2024 Edexcel IAS-Level Physics (XPH11) 模擬試題連答案詳解

The maximum height \(H\) is given by \(H = \frac{u^2 \sin^2 \theta}{2g}\). The horizontal range \(R\) is given by \(R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}\). The ratio of maximum height to horizontal range is: \(\frac{H}{R} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{2u^2 \sin\theta \cos\theta}{g}} = \frac{\sin^2\theta}{4\sin\theta\cos\theta} = \frac{1}{4}\tan\theta\). Hence, the correct option is A.

1 mark for the correct option (A).

The Young modulus \(E\) is given by \(E = \frac{FL}{A\Delta x}\), which gives extension \(\Delta x = \frac{FL}{AE}\). Since the cross-sectional area is \(A = \frac{\pi d^2}{4}\), the extension is proportional to \(\frac{FL}{d^2}\). For the second wire, the new extension \(\Delta x_2\) is proportional to \(\frac{(2F)(2L)}{(2d)^2} = \frac{4FL}{4d^2} = \frac{FL}{d^2}\). Therefore, the extension remains \(\Delta x\).

1 mark for the correct option (B).

For the combined system of mass \(3m\) moving up the incline with acceleration \(a\): \(F - 3mg\sin\theta = 3ma\), which simplifies to \(ma = \frac{F}{3} - mg\sin\theta\). For the trailing block of mass \(m\), the only force pulling it up the incline is the tension \(T\) in the string. The equation of motion for this block is: \(T - mg\sin\theta = ma\). Substituting \(ma\) into this equation gives: \(T - mg\sin\theta = \frac{F}{3} - mg\sin\theta\), which simplifies directly to \(T = \frac{F}{3}\).

1 mark for the correct option (A).

At terminal velocity, the upward forces (viscous drag and upthrust) equal the downward force (weight). Thus, \(6\pi\eta r v + \frac{4}{3}\pi r^3 \rho_l g = \frac{4}{3}\pi r^3 \rho_s g\). Rearranging for viscous drag gives \(6\pi\eta r v = \frac{4}{3}\pi r^3 g (\rho_s - \rho_l)\). Solving for \(v\) yields \(v = \frac{2r^2 g(\rho_s - \rho_l)}{9\eta}\).

1 mark for the correct option (A).

The mass of water discharged per second is \(\frac{dm}{dt} = \rho A v\). The pump must supply gravitational potential energy per second to lift the water: \(P_{\text{potential}} = \frac{dm}{dt} g h = \rho A v g h\). It must also supply kinetic energy per second to discharge the water: \(P_{\text{kinetic}} = \frac{1}{2} \frac{dm}{dt} v^2 = \frac{1}{2} \rho A v^3\). The total minimum useful power is the sum of these two: \(P = \rho A v (gh + \frac{1}{2}v^2)\).

1 mark for the correct option (D).

According to Newton's second law, the average force on the ball is equal to the rate of change of momentum: \(F_{\text{ball}} = \frac{\Delta p}{\Delta t}\). Taking the initial direction towards the wall as positive, the initial momentum is \(mu\) and the final momentum is \(-mv\). The change in momentum of the ball is \(\Delta p = -mv - mu = -m(u + v)\). By Newton's third law, the force exerted on the wall is equal in magnitude and opposite in direction: \(F = \frac{m(u + v)}{\Delta t}\).

1 mark for the correct option (B).

The work done in stretching the rubber is the area under the force-extension graph, which is found by integration: \(W = \int_{0}^{e} F \, dx = \int_{0}^{e} k x^2 \, dx = \left[ \frac{1}{3} k x^3 \right]_{0}^{e} = \frac{1}{3} k e^3\).

1 mark for the correct option (B).

For rotational equilibrium, take moments about the pivot. The weight \(W\) acts at the midpoint of the uniform beam, creating a clockwise moment of \(W \times \frac{L}{2}\). The vertical component of the tension in the cable is \(T \sin\theta\), which acts at the far end of the beam (distance \(L\)), creating a counter-clockwise moment of \(T \sin\theta \times L\). Equating the moments: \(T \sin\theta \times L = W \times \frac{L}{2}\). Solving for tension gives \(T = \frac{W}{2\sin\theta}\).

1 mark for the correct option (B).

Since the box moves at a constant speed, the resultant force acting on it parallel to the slope is zero. Resolving forces parallel to the slope: \(F - f - mg \sin\theta = 0\), where \(f\) is the force of friction. Rearranging for \(f\) gives \(f = F - mg \sin\theta\). The rate of work done against friction is the power dissipated by friction, given by: \(P = f v = (F - mg \sin\theta)v\). Therefore, B is the correct choice.

1 mark for the correct option B.

According to Stokes' Law, the viscous drag force on a sphere of radius \(r\) moving at velocity \(v\) through a fluid of viscosity \(\eta\) in laminar flow is given by \(D = 6\pi \eta r v\). At terminal velocity, the forces are in equilibrium: \(W = U + D\). Substituting the expressions for weight and upthrust gives: \(\frac{4}{3}\pi r^3 \rho_s g = \frac{4}{3}\pi r^3 \rho_f g + 6\pi \eta r v\). Rearranging for terminal velocity \(v\) yields \(v = \frac{2r^2(\rho_s - \rho_f)g}{9\eta}\). Since both spheres are made of the same material and fall through the same liquid, the terminal velocity is directly proportional to the square of the radius: \(v \propto r^2\). Since sphere X has twice the radius of sphere Y, its terminal velocity is \(2^2 = 4\) times larger. Thus, B is the correct option.

1 mark for the correct option B.

(a) For vertical motion: \(s_y = u_y t + \frac{1}{2} g t^2\). Since \(u_y = 0\), we have \(45.0 = \frac{1}{2} \times 9.81 \times t^2\). Rearranging gives \(t = \sqrt{\frac{2 \times 45.0}{9.81}} = 3.03\text{ s}\), which is about \(3\text{ s}\). (b) For horizontal motion: \(s_x = v_x \times t = 15.0 \times 3.03 = 45.5\text{ m}\). (c) Vertical component of velocity just before impact: \(v_y = u_y + g t = 0 + (9.81 \times 3.03) = 29.7\text{ m s}^{-1}\). Horizontal component of velocity: \(v_x = 15.0\text{ m s}^{-1}\). Magnitude of velocity: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15.0^2 + 29.7^2} = 33.3\text{ m s}^{-1}\). Direction: \(\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{29.7}{15.0}\right) = 63.2^\circ\) below the horizontal.

