k = \frac{mg}{\Delta x} = \frac{0.40 \text{ kg} \times 9.81 \text{ m s}^{-2}}{0.098 \text{ m}} = 40.04 \text{ N m}^{-1} \approx 40 \text{ N m}^{-1} \).

\textbf{(b)} The time period \( T \) of a mass-spring system in simple harmonic motion is:
\( T = 2\pi \sqrt{\frac{m}{k}} \)
Using the calculated spring constant \( k = 40.04 \text{ N m}^{-1} \):
\( T = 2\pi \sqrt{\frac{0.40 \text{ kg}}{40.04 \text{ N m}^{-1}}} = 2\pi \sqrt{0.00999} \approx 0.628 \text{ s} \approx 0.63 \text{ s} \).

\textbf{(c)} Since the mass is released from rest at the maximum displacement \( A = 0.050 \text{ m} \) at time \( t = 0 \), displacement is modelled by a cosine function:
\( x = A \cos(\omega t) \)

The angular frequency is:
\( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{40.04}{0.40}} = 10.0 \text{ rad s}^{-1} \)

Therefore, the displacement equation is:
\( x = 0.050 \cos(10 t) \)

The maximum kinetic energy equals the maximum elastic potential energy stored in the spring during oscillation:
\( E_{k,\text{max}} = \frac{1}{2} k A^2 \)
\( E_{k,\text{max}} = 0.5 \times 40.04 \text{ N m}^{-1} \times (0.050 \text{ m})^2 = 0.050 \text{ J} \).

Alternatively, using the maximum velocity:
\( v_{\text{max}} = \omega A = 10.0 \text{ rad s}^{-1} \times 0.050 \text{ m} = 0.50 \text{ m s}^{-1} \)
\( E_{k,\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = 0.5 \times 0.40 \text{ kg} \times (0.50 \text{ m s}^{-1})^2 = 0.050 \text{ J} \).

**(a) (Max 2 marks)**
- States or uses Hooke's law in equilibrium: \( k \Delta x = mg \) (1)
- Substitutes \( g = 9.81 \text{ m s}^{-2} \) and \( \Delta x = 0.098 \text{ m} \) correctly to obtain \( 40 \text{ N m}^{-1} \) (1)

**(b) (Max 2 marks)**
- Recalls or states \( T = 2\pi \sqrt{\frac{m}{k}} \) (1)
- Correctly calculates \( T = 0.63 \text{ s} \) (accept range \( 0.62 \text{ s} \) to \( 0.63 \text{ s} \)) (1)

**(c) (Max 3 marks)**
- Calculates \( \omega = 10 \text{ rad s}^{-1} \) and writes equation \( x = 0.050 \cos(10 t) \) (or equivalent with sine/phase shift) (1)
- Uses \( E_{k,\text{max}} = \frac{1}{2} k A^2 \) or \( E_{k,\text{max}} = \frac{1}{2} m (\omega A)^2 \) (1)
- Obtains kinetic energy of \( 0.050 \text{ J} \) (1)

(a) To ensure accurate and uniform temperature measurements:
1. The flask must be fully submerged in the water bath so that all of the enclosed air is at the same temperature as the water.
2. The water bath must be stirred continuously before taking a reading to eliminate temperature gradients within the water.
3. After changing the heating level, the student must wait for a few minutes to allow the air inside the flask to reach thermal equilibrium with the surrounding water.
4. Use a high-resolution digital thermometer placed close to the flask to measure the temperature.

(b) Assuming a linear relationship of the form \(p = m\theta + C\):
* Gradient \(m = \frac{p_2 - p_1}{\theta_2 - \theta_1} = \frac{122 - 101}{80.0 - 20.0} = \frac{21}{60.0} = 0.35\text{ kPa }^\circ\text{C}^{-1}\).
* Find the intercept \(C\) using \(p_1 = m\theta_1 + C\):
\(C = 101 - (0.35 \times 20.0) = 101 - 7.0 = 94.0\text{ kPa}\).
* At absolute zero (\(\theta_0\)), the pressure of an ideal gas extrapolates to zero (\(p = 0\)):
\(0 = 0.35\theta_0 + 94.0 \implies \theta_0 = -\frac{94.0}{0.35} = -268.57^\circ\text{C}\).
To appropriate significant figures, this is \(-269^\circ\text{C}\).

(c) The temperature change is \(\Delta\theta = 80.0 - 20.0 = 60.0^\circ\text{C}\).
* When subtracting values, their absolute uncertainties are added:
\(\text{Absolute uncertainty in } \Delta\theta = 0.5^\circ\text{C} + 0.5^\circ\text{C} = 1.0^\circ\text{C}\).
* Percentage uncertainty = \(\frac{1.0}{60.0} \times 100\% \approx 1.67\%\) (or \(1.7\%\)).

