2024 AQA IAL Physics (9630) 模擬試題連答案詳解

First, resolve the initial velocity into horizontal and vertical components: Horizontal component \(u_x = 22.0 \cos(35.0^\circ) = 18.02 \text{ m/s}\). Vertical component \(u_y = 22.0 \sin(35.0^\circ) = 12.62 \text{ m/s}\) upwards. Next, use the equation of motion for vertical displacement \(s_y = u_y t - \frac{1}{2} g t^2\). Setting \(s_y = -45.0 \text{ m}\) (taking downwards as negative) and \(g = 9.81 \text{ m/s}^2\), we get: \(-45.0 = 12.62 t - 4.905 t^2\). Rearranging gives the quadratic equation: \(4.905 t^2 - 12.62 t - 45.0 = 0\). Solving for positive time \(t\): \(t = \frac{12.62 + \sqrt{(-12.62)^2 - 4(4.905)(-45.0)}}{2 \times 4.905} = 4.58 \text{ s}\). Finally, calculate the horizontal distance \(x = u_x \times t = 18.02 \text{ m/s} \times 4.58 \text{ s} = 82.5 \text{ m}\).

1 mark: Correct horizontal velocity component \(18.02 \text{ m/s}\). 1 mark: Correct vertical velocity component \(12.62 \text{ m/s}\). 1 mark: Correct formulation of quadratic equation for vertical displacement. 1 mark: Calculating correct time of flight \(t = 4.58 \text{ s}\). 1 mark: Correct formula for horizontal distance \(x = u_x t\). 1 mark: Correct final answer with unit \(82.5 \text{ m}\) (accept range \(82.4 \text{ m}\) to \(82.6 \text{ m}\)).

At the peak of the bridge, the centripetal force is provided by the difference between the gravitational force and the normal reaction force: \(mg - R = \frac{m v^2}{r}\). The car loses contact when \(R = 0\), which gives \(mg = \frac{m v_{max}^2}{r} \implies v_{max} = \sqrt{gr}\). Substituting the values: \(v_{max} = \sqrt{9.81 \times 35.0} = 18.53 \text{ m/s} \approx 18.5 \text{ m/s}\). For the second part, at \(v = 15.0 \text{ m/s}\): \(R = mg - \frac{m v^2}{r} = 1200 \times 9.81 - \frac{1200 \times 15.0^2}{35.0} = 11772 - 7714.3 = 4057.7 \text{ N} \approx 4060 \text{ N}\) (or \(4.06 \times 10^3 \text{ N}\)).

1 mark: Identifies \(R = 0\) as condition for losing contact. 1 mark: Derives or states \(v_{max} = \sqrt{gr}\). 1 mark: Calculates \(v_{max} = 18.5 \text{ m/s}\) (accept \(19 \text{ m/s}\) if \(g=9.8\text{ m/s}^2\) is used). 1 mark: States the force equation \(mg - R = \frac{m v^2}{r}\). 1 mark: Correct substitution of values for \(v = 15.0 \text{ m/s}\). 1 mark: Calculates \(R = 4060 \text{ N}\) with unit (accept range \(4050\text{ N}\) to \(4070\text{ N}\)).

First, calculate the angular frequency \(\omega = \frac{2\pi}{T} = \frac{2\pi}{1.20} = 5.236 \text{ rad/s}\). The maximum acceleration occurs at maximum displacement \(A = 0.0800 \text{ m}\): \(a_{max} = \omega^2 A = (5.236)^2 \times 0.0800 = 27.42 \times 0.0800 = 2.19 \text{ m/s}^2\). The total mechanical energy is given by \(E = \frac{1}{2} m \omega^2 A^2\) or \(E = \frac{1}{2} k A^2\) where \(k = m \omega^2\). Substituting the values: \(E = 0.5 \times 0.450 \times (5.236)^2 \times (0.0800)^2 = 0.5 \times 0.450 \times 27.42 \times 0.00640 = 0.0395 \text{ J}\) (or \(3.95 \times 10^{-2} \text{ J}\)).

1 mark: Calculates \(\omega = 5.24 \text{ rad/s}\). 1 mark: Uses \(a_{max} = \omega^2 A\) with amplitude converted to meters (\(0.0800\text{ m}\)). 1 mark: Calculates \(a_{max} = 2.19 \text{ m/s}^2\) (accept \(2.2 \text{ m/s}^2\)). 1 mark: States the energy formula \(E = \frac{1}{2} m \omega^2 A^2\) or equivalent. 1 mark: Substitutes values correctly into the energy equation. 1 mark: Calculates \(E = 0.0395 \text{ J}\) (accept \(0.039 \text{ J}\) to \(0.040 \text{ J}\)) with correct unit.