Part (a): [1M] Use of \(s = \frac{1}{2} g t^2\) or equivalent. [1M] Correct substitution of \(s = 45.0\) and \(g = 9.81\) to show \(t \approx 3.0\text{ s}\). Part (b): [1M] Use of \(d = v_x t\). [1M] Correct calculation of distance to 3 s.f. (accept \(45.0\text{ m}\) to \(45.5\text{ m}\)). Part (c): [1M] Calculation of \(v_y = 29.7\text{ m s}^{-1}\). [1M] Use of Pythagoras' theorem to find magnitude. [1M] Correct magnitude \(33.3\text{ m s}^{-1}\) (accept range \(33.2\) - \(33.4\)). [0.7M] Calculation of angle \(63.2^\circ\) below horizontal.

(a) The three forces acting on the ball bearing are: Weight acting vertically downwards, Upthrust acting vertically upwards, and Viscous drag acting vertically upwards. (b) At terminal velocity, the upward forces equal the downward force: \(W = U + F\), where \(W\) is Weight, \(U\) is Upthrust, and \(F\) is Viscous drag. \(W = \rho_s V g\), \(U = \rho_f V g\), and \(F = 6 \pi \eta r v\). Volume of a sphere \(V = \frac{4}{3} \pi r^3\). Therefore, \(\rho_s \left(\frac{4}{3} \pi r^3\right) g = \rho_f \left(\frac{4}{3} \pi r^3\right) g + 6 \pi \eta r v\). Simplifying this yields: \(v = \frac{2 r^2 g (\rho_s - \rho_f)}{9 \eta}\). Substituting the given values: \(v = \frac{2 \times (2.0 \times 10^{-3})^2 \times 9.81 \times (7800 - 920)}{9 \times 0.25} = \frac{2 \times 4.0 \times 10^{-6} \times 9.81 \times 6880}{2.25} = 0.24\text{ m s}^{-1}\).

Part (a): [1.7M] Identification of weight downwards, upthrust upwards, and drag upwards (all three needed for full marks, partial credit of 1 mark for two). Part (b): [1M] Statement of equilibrium condition: \(W = U + F\). [1M] Correct substitution of formulas for \(W\), \(U\), and Stokes' law \(F\). [1M] Rearranging formula to make \(v\) the subject. [1M] Substitution of given numerical values. [1M] Correct final answer with units: \(0.24\text{ m s}^{-1}\).

(a) Tensile stress: \(\sigma = \frac{F}{A} = \frac{120\text{ N}}{1.5 \times 10^{-6}\text{ m}^2} = 8.0 \times 10^7\text{ Pa}\). (b) Young Modulus \(E = \frac{\text{Stress}}{\text{Strain}}\). Strain \(\epsilon = \frac{\sigma}{E} = \frac{8.0 \times 10^7}{2.0 \times 10^{11}} = 4.0 \times 10^{-4}\). Since strain \(\epsilon = \frac{\Delta L}{L}\), extension \(\Delta L = \epsilon \times L = 4.0 \times 10^{-4} \times 2.5\text{ m} = 1.0 \times 10^{-3}\text{ m} = 1.0\text{ mm}\). (c) Elastic strain energy: \(E_{el} = \frac{1}{2} F \Delta L = 0.5 \times 120\text{ N} \times 1.0 \times 10^{-3}\text{ m} = 0.060\text{ J}\).

Part (a): [1M] Use of \(\sigma = F/A\). [1M] Correct calculation of stress as \(8.0 \times 10^7\text{ Pa}\) or \(\text{N m}^{-2}\). Part (b): [1M] Use of Young modulus formula \(E = \frac{\sigma}{\epsilon}\) or equivalent. [1M] Use of strain definition \(\epsilon = \Delta L / L\). [1.7M] Correct calculation of extension as \(1.0 \times 10^{-3}\text{ m}\) or \(1.0\text{ mm}\). Part (c): [1M] Use of \(E_{el} = \frac{1}{2} F \Delta L\). [1M] Correct value \(0.060\text{ J}\) (accept \(6.0 \times 10^{-2}\text{ J}\)).

(a) Height gained: \(h = 8.50 \times \sin(20.0^\circ) = 2.907\text{ m}\). Work done against gravity: \(W_g = m g h = 18.0 \times 9.81 \times 2.907 = 513.3\text{ J} \approx 513\text{ J}\). (b) Work done against friction: \(W_f = f \times d = 35.0 \times 8.50 = 297.5\text{ J}\). Total work done: \(W_{\text{total}} = W_g + W_f = 513.3 + 297.5 = 810.8\text{ J} \approx 811\text{ J}\). (c) Time taken: \(t = \frac{d}{v} = \frac{8.50}{1.20} = 7.083\text{ s}\). Power required: \(P = \frac{W_{\text{total}}}{t} = \frac{810.8}{7.083} = 114.5\text{ W} \approx 115\text{ W}\). Alternatively, using \(P = F v\): Total parallel force \(F = m g \sin(20.0^\circ) + f = (18.0 \times 9.81 \times \sin 20.0^\circ) + 35.0 = 60.39 + 35.0 = 95.39\text{ N}\). Power \(P = F \times v = 95.39 \times 1.20 = 114.5\text{ W}\).

Part (a): [1M] Calculation of height \(h = 8.50 \sin(20^\circ)\). [1.7M] Calculation of gravitational work done \(513\text{ J}\) (accept range \(510\text{ J}\) to \(514\text{ J}\)). Part (b): [1M] Calculation of friction work done \(298\text{ J}\). [1M] Adding friction work to gravity work to get \(811\text{ J}\) (accept range \(808\text{ J}\) to \(812\text{ J}\)). Part (c): [1M] Calculation of time \(t = 7.08\text{ s}\) or formula \(P = F v\). [1M] Use of total force or total work divided by time. [1M] Correct final answer of \(114\text{ W}\) or \(115\text{ W}\).