(d) If all pressure readings are systematically \(5\text{ kPa}\) too high:
* The change in pressure \(\Delta p = p_2 - p_1\) remains the same, so the gradient \(m\) is unchanged.
* The measured pressure intercept \(C'\) becomes larger by \(5\text{ kPa}\) (i.e., \(C' = 99\text{ kPa}\)).
* Since \(\theta_0 = -\frac{C'}{m}\), a larger value of \(C'\) with an unchanged \(m\) results in a larger magnitude of absolute zero, making the calculated absolute zero more negative (e.g., \(-\frac{99}{0.35} = -283^\circ\text{C}\)).

(a) [Total: 4 marks]
* MP1: Fully submerge the flask in the water bath [1]
* MP2: Stir the water bath continuously [1]
* MP3: Wait for a period of time after heating to achieve thermal equilibrium [1]
* MP4: Position the thermometer close to the flask / use high-resolution probe [1]

(b) [Total: 3.5 marks]
* MP1: Calculate gradient \(m = 0.35\text{ kPa }^\circ\text{C}^{-1}\) [1]
* MP2: Use \(y = mx + c\) to find pressure intercept \(C = 94.0\text{ kPa}\) [1]
* MP3: Set \(p = 0\) and solve for temperature [1]
* MP4: State final answer of \(-269^\circ\text{C}\) (or \(-268.6^\circ\text{C}\)) with correct unit [0.5]

(c) [Total: 2 marks]
* MP1: Add absolute uncertainties to find absolute uncertainty of \(1.0^\circ\text{C}\) [1]
* MP2: Correct percentage uncertainty of \(1.7\%\) (accept \(1.67\%\)) [1]

(d) [Total: 3 marks]
* MP1: State that the gradient \(m\) remains unchanged because both pressures increase by the same amount [1]
* MP2: State that the pressure intercept \(C\) increases [1]
* MP3: Conclude that the calculated absolute zero value becomes more negative / further from \(0^\circ\text{C}\) [1]

(a) To measure the period \(T\) accurately:
* Place a fiducial marker at the equilibrium position of the oscillations to provide a clear, repeatable reference point.
* Attach a pointer to the mass to align with the marker.
* Time a large number of oscillations (typically 20 oscillations) to reduce the relative uncertainty due to human reaction time, repeat this measurement, and divide the mean total time by the number of oscillations to find \(T\).

(b) Squaring the equation:
\[T^2 = \frac{4\pi^2(m + m_{\text{eff}})}{k} = \left(\frac{4\pi^2}{k}\right)m + \frac{4\pi^2 m_{\text{eff}}}{k}\]
Comparing this to \(y = mx + c\):
* Gradient \(G = \frac{4\pi^2}{k}\)
* Intercept \(C = \frac{4\pi^2 m_{\text{eff}}}{k}\)
This is a straight-line equation because both \(k\) and \(m_{\text{eff}}\) are constants.

(c) Calculation of \(k\):
\[G = 1.58 \implies k = \frac{4\pi^2}{G} = \frac{4\pi^2}{1.58} = 24.99 \approx 25.0\text{ N m}^{-1}\]
Uncertainty in \(k\):
* Percentage uncertainty in \(G = \frac{0.05}{1.58} \times 100\% \approx 3.16\%\).
* Since \(k = 4\pi^2 G^{-1}\), the percentage uncertainty in \(k\) is also \(3.16\%\).
* Absolute uncertainty in \(k = 3.16\% \times 24.99\text{ N m}^{-1} \approx 0.79\text{ N m}^{-1}\).
* Final value: \(k = 25.0 \pm 0.8\text{ N m}^{-1}\).

(d) Calculation of \(m_{\text{eff}}\):
\[C = \frac{4\pi^2 m_{\text{eff}}}{k} = G \cdot m_{\text{eff}} \implies m_{\text{eff}} = \frac{C}{G} = \frac{0.038}{1.58} = 0.024\text{ kg} \text{ (or } 24\text{ g)}\]
To check if this value is reasonable:
* Measure the total mass of the spring alone using a top-pan balance.
* Compare \(m_{\text{eff}}\) to the spring's actual mass. Theoretically, \(m_{\text{eff}}\) should be approximately \(\frac{1}{3}\) of the actual mass of the spring.