First, calculate the swept area of the turbine blades: \(A = \pi r^2 = \pi \times (38.0)^2 = 4536.5 \text{ m}^2\). The mass of air passing through per second is \(\frac{dm}{dt} = \rho A v = 1.20 \times 4536.5 \times 12.0 = 65326 \text{ kg/s} \approx 6.53 \times 10^4 \text{ kg/s}\). The kinetic energy of this mass of air per second (total wind power) is \(P_{wind} = \frac{1}{2} \frac{dm}{dt} v^2 = 0.5 \times 65326 \times 12.0^2 = 4.703 \times 10^6 \text{ W}\) (or \(4.70 \text{ MW}\)). The electrical power output is \(P_{elec} = 0.340 \times P_{wind} = 0.340 \times 4.703 \times 10^6 = 1.599 \times 10^6 \text{ W} \approx 1.60 \text{ MW}\).

1 mark: Calculates swept area \(A = \pi \times (38.0)^2 = 4536.5 \text{ m}^2\). 1 mark: States or uses formula \(\frac{dm}{dt} = \rho A v\). 1 mark: Calculates mass flow rate \(\approx 6.53 \times 10^4 \text{ kg/s}\) (accept \(6.5 \times 10^4 \text{ kg/s}\)). 1 mark: Uses \(P_{wind} = \frac{1}{2} \frac{dm}{dt} v^2\) or \(\frac{1}{2} \rho A v^3\). 1 mark: Applies the \(34.0\%\) efficiency factor to the total wind power. 1 mark: Calculates electrical power output \(1.60 \text{ MW}\) (accept range \(1.58 \text{ MW}\) to \(1.62 \text{ MW}\)) with unit.

First, convert the half-life of Cobalt-60 into seconds: \(T_{1/2} = 5.27 \text{ years} \times 365.25 \text{ days/year} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} = 1.663 \times 10^8 \text{ s}\). The decay constant is \(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.6931}{1.663 \times 10^8 \text{ s}} = 4.167 \times 10^{-9} \text{ s}^{-1} \approx 4.17 \times 10^{-9} \text{ s}^{-1}\). Next, find the initial number of nuclei \(N_0\) using \(A_0 = \lambda N_0\): \(N_0 = \frac{A_0}{\lambda} = \frac{3.40 \times 10^5}{4.167 \times 10^{-9}} = 8.159 \times 10^{13}\) nuclei. The number of remaining nuclei after \(t = 12.0 \text{ years}\) is given by \(N = N_0 e^{-\lambda t}\). Convert \(12.0 \text{ years}\) to seconds: \(t = 12.0 \times 365.25 \times 24 \times 3600 = 3.787 \times 10^8 \text{ s}\). Then \(N = 8.159 \times 10^{13} \times e^{-(4.167 \times 10^{-9} \times 3.787 \times 10^8)} = 8.159 \times 10^{13} \times e^{-1.578} = 8.159 \times 10^{13} \times 0.2064 = 1.684 \times 10^{13} \approx 1.68 \times 10^{13}\) nuclei.

1 mark: Converts half-life to seconds correctly (\(1.66 \times 10^8 \text{ s}\)). 1 mark: Uses \(\lambda = \frac{\ln 2}{T_{1/2}}\). 1 mark: Calculates decay constant \(\lambda = 4.17 \times 10^{-9} \text{ s}^{-1}\). 1 mark: Calculates initial number of nuclei \(N_0 = 8.16 \times 10^{13}\) using \(N_0 = \frac{A_0}{\lambda}\). 1 mark: Uses decay equation \(N = N_0 e^{-\lambda t}\) or \(N = N_0 (0.5)^{t / T_{1/2}}\). 1 mark: Calculates remaining nuclei \(N = 1.68 \times 10^{13}\) (accept range \(1.65 \times 10^{13}\) to \(1.71 \times 10^{13}\)).

First, calculate the mass defect: \(\Delta m = (m_{\text{D}} + m_{\text{T}}) - (m_{\text{He}} + m_{\text{n}}) = (2.01355 + 3.01550) - (4.00150 + 1.00867) = 5.02905 - 5.01017 = 0.01888 \text{ u}\). Convert mass defect to energy in MeV: \(E = 0.01888 \times 931.5 \text{ MeV} = 17.587 \text{ MeV} \approx 17.6 \text{ MeV}\). Convert this single reaction energy to Joules: \(E_{\text{J}} = 17.587 \times 10^6 \times 1.60 \times 10^{-19} \text{ J} = 2.814 \times 10^{-12} \text{ J}\). To release \(1.50 \times 10^{11} \text{ J}\), the number of reactions (and thus the number of deuterium nuclei) required is: \(N = \frac{1.50 \times 10^{11} \text{ J}}{2.814 \times 10^{-12} \text{ J}} = 5.330 \times 10^{22}\) nuclei. The number of moles of deuterium is \(n_{\text{moles}} = \frac{N}{N_A} = \frac{5.330 \times 10^{22}}{6.02 \times 10^{23}} = 0.08854 \text{ mol}\). The mass of deuterium required is \(m = n_{\text{moles}} \times 2.00 \text{ g/mol} = 0.08854 \times 2.00 = 0.177 \text{ g}\) (or \(1.77 \times 10^{-4} \text{ kg}\)).