(a) The principle of moments states that for a body in rotational equilibrium, the sum of the clockwise moments about any pivot point is equal to the sum of the anticlockwise moments about that same pivot point. (b) Let's take moments about pivot \(A\). The board is uniform, so its weight acts at its center of gravity, which is \(2.0\text{ m}\) from \(A\). Board weight \(W_b = m_b g = 25 \times 9.81 = 245.25\text{ N}\). Diver weight \(W_d = m_d g = 70 \times 9.81 = 686.7\text{ N}\), acting at \(4.0\text{ m}\) from \(A\). Sum of clockwise moments = \((245.25 \times 2.0) + (686.7 \times 4.0) = 490.5 + 2746.8 = 3237.3\text{ N m}\). Sum of anticlockwise moments is due to the upward support force at \(B\), \(F_B\), acting at \(1.2\text{ m}\): Moment = \(F_B \times 1.2\). Equating moments: \(1.2 \times F_B = 3237.3 \Rightarrow F_B = 2697.75\text{ N} \approx 2700\text{ N}\). (c) For vertical equilibrium: \(\sum F_{\text{up}} = \sum F_{\text{down}}\). Assuming force at \(A\) is upward: \(F_A + F_B = W_b + W_d\). \(F_A + 2697.75 = 245.25 + 686.7 = 932\text{ N}\). \(F_A = 932 - 2697.75 = -1765.75\text{ N}\). The negative sign indicates that the force acts in the opposite direction to our assumption. Thus, the force at \(A\) is \(1770\text{ N}\) acting downwards.

Part (a): [1M] Statement of sum of clockwise moments equals sum of anticlockwise moments. [1M] Mention of 'equilibrium' or 'about a point'. Part (b): [1M] Identification of board weight acting at \(2.0\text{ m}\). [1M] Use of moment formula \(\text{moment} = Fd\) for board weight and diver weight. [1M] Setting up equation \(F_B \times 1.2 = (25 \times 9.81 \times 2.0) + (70 \times 9.81 \times 4.0)\). [0.7M] Correct final answer: \(2700\text{ N}\) (accept \(2.7 \times 10^3\text{ N}\)). Part (c): [1M] Use of vertical equilibrium condition \(F_A + F_B = W_b + W_d\). [1M] Correct magnitude \(1770\text{ N}\) (accept \(1.77 \times 10^3\text{ N}\) or \(1.8 \times 10^3\text{ N}\)) and stating direction is 'downward'.

(a) Plastic deformation is a type of deformation in which the material does not return to its original shape or dimensions when the deforming force or stress is removed; it remains permanently stretched. (b) Stiffness is given by Hooke's Law constant in the linear region: \(k = \frac{F}{\Delta x} = \frac{90\text{ N}}{3.0 \times 10^{-3}\text{ m}} = 3.0 \times 10^4\text{ N m}^{-1}\). (c) Total work done is the sum of the work done in the elastic (linear) region and the plastic region (area under the graph). Elastic region work: \(W_e = \frac{1}{2} \times F \times \Delta x_e = 0.5 \times 90\text{ N} \times 3.0 \times 10^{-3}\text{ m} = 0.135\text{ J}\). Plastic region work: \(W_p = F_{\text{avg}} \times \Delta x_p = 110\text{ N} \times (12.0 - 3.0) \times 10^{-3}\text{ m} = 110 \times 9.0 \times 10^{-3} = 0.99\text{ J}\). Total work done: \(W_{\text{total}} = 0.135 + 0.99 = 1.125\text{ J} \approx 1.1\text{ J}\).

Part (a): [1M] Stating that the deformation is permanent. [1M] Specifying that it does not return to its original shape/length when the force/load is removed. Part (b): [1M] Use of stiffness formula \(k = F/x\). [1.7M] Correct substitution and calculation to yield \(3.0 \times 10^4\text{ N m}^{-1}\). Part (c): [1M] Area under the linear part: \(\frac{1}{2} F_e x_e = 0.135\text{ J}\). [1M] Area under the plastic part: \(F_{\text{avg}} \times (x_p - x_e) = 0.99\text{ J}\). [1M] Sum of both areas to give total work of \(1.1\text{ J}\) (accept \(1.12\text{ J}\) or \(1.13\text{ J}\)).

(a) Mass flow rate: \(\frac{\Delta m}{\Delta t} = \rho A v\). Rearranging for velocity: \(v = \frac{1}{\rho A} \frac{\Delta m}{\Delta t} = \frac{15.0}{1000 \times 3.00 \times 10^{-4}} = \frac{15.0}{0.300} = 50.0\text{ m s}^{-1}\). (b) Rate of change of momentum: \(\frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t} \times \Delta v\). Since the water comes to rest on striking the wall, \(\Delta v = v - 0 = 50.0\text{ m s}^{-1}\). Therefore, \(\frac{\Delta p}{\Delta t} = 15.0\text{ kg s}^{-1} \times 50.0\text{ m s}^{-1} = 750\text{ kg m s}^{-2}\) (or \(750\text{ N}\)). (c) According to Newton's second law, the force required to stop the water is equal to the rate of change of momentum of the water, which is \(750\text{ N}\). The wall exerts a force of \(750\text{ N}\) on the water directed away from the wall. By Newton's third law, the water exerts an equal and opposite force on the wall. Hence, the force exerted by the water on the wall has a magnitude of \(750\text{ N}\) and is directed horizontally into the wall (in the direction of the initial flow).

Part (a): [1M] Statement or use of \(\frac{\Delta m}{\Delta t} = \rho A v\). [1.7M] Correct substitution showing \(v = 50.0\text{ m s}^{-1}\). Part (b): [1M] Use of rate of change of momentum formula \(\frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t} \Delta v\). [1M] Correct calculation yielding \(750\text{ N}\) (or \(\text{kg m s}^{-2}\)). Part (c): [1M] Stating force magnitude is \(750\text{ N}\) (from Newton's second law). [1M] Reference to Newton's third law (action and reaction are equal and opposite). [1M] Stating the direction of force on the wall is into the wall / same direction as water flow.