(a) [Total: 3 marks]
* MP1: Position fiducial marker at the center (equilibrium position) of oscillation [1]
* MP2: Time a large number of oscillations (e.g., 20) and divide total time by number of oscillations [1]
* MP3: Justify timing multiple oscillations to reduce the effect of reaction time on uncertainty [1]

(b) [Total: 3 marks]
* MP1: Show algebraic squaring to obtain \(T^2 = \left(\frac{4\pi^2}{k}\right)m + \frac{4\pi^2 m_{\text{eff}}}{k}\) [1]
* MP2: State Gradient \(= \frac{4\pi^2}{k}\) [1]
* MP3: State y-intercept \(= \frac{4\pi^2 m_{\text{eff}}}{k}\) [1]

(c) [Total: 3.5 marks]
* MP1: Calculate \(k = 25.0\text{ N m}^{-1}\) (or \(24.99\)) [1]
* MP2: Calculate percentage uncertainty in gradient \(= 3.16\%\) [1]
* MP3: Calculate absolute uncertainty in \(k = 0.8\text{ N m}^{-1}\) (or \(0.79\)) [1]
* MP4: Express final answer with matching sig figs: \(25.0 \pm 0.8\text{ N m}^{-1}\) [0.5]

(d) [Total: 3 marks]
* MP1: Calculate \(m_{\text{eff}} = 0.024\text{ kg}\) (accept \(24\text{ g}\)) [1]
* MP2: Suggest measuring the total mass of the spring with a balance [1]
* MP3: Compare experimental \(m_{\text{eff}}\) with the theoretical expectation of \(\approx \frac{1}{3}\) of the spring mass [1]

(a) The circuit diagram must contain:
* A d.c. power supply, a capacitor, a resistor, a voltmeter, and a single-pole double-throw (two-way) switch.
* Position 1 of the switch connects the power supply directly across the capacitor to charge it.
* Position 2 of the switch disconnects the power supply and connects the capacitor in a closed loop containing the resistor.
* The voltmeter must be connected in parallel across the capacitor.

(b) Taking natural logarithms on both sides of \(V = V_0 e^{-t/RC}\):
\[\ln(V) = \ln(V_0) - \frac{t}{RC}\]
This can be written in the straight-line form \(y = mx + c\):
\[\ln(V) = -\left(\frac{1}{RC}\right)t + \ln(V_0)\]
* Plotting \(\ln(V/\text{V})\) on the y-axis against \(t\) on the x-axis gives a straight line.
* The gradient \(m = -\frac{1}{RC}\).
* Therefore, capacitance \(C = -\frac{1}{m R}\).

(c) Calculating \(C\):
\[C = -\frac{1}{m R} = -\frac{1}{(-0.0425\text{ s}^{-1}) \times (47.0 \times 10^3\text{ }\Omega)} = \frac{1}{1997.5} = 5.01 \times 10^{-4}\text{ F} \text{ (or } 501\text{ }\mu\text{F)}\]
Percentage uncertainty in \(C\):
* For products/quotients, percentage uncertainties are added:
\(\% \text{ uncertainty in } C = \% \text{ uncertainty in } R + \% \text{ uncertainty in gradient}\)
\(\% \text{ uncertainty in } C = 5\% + 3\% = 8\%\).

(d) Effect of the analogue voltmeter:
* The voltmeter is in parallel with the capacitor during discharge. Its internal resistance \(R_{\text{V}}\) acts in parallel with the fixed resistor \(R\).
* This decreases the total resistance of the discharge path: \(R_{\text{eff}} = \frac{R \cdot R_{\text{V}}}{R + R_{\text{V}}}\).
* A lower resistance causes the capacitor to discharge more rapidly, increasing the rate of decay. This makes the gradient of \(\ln(V)\) against \(t\) steeper (more negative).
* Since \(C\) is calculated assuming only the nominal resistance \(R\) is present, using a larger-than-actual resistance value in \(C = -\frac{1}{m R}\) results in an underestimation of the actual capacitance \(C\).

(a) [Total: 3 marks]
* MP1: Correct symbol for power supply, capacitor, voltmeter, and two-way switch [1]
* MP2: Voltmeter connected in parallel across the capacitor [1]
* MP3: Switch arrangement allows charging from supply, then discharging through the resistor [1]

(b) [Total: 3 marks]
* MP1: Take natural logarithms correctly to get \(\text{ln}(V) = \text{ln}(V_0) - \frac{t}{RC}\) [1]
* MP2: State that a graph of \(\text{ln}(V)\) vs \(t\) is linear with gradient \(m = -1/RC\) [1]
* MP3: Show rearranging to express \(C = -1/(mR)\) [1]