(a) The forces acting on the sledge are: Weight acting vertically downwards, Normal contact force acting perpendicular to the slope, and Friction acting up the slope (opposing the motion). (b) The component of weight acting parallel to the slope (downwards) is \(W_{||} = m g \sin(\theta) = 12.0 \times 9.81 \times \sin(15.0^\circ) = 30.46\text{ N}\). The net force down the slope is \(F_{\text{net}} = W_{||} - f = 30.46 - 18.0 = 12.46\text{ N}\). Using Newton's second law, \(a = \frac{F_{\text{net}}}{m} = \frac{12.46}{12.0} = 1.04\text{ m s}^{-2}\), which is approximately \(1.0\text{ m s}^{-2}\). (c) Using the equation of motion: \(v^2 = u^2 + 2 a s\). Since the sledge starts from rest, \(u = 0\). \(v^2 = 0 + 2 \times 1.04 \times 15.0 = 31.2\text{ m}^2\text{ s}^{-2}\). Thus, \(v = \sqrt{31.2} = 5.59\text{ m s}^{-1}\).

Part (a): [1.7M] List of weight (vertically down), normal reaction (perpendicular to slope), and friction (parallel to slope upwards). Deduct 0.7 marks if directions are omitted or incorrect. Part (b): [1M] Calculation of weight component down the slope as \(12.0 \times 9.81 \times \sin(15^\circ) = 30.5\text{ N}\). [1M] Use of \(F = ma\) with net force. [1M] Correctly obtaining \(a = 1.04\text{ m s}^{-2}\). Part (c): [1M] Use of \(v^2 = u^2 + 2as\). [1M] Correct calculation of velocity as \(5.59\text{ m s}^{-1}\) (accept \(5.5\) to \(5.6\text{ m s}^{-1}\)).

**(a)**
First, calculate the elastic potential energy stored in the compressed spring:
\(E_{ep} = \frac{1}{2} k x^2\)
\(E_{ep} = 0.5 \times 320\text{ N m}^{-1} \times (0.055\text{ m})^2 = 0.484\text{ J}\)

Since energy transfer is \(100\%\) efficient, the kinetic energy \(E_k\) of the ball bearing immediately upon release equals the stored elastic potential energy:
\(E_k = \frac{1}{2} m v^2 = E_{ep}\)
\(0.5 \times 0.025\text{ kg} \times v^2 = 0.484\text{ J}\)
\(v^2 = \frac{2 \times 0.484\text{ J}}{0.025\text{ kg}} = 38.72\text{ m}^2\text{ s}^{-2}\)
\(v = \sqrt{38.72} = 6.22\text{ m s}^{-1}\)
This is approximately \(6.2\text{ m s}^{-1}\).

**(b)**
Resolve the initial launch velocity into horizontal and vertical components:
* Vertical component: \(u_y = v \sin(30.0^\circ)\)
* Using \(v = 6.22\text{ m s}^{-1}\): \(u_y = 6.22 \times \sin(30.0^\circ) = 3.11\text{ m s}^{-1}\) (upwards)
* Using \(v = 6.2\text{ m s}^{-1}\): \(u_y = 6.2 \times \sin(30.0^\circ) = 3.10\text{ m s}^{-1}\) (upwards)
* Horizontal component: \(u_x = v \cos(30.0^\circ)\)
* Using \(v = 6.22\text{ m s}^{-1}\): \(u_x = 6.22 \times \cos(30.0^\circ) = 5.39\text{ m s}^{-1}\)
* Using \(v = 6.2\text{ m s}^{-1}\): \(u_x = 6.2 \times \cos(30.0^\circ) = 5.37\text{ m s}^{-1}\)

Use the equation of motion for vertical displacement \(s_y = -1.30\text{ m}\) (taking upwards as positive, so \(a = -9.81\text{ m s}^{-2}\)) to find the time of flight, \(t\):
\(s_y = u_y t + \frac{1}{2} a t^2\)

Using \(v = 6.22\text{ m s}^{-1}\):
\(-1.30 = 3.11 t - 4.905 t^2\)
\(4.905 t^2 - 3.11 t - 1.30 = 0\)

Solving the quadratic equation:
\(t = \frac{3.11 \pm \sqrt{(-3.11)^2 - 4(4.905)(-1.30)}}{2(4.905)}\)
\(t = \frac{3.11 \pm \sqrt{9.672 + 25.506}}{9.81} = \frac{3.11 + 5.931}{9.81} = 0.922\text{ s}\)

Using the show-that value \(v = 6.2\text{ m s}^{-1}\):
\(-1.30 = 3.10 t - 4.905 t^2\)
\(4.905 t^2 - 3.10 t - 1.30 = 0\)
\(t = \frac{3.10 \pm \sqrt{(-3.10)^2 - 4(4.905)(-1.30)}}{2(4.905)} = \frac{3.10 + 5.926}{9.81} = 0.920\text{ s}\)

Calculate the horizontal distance \(d\):
\(d = u_x \times t\)
* Using calculated value (\(v = 6.22\text{ m s}^{-1}\)): \(d = 5.39\text{ m s}^{-1} \times 0.922\text{ s} = 4.97\text{ m}\)
* Using show-that value (\(v = 6.2\text{ m s}^{-1}\)): \(d = 5.37\text{ m s}^{-1} \times 0.920\text{ s} = 4.94\text{ m}\)

**(a)**
* **MP1**: Use of \(E_{ep} = \frac{1}{2} k x^2\) (1)
* **MP2**: Equating elastic potential energy to kinetic energy \(\frac{1}{2} m v^2\) (1)
* **MP3**: Correctly calculates \(v = 6.22\text{ m s}^{-1}\) (must show at least 3 s.f. to support the "show that" value of \(6.2\text{ m s}^{-1}\)) (1)

**(b)**
* **MP1**: Calculates both vertical and horizontal components of the launch velocity (1)
* \(u_y = 3.11\text{ m s}^{-1}\) (or \(3.10\text{ m s}^{-1}\))
* \(u_x = 5.39\text{ m s}^{-1}\) (or \(5.37\text{ m s}^{-1}\))
* **MP2**: Use of \(s = u t + \frac{1}{2} a t^2\) with vertical values to set up a quadratic equation or separate kinematic equations to find total vertical travel time (1)
* **MP3**: Correct calculation of time of flight: \(t = 0.92\text{ s}\) (accept \(0.920\text{ s}\) or \(0.922\text{ s}\)) (1)
* **MP4**: Use of \(d = u_x t\) (1)
* **MP5**: Correct final value of horizontal distance between \(4.94\text{ m}\) and \(4.97\text{ m}\) (accept \(4.9\text{ m}\) or \(5.0\text{ m}\) with consistent rounding to 2 s.f.) (1)

*Accept alternative correct multi-stage methods (e.g., finding peak height first and then time to fall).*
*Do not penalize early rounding if working is clearly shown.*

When unpolarised light of intensity \(I_0\) passes through the first polariser, its intensity is halved to \(I_1 = \frac{1}{2}I_0\). When this polarised light passes through the second polariser, we apply Malus's Law: \(I_2 = I_1 \cos^2(\theta)\). Here, \(\theta = 30^\circ\), so \(I_2 = \frac{1}{2}I_0 \cos^2(30^\circ) = \frac{1}{2}I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{8}I_0 = 0.375 I_0\).

B is the correct answer. (1 mark for selecting B). Option A is incorrect because it misses the initial division of unpolarised light intensity by 2. Option C is incorrect because it uses \(\cos(30^\circ)\) instead of \(\cos^2(30^\circ)\). Option D is incorrect because it does not account for the first polariser reducing the intensity of unpolarised light by half.

Resistance is given by \(R = \rho \frac{l}{A}\), where area \(A = \pi \frac{d^2}{4}\). Thus \(R \propto \frac{l}{d^2}\). For the second wire, \(R_2 \propto \frac{2l}{(0.5d)^2} = \frac{2l}{0.25d^2} = 8 \frac{l}{d^2}\). Therefore, \(R_2 = 8R\).

C is the correct answer. (1 mark for selecting C). Option A is incorrect because it only accounts for the change in length. Option B is incorrect because it scales inversely with diameter rather than diameter squared. Option D is incorrect due to a calculation error in squaring \(0.5\).

Using the formula for the critical angle at the boundary between two media: \(\sin(\theta_c) = \frac{n_{\text{liquid}}}{n_{\text{glass}}}\). Rearranging gives \(n_{\text{liquid}} = n_{\text{glass}} \sin(\theta_c) = 1.52 \times \sin(58^\circ) \approx 1.52 \times 0.848 = 1.29\).

C is the correct answer. (1 mark for selecting C). Option A is incorrect because it is calculated as \(\sin(58^\circ)/1.52\). Option B is incorrect because it uses \(\cos\) instead of \(\sin\). Option D is incorrect because it is calculated as \(1.52/\sin(58^\circ)\).

The relationship between terminal potential difference and current is \(V = E - Ir\), which can be written as \(V = -rI + E\). This represents a straight line with a negative gradient of \(-r\) and a y-intercept equal to the e.m.f. \(E\).

B is the correct answer. (1 mark for selecting B). Option A is incorrect because it describes a pure ohmic component. Options C and D are incorrect because the relationship is linear, not curved.

From Einstein's photoelectric equation, \(hf = \phi + E_k\), where \(\phi\) is the work function. When the frequency is \(2f\), the equation becomes \(h(2f) = \phi + E_{k,\text{new}}\). Substituting \(hf\) yields \(2(E_k + \phi) = \phi + E_{k,\text{new}} \implies E_{k,\text{new}} = 2E_k + \phi\). Since the work function \(\phi\) is positive, \(E_{k,\text{new}}\) must be greater than \(2E_k\).

C is the correct answer. (1 mark for selecting C). Option A is incorrect because it fails to account for the work function term staying constant. Option B is incorrect because the threshold frequency effect means the energy increases by more than double. Option D is incorrect because kinetic energy must change when frequency changes.

Since the wires are in series, the current \(I\) is the same in both. Drift velocity is \(v = \frac{I}{nAe}\). Since both are copper, they have the same number density \(n\). Therefore, \(v \propto \frac{1}{A} \propto \frac{1}{d^2}\). Since \(d_X = 2d_Y\), the area is \(A_X = 4A_Y\). Thus, the drift velocity in X is a quarter of that in Y, so \(\frac{v_X}{v_Y} = 0.25\).

A is the correct answer. (1 mark for selecting A). Option B is incorrect as it uses a linear diameter ratio instead of area. Option C is the inverse of the diameter ratio. Option D is the inverse of the area ratio.

For a string fixed at both ends, the \(n\)-th harmonic has \(n\) antinodes and \(n+1\) nodes. Therefore, the third harmonic (\(n=3\)) has 3 antinodes and 4 nodes.

C is the correct answer. (1 mark for selecting C). Option A is incorrect as it describes the second harmonic. Option B is incorrect because the fixed ends must be nodes, making the number of nodes different from antinodes. Option D is incorrect as the number of nodes must exceed antinodes by one.

The grating equation is \(d \sin\theta = n\lambda\). For the first-order maximum (\(n=1\)): \(d \sin(15.0^\circ) = \lambda\). For the second-order maximum (\(n=2\)): \(d \sin\theta_2 = 2\lambda\). Dividing the two equations gives \(\sin\theta_2 = 2 \sin(15.0^\circ) = 2 \times 0.2588 = 0.5176\). Thus, \(\theta_2 = \sin^{-1}(0.5176) \approx 31.2^\circ\).

B is the correct answer. (1 mark for selecting B). Option A is incorrect because it assumes a linear relationship between angle and order. Option C and D are incorrect due to trigonometric calculation errors.