(c) [Total: 3.5 marks]
* MP1: Correct substitution of values into the formula [1]
* MP2: Calculate \(C = 5.01 \times 10^{-4}\text{ F}\) (or \(501\text{ }\mu\text{F}\)) [1]
* MP3: Add percentage uncertainties to get \(8\%\) [1]
* MP4: Correct units (F or \(\mu\text{F}\)) [0.5]

(d) [Total: 3 marks]
* MP1: Identify that the voltmeter internal resistance is in parallel with the resistor, lowering the effective resistance [1]
* MP2: Conclude that the discharge rate increases, making the gradient steeper / more negative [1]
* MP3: Explain that this leads to an underestimation of the calculated value of \(C\) [1]

(a) Controlled variables:
1. Length of the wire in the magnetic field (\(L\)): Ensure the magnet assembly is not moved relative to the wire, and the wire is kept straight and perpendicular to the magnetic field lines.
2. Magnetic flux density (\(B\)): Use the same magnet assembly throughout and ensure no other magnetic objects are brought near the apparatus.
3. Temperature of the wire: Keep current values low and turn off the power supply between readings to prevent heating, which could cause resistance changes or air convection currents.

(b) Derivation:
* By Newton's third law, the force exerted by the wire on the magnets is equal in magnitude and opposite in direction to the force exerted by the magnets on the wire:
\(F_{\text{magnets}} = F_{\text{wire}} = BIL\).
* This downward/upward magnetic force on the magnet assembly changes the normal contact force on the balance, leading to a change in the mass reading \(\Delta m\):
\(F_{\text{magnets}} = \Delta m \cdot g\).
* Equating the forces:
\(\Delta m \cdot g = BIL\).
* Rearranging for \(\Delta m\):
\(\Delta m = \left(\frac{BL}{g}\right) I\).

(c) Calculation of \(B\):
* Gradient \(G = \frac{BL}{g} = 1.25 \times 10^{-3}\text{ kg A}^{-1}\).
* \(B = \frac{G \cdot g}{L} = \frac{(1.25 \times 10^{-3}\text{ kg A}^{-1}) \times 9.81\text{ m s}^{-2}}{0.052\text{ m}} = \frac{0.0122625}{0.052} \approx 0.2358\text{ T}\).
* Percentage uncertainty in \(L = \frac{0.1\text{ cm}}{5.2\text{ cm}} \times 100\% \approx 1.92\%\).
* Total percentage uncertainty in \(B = 4\% \text{ (from gradient)} + 1.92\% \text{ (from } L\text{)} = 5.92\%\).
* Absolute uncertainty in \(B = 5.92\% \times 0.2358\text{ T} \approx 0.014\text{ T}\).
* Final value: \(B = 0.236 \pm 0.014\text{ T}\).

(d) Errors and minimization:
* Systematic error: The magnetic field from the wire or magnets might interact with nearby iron clamps or the metallic body of the balance. Minimization: Ensure all retort stands and clamps near the magnet assembly are made of non-magnetic materials (e.g., brass or aluminum) and isolate the balance.
* Random error: Fluctuation in the balance readings due to air currents or electrical noise in the power supply. Minimization: Place a draft shield around the balance and take multiple readings for each current value, then calculate the average.

(a) [Total: 4 marks]
* MP1: Identify length of wire in magnetic field as a control variable and state how to keep it constant [1]
* MP2: Identify magnetic field strength as a control variable [1]
* MP3: Identify temperature of wire as a control variable [1]
* MP4: State method to control temperature (e.g., small current, switch off between readings) [1]

(b) [Total: 3 marks]
* MP1: State that the force on the balance is equal to the magnetic force via Newton's Third Law [1]
* MP2: Link force to mass change: \(F = \Delta m \cdot g\) [1]
* MP3: Equate and show clear algebraic steps to arrive at \(\Delta m = \left(\frac{BL}{g}\right)I\) [1]

(c) [Total: 2.5 marks]
* MP1: Calculate \(B = 0.236\text{ T}\) (or \(0.24\text{ T}\)) [1]
* MP2: Determine percentage uncertainty in \(L\) (approx. \(1.9\%\)) and total percentage uncertainty in \(B\) (approx. \(5.9\%\)) [1]
* MP3: State absolute uncertainty of \(\pm 0.014\text{ T}\) (or \(\pm 0.01\text{ T}\) if rounded to 2 s.f.) [0.5]

(d) [Total: 3 marks]
* MP1: Correctly identify a systematic error (e.g., balance calibration or magnetic interference from nearby steel objects) and its minimization [1]
* MP2: Correctly identify a random error (e.g., draughts / air currents or current fluctuation) [1]
* MP3: Suggest an appropriate minimization method for the random error (e.g., draught shield or taking repeat readings) [1]