The current \(I\) is the same in both wires because they are connected in series. The drift velocity \(v\) is given by the equation \(I = nAvq\), where \(n\) is the charge carrier density, \(A\) is the cross-sectional area of the wire, and \(q\) is the elementary charge. Since both wires are made of copper, \(n\) and \(q\) are the same. This means \(v \propto \frac{1}{A}\). The cross-sectional area \(A\) is proportional to the square of the diameter, \(A \propto d^2\). Therefore, \(v \propto \frac{1}{d^2}\). This gives \(\frac{v_X}{v_Y} = \left(\frac{d_Y}{d_X}\right)^2 = \left(\frac{3d}{d}\right)^2 = 9\).

D is the correct answer. 1 mark for identifying that the ratio is 9.

The diffraction grating equation is given by \(d \sin\theta = n \lambda\), where \(d\) is the grating spacing and \(n\) is the order of the maximum. For the initial setup: \(d \sin\theta = 3\lambda\). For the second setup with wavelength \(\frac{\lambda}{2}\) and the sixth-order maximum (\(n = 6\)): \(d \sin\theta_{\text{new}} = 6 \left(\frac{\lambda}{2}\right) = 3\lambda\). Since both expressions equal \(3\lambda\), we have \(\sin\theta_{\text{new}} = \sin\theta\), which means \(\theta_{\text{new}} = \theta\).

B is the correct answer. 1 mark for identifying that the angle remains \(\theta\).

(a) The total resistance of the series circuit is:
\(R_{\text{total}} = R_{\text{fixed}} + R_{\text{thermistor}} = 820\ \Omega + 1200\ \Omega = 2020\ \Omega\)

The current in the circuit is:
\(I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{9.0\text{ V}}{2020\ \Omega} \approx 4.455 \times 10^{-3}\text{ A}\)

The potential difference across the fixed resistor is:
\(V_{\text{fixed}} = I \times R_{\text{fixed}} = 4.455 \times 10^{-3}\text{ A} \times 820\ \Omega = 3.65\text{ V}\)

(b) As the temperature of the semiconductor thermistor increases, the thermal energy of the lattice increases, which liberates more charge carriers (conduction electrons) from the valence band to the conduction band. This significantly increases the charge carrier density \(n\). According to \(I = nAve\), the resistance of the thermistor decreases. Since the thermistor's resistance decreases, it takes a smaller share of the total potential difference, causing the potential difference across the fixed resistor to increase.

(c) At the higher temperature, the potential difference across the fixed resistor is \(5.4\text{ V}\).
Therefore, the potential difference across the thermistor is:
\(V_{\text{thermistor}} = 9.0\text{ V} - 5.4\text{ V} = 3.6\text{ V}\)

Using the potential divider ratio:
\(\frac{V_{\text{thermistor}}}{V_{\text{fixed}}} = \frac{R_{\text{thermistor}}}{R_{\text{fixed}}}\)

\(\frac{3.6\text{ V}}{5.4\text{ V}} = \frac{R_{\text{thermistor}}}{820\ \Omega}\)

\(R_{\text{thermistor}} = \frac{3.6}{5.4} \times 820\ \Omega = 547\ \Omega\) (or \(550\ \Omega\) to 2 s.f.)

(a) [3 Marks]
- Calculate total resistance: \(R_{\text{total}} = 2020\ \Omega\) (1)
- Calculate current: \(I = 4.46 \times 10^{-3}\text{ A}\) or use potential divider formula (1)
- Final value: \(3.65\text{ V}\) (accept \(3.6\text{ V}\) to \(3.7\text{ V}\)) (1)

(b) [3 Marks]
- Increased temperature releases more charge carriers / increases number density \(n\) (1)
- This causes the resistance of the thermistor to decrease (1)
- Therefore, the fixed resistor gets a larger share of the total p.d. (1)

(c) [2.75 Marks]
- Calculate potential difference across thermistor as \(3.6\text{ V}\) OR calculate current \(I = 6.59 \times 10^{-3}\text{ A}\) (1)
- Correct substitution into ratio or Ohm's law (1)
- Final answer: \(547\ \Omega\) (accept \(550\ \Omega\)) (0.75)

(a) First, find the grating spacing \(d\):
\(d = \frac{1 \times 10^{-3}\text{ m}}{500} = 2.0 \times 10^{-6}\text{ m}\)

Using the grating equation:
\(d \sin\theta = n\lambda\)

For the maximum possible order, the angle of diffraction \(\theta\) cannot exceed \(90^\circ\). Therefore, \(\sin\theta \le 1\):
\(n \le \frac{d}{\lambda} = \frac{2.0 \times 10^{-6}\text{ m}}{632.8 \times 10^{-9}\text{ m}} \approx 3.16\)

Since the spectral order \(n\) must be an integer, the highest spectral order that can be observed is \(n = 3\).

(b) For the first-order maxima (\(n = 1\)):
For \(\lambda_1 = 589.0\text{ nm}\):
\(\sin\theta_1 = \frac{1 \times 589.0 \times 10^{-9}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 0.2945\)
\(\theta_1 = \sin^{-1}(0.2945) \approx 17.126^\circ\)

For \(\lambda_2 = 589.6\text{ nm}\):
\(\sin\theta_2 = \frac{1 \times 589.6 \times 10^{-9}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 0.2948\)
\(\theta_2 = \sin^{-1}(0.2948) \approx 17.144^\circ\)

The angular separation is:
\(\Delta\theta = \theta_2 - \theta_1 = 17.144^\circ - 17.126^\circ = 0.018^\circ\) (or \(3.1 \times 10^{-4}\text{ rad}\))

(a) [3 Marks]
- Calculate grating spacing \(d = 2.0 \times 10^{-6}\text{ m}\) (1)
- Use of \(d\sin\theta = n\lambda\) with \(\sin\theta = 1\) (1)
- Value \(3.16\) leading to integer answer \(n = 3\) (1)

(b) [5.75 Marks]
- Correct calculation of \(\theta_1 = 17.13^\circ\) (2)
- Correct calculation of \(\theta_2 = 17.14^\circ\) (1.75)
- Subtraction of angles to get angular separation \(\Delta\theta = 0.018^\circ\) (accept \(0.017^\circ\) to \(0.020^\circ\)) (2)

(a) The total electrical energy per unit charge supplied by the cell is the e.m.f. \(\varepsilon\). By conservation of energy, this must equal the electrical energy per unit charge converted into other forms in the circuit. This consists of:
1) Energy per unit charge delivered to the external load, which is the terminal potential difference \(V\).
2) Wasted energy per unit charge lost as heat in the internal resistance, which is \(Ir\).

Therefore:
\(\varepsilon = V + Ir\)

Rearranging for \(V\) yields:
\(V = -rI + \varepsilon\)

(b) Using the given values, we set up two simultaneous equations:
1) \(1.38 = \varepsilon - 0.40 r\)
2) \(1.11 = \varepsilon - 0.85 r\)

Subtracting equation (1) from equation (2):
\(1.11 - 1.38 = (\varepsilon - 0.85 r) - (\varepsilon - 0.40 r)\)
\(-0.27 = -0.45 r\)
\(r = \frac{-0.27}{-0.45} = 0.60\ \Omega\)

Substitute \(r = 0.60\ \Omega\) back into equation (1):
\(1.38 = \varepsilon - 0.40(0.60)\)
\(1.38 = \varepsilon - 0.24\)
\(\varepsilon = 1.38 + 0.24 = 1.62\text{ V}\)

(a) [3 Marks]
- Statement that total energy supplied per unit charge (e.m.f.) equals useful external energy plus wasted internal energy per unit charge (1)
- Identification of terminal p.d. \(V\) as external work done per unit charge and \(Ir\) as work done per unit charge against internal resistance (1)
- Correct algebraic steps to obtain \(V = -rI + \varepsilon\) (1)

(b) [5.75 Marks]
- Setup of two simultaneous equations based on \(V = \varepsilon - Ir\) (2)
- Eliminate \(\varepsilon\) to find \(r\) (1)
- Value of \(r = 0.60\ \Omega\) (1)
- Value of \(\varepsilon = 1.62\text{ V}\) (1.75)

(a) The energy of a photon is given by:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{380 \times 10^{-9}\text{ m}}\)
\(E \approx 5.234 \times 10^{-19}\text{ J}\), which is approximately \(5.2 \times 10^{-19}\text{ J}\).

(b) First convert the work function from eV to Joules:
\(\Phi = 2.28\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.648 \times 10^{-19}\text{ J}\)

According to Einstein's photoelectric equation:
\(E_{k,\text{max}} = E - \Phi\)
\(E_{k,\text{max}} = 5.234 \times 10^{-19}\text{ J} - 3.648 \times 10^{-19}\text{ J} = 1.586 \times 10^{-19}\text{ J}\) (or \(1.59 \times 10^{-19}\text{ J}\))

(c) The maximum kinetic energy is related to speed by:
\(E_{k,\text{max}} = \frac{1}{2} m_e v^2\)
\(v = \sqrt{\frac{2 E_{k,\text{max}}}{m_e}} = \sqrt{\frac{2 \times 1.586 \times 10^{-19}\text{ J}}{9.11 \times 10^{-31}\text{ kg}}} = \sqrt{3.482 \times 10^{11}}\)
\(v \approx 5.90 \times 10^5\text{ m s}^{-1}\)

(a) [2 Marks]
- Use of \(E = hc/\lambda\) (1)
- Correct calculation leading to \(5.2 \times 10^{-19}\text{ J}\) (must show at least 3 s.f., e.g. \(5.23 \times 10^{-19}\text{ J}\)) (1)

(b) [2.75 Marks]
- Conversion of work function into joules (\(3.65 \times 10^{-19}\text{ J}\)) (1)
- Application of Einstein's photoelectric equation (1)
- Correct calculation of \(1.59 \times 10^{-19}\text{ J}\) (accept \(1.55 \times 10^{-19}\text{ J}\) to \(1.60 \times 10^{-19}\text{ J}\)) (0.75)

(c) [4 Marks]
- Use of \(E_k = \frac{1}{2}mv^2\) (1)
- Rearrangement for \(v\) (1)
- Substitution of correct mass and kinetic energy (1)
- Final value \(5.9 \times 10^5\text{ m s}^{-1}\) (accept \(5.8 \times 10^5\) to \(6.0 \times 10^5\)) (1)

(a) Microwaves emitted by the transmitter travel towards the vertical metal plate and are reflected.
The incident waves and the reflected waves travel in opposite directions with the same frequency and amplitude.
These waves superpose (interfere). At points where they meet in phase, constructive interference occurs, producing antinodes (maxima). At points where they meet in antiphase (phase difference of \(\pi\) rad or \(180^\circ\)), destructive interference occurs, producing nodes (minima). This forms a stationary (standing) wave.

(b) In a standing wave, the distance between adjacent nodes (minima) is equal to half a wavelength (\(\lambda/2\)).
The distance from the 1st minimum to the 5th minimum represents 4 adjacent half-wavelength intervals:
\(4 \times \frac{\lambda}{2} = 2\lambda = 5.6\text{ cm}\)

Therefore, the wavelength \(\lambda\) is:
\(\lambda = \frac{5.6\text{ cm}}{2} = 2.8\text{ cm} = 0.028\text{ m}\)

Using the wave equation:
\(c = f\lambda\)
\(f = \frac{c}{\lambda} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{0.028\text{ m}} \approx 1.07 \times 10^{10}\text{ Hz}\) (or \(10.7\text{ GHz}\))

(a) [4 Marks]
- Reflection of waves from the metal plate (1)
- Two waves of same frequency/wavelength traveling in opposite directions superpose (1)
- Nodes formed where waves are in antiphase / destructively interfere (1)
- Antinodes formed where waves are in phase / constructively interfere (1)

(b) [4.75 Marks]
- Recognition that distance between adjacent minima is \(\lambda/2\) (1)
- Equation relating distance to wavelength: \(2\lambda = 5.6\text{ cm}\) (1)
- Correct calculation of \(\lambda = 0.028\text{ m}\) (1)
- Calculation of \(f = 1.07 \times 10^{10}\text{ Hz}\) (accept \(1.1 \times 10^{10}\text{ Hz}\)) (1.75)

(a) The cross-sectional area \(A\) is given by:
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.80 \times 10^{-3}\text{ m})^2}{4} \approx 5.027 \times 10^{-7}\text{ m}^2\), which is approximately \(5.0 \times 10^{-7}\text{ m}^2\).

(b) Using the drift velocity formula:
\(I = nAve\)

Rearranging for drift velocity \(v\):
\(v = \frac{I}{nAe}\)
\(v = \frac{2.5\text{ A}}{(8.5 \times 10^{28}\text{ m}^{-3}) \times (5.027 \times 10^{-7}\text{ m}^2) \times (1.60 \times 10^{-19}\text{ C})}\)
\(v = \frac{2.5}{6836.7} \approx 3.66 \times 10^{-4}\text{ m s}^{-1}\) (or \(3.65 \times 10^{-4}\text{ m s}^{-1}\) if using \(5.0 \times 10^{-7}\text{ m}^2\)).

(c) From the equation \(v = \frac{I}{nAe}\):
Since the current \(I\) is held constant, the number density \(n\) (a constant for copper), the cross-sectional area \(A\), and the elementary charge \(e\) remain unchanged. Therefore, there is no effect on the drift velocity of the conduction electrons (it remains constant).

(a) [2 Marks]
- Correct use of area formula \(A = \pi r^2\) or \(A = \frac{\pi d^2}{4}\) (1)
- Correct calculation leading to \(5.0 \times 10^{-7}\text{ m}^2\) (must show at least 3 s.f.) (1)

(b) [3.75 Marks]
- Recall of formula \(I = nAve\) (1)
- Correct rearrangement to \(v = I / nAe\) (1)
- Substitution of correct values including \(e = 1.6 \times 10^{-19}\text{ C}\) (1)
- Final value \(3.65 \times 10^{-4}\text{ m s}^{-1}\) or \(3.66 \times 10^{-4}\text{ m s}^{-1}\) (0.75)

(c) [3 Marks]
- State that drift velocity is determined by \(v = I / nAe\) (1)
- State that \(n\), \(A\), and \(e\) are independent of temperature (or remain constant) (1)
- Conclude that since current \(I\) is constant, there is no change in the drift velocity (1)

(a) The critical angle \(\theta_c\) is given by:
\(\sin\theta_c = \frac{n_{\text{air}}}{n_{\text{glass}}} = \frac{1.00}{1.52} \approx 0.6579\)
\(\theta_c = \sin^{-1}(0.6579) \approx 41.1^\circ\)

(b) For a boundary between two medium of refractive indices \(n_1\) (glass) and \(n_2\) (liquid), where \(n_1 > n_2\):
\(\sin\theta_c = \frac{n_2}{n_1} = \frac{n_{\text{liquid}}}{1.52}\)

Substitute \(\theta_c = 62.0^\circ\):
\(\sin(62.0^\circ) = \frac{n_{\text{liquid}}}{1.52}\)
\(n_{\text{liquid}} = 1.52 \times \sin(62.0^\circ) = 1.52 \times 0.8829 \approx 1.34\)

(c) The angle of incidence is \(\theta_i = 65.0^\circ\).
Since the angle of incidence (\(65.0^\circ\)) is greater than the critical angle for the glass-liquid boundary (\(62.0^\circ\)), the ray cannot be refracted into the liquid. It undergoes total internal reflection (TIR) and is completely reflected back into the glass prism.

(a) [3 Marks]
- Use of \(\sin\theta_c = 1/n\) (1)
- Correct substitution (1)
- Final value \(41.1^\circ\) (accept \(41^\circ\)) (1)

(b) [3.75 Marks]
- Use of \(\sin\theta_c = n_2 / n_1\) (1)
- Rearrangement to make \(n_2\) the subject (1)
- Substitution of correct values (1)
- Final value \(1.34\) (0.75)

(c) [2 Marks]
- Comparison of incidence angle (\(65.0^\circ\)) with critical angle (\(62.0^\circ\)) (1)
- Correctly identifies that total internal reflection occurs (1)

(a) In plane-polarized light, the oscillations of the electric field vector (or wave vibrations) are restricted to a single plane which contains the direction of wave propagation.

(b) For unpolarized light passing through an ideal polarizing filter, the transmitted intensity is exactly half of the incident intensity:
\(I_1 = \frac{1}{2} I_0\)

(c) According to Malus's Law:
\(I = I_1 \cos^2\theta\)

We require the transmitted intensity \(I\) to be \(25\%\) of \(I_1\), which means:
\(I = 0.25 I_1\)

Therefore:
\(0.25 I_1 = I_1 \cos^2\theta\)
\(\cos^2\theta = 0.25\)
\(\cos\theta = \sqrt{0.25} = 0.5\)
\(\theta = \cos^{-1}(0.5) = 60^\circ\) (as shown).

(d) Sunlight reflected off horizontal surfaces, such as a wet road, becomes partially or fully horizontally polarized. Polarizing sunglasses are designed with a vertical transmission axis. This vertical transmission axis completely absorbs/blocks the horizontally polarized reflected light (glare) while letting vertically polarized light pass through, thereby reducing the intensity of the glare.

(a) [2 Marks]
- Oscillations are restricted to a single plane (1)
- This plane is perpendicular to the direction of wave travel / energy transfer (1)

(b) [1 Mark]
- Correct relationship: \(I_1 = 0.5 I_0\) or \(I_1 = I_0 / 2\) (1)

(c) [2.75 Marks]
- Recall and state Malus's law \(I = I_1 \cos^2\theta\) (1)
- Substitute \(I = 0.25 I_1\) to get \(\cos^2\theta = 0.25\) (1)
- Take square root and find \(\theta = 60^\circ\) (0.75)

(d) [3 Marks]
- Reflected light from a horizontal surface is horizontally polarized (1)
- Sunglasses have a vertical transmission axis / vertical polaroid (1)
- The sunglasses block/absorb the horizontally polarized light (reducing glare) (1)