1 mark: Calculates mass defect \(\Delta m = 0.01888 \text{ u}\). 1 mark: Calculates reaction energy \(17.6 \text{ MeV}\) (accept \(17.58 \text{ MeV}\) to \(17.60 \text{ MeV}\)). 1 mark: Converts energy to Joules \(2.81 \times 10^{-12} \text{ J}\). 1 mark: Calculates number of deuterium nuclei needed \(N = 5.33 \times 10^{22}\). 1 mark: Uses Avogadro constant and molar mass of deuterium to set up mass calculation. 1 mark: Calculates mass of deuterium as \(0.177 \text{ g}\) (or \(1.77 \times 10^{-4} \text{ kg}\)) with unit.

The time constant \(\tau\) is given by \(\tau = RC\). Substituting the values: \(\tau = 150 \times 10^3 \; \Omega \times 47.0 \times 10^{-6} \text{ F} = 7.05 \text{ s}\). The potential difference \(V\) at time \(t\) during discharge is given by \(V = V_0 e^{-t/RC}\). Substituting \(V = 2.50 \text{ V}\) and \(V_0 = 12.0 \text{ V}\): \(2.50 = 12.0 e^{-t/7.05}\). Dividing by \(12.0\): \(0.2083 = e^{-t/7.05}\). Taking the natural logarithm of both sides: \(\ln(0.2083) = -1.569 = -\frac{t}{7.05}\). Rearranging to solve for \(t\): \(t = 1.569 \times 7.05 = 11.06 \text{ s} \approx 11.1 \text{ s}\).

1 mark: States the formula \(\tau = RC\). 1 mark: Calculates the time constant \(\tau = 7.05 \text{ s}\) (or \(7.1 \text{ s}\)). 1 mark: States the discharge equation \(V = V_0 e^{-t/RC}\) (or uses \(V = V_0 e^{-t/\tau}\)). 1 mark: Correct substitution of values into the decay equation. 1 mark: Correct logarithmic rearrangement to make \(t\) the subject. 1 mark: Calculates \(t = 11.1 \text{ s}\) (accept \(11 \text{ s}\) or \(11.0 \text{ s}\)) with unit.

The total energy released from the rest mass of the pion is given by Einstein's mass-energy equation: \(E = m c^2 = 2.40 \times 10^{-28} \text{ kg} \times (3.00 \times 10^8 \text{ m/s})^2 = 2.16 \times 10^{-11} \text{ J}\). Convert this total energy into MeV: \(E_{\text{MeV}} = \frac{2.16 \times 10^{-11} \text{ J}}{1.60 \times 10^{-13} \text{ J/MeV}} = 135 \text{ MeV}\). Since two identical photons are created and momentum must be conserved, the two photons share the energy equally. Thus, the energy of each photon is: \(E_{\gamma} = \frac{E}{2} = 1.08 \times 10^{-11} \text{ J}\) (or \(67.5 \text{ MeV}\)). Use the equation \(E_{\gamma} = h f\) to find the frequency \(f\): \(f = \frac{E_{\gamma}}{h} = \frac{1.08 \times 10^{-11} \text{ J}}{6.63 \times 10^{-34} \text{ J s}} = 1.629 \times 10^{22} \text{ Hz} \approx 1.63 \times 10^{22} \text{ Hz}\).

1 mark: States or uses \(E = mc^2\). 1 mark: Calculates total energy in Joules \(2.16 \times 10^{-11} \text{ J}\). 1 mark: Converts total energy to MeV \(135 \text{ MeV}\) (accept \(140 \text{ MeV}\)). 1 mark: States or uses that the energy of each photon is half of the total energy (\(1.08 \times 10^{-11} \text{ J}\) or \(67.5 \text{ MeV}\)). 1 mark: Uses \(E = hf\). 1 mark: Calculates frequency of each photon \(1.63 \times 10^{22} \text{ Hz}\) (accept range \(1.61 \times 10^{22} \text{ Hz}\) to \(1.65 \times 10^{22} \text{ Hz}\)) with unit.

1. Nuclei are positively charged (containing protons) and therefore experience a strong electrostatic (Coulomb) repulsive force when they approach one another.
2. For fusion to occur, the nuclei must get close enough (within approximately \(10^{-15}\text{ m}\)) for the short-range strong nuclear force to overcome this electrostatic repulsion and bind them together.
3. High temperature corresponds to extremely high mean kinetic energy of the nuclei. This energy enables them to overcome the Coulomb barrier during collisions.
4. High density ensures that there is a large number of nuclei per unit volume, which maximizes the collision frequency. This maintains a reaction rate high enough to release more energy than is lost to the surroundings, thereby sustaining the reaction.

- **M1:** State that nuclei are positively charged and experience electrostatic (Coulomb) repulsion when close. [1 mark]
- **M2:** Explain that high temperature provides high kinetic energy, which is necessary to overcome this electrostatic barrier so they can get close enough for the strong nuclear force to act. [1 mark]
- **M3:** State that high density increases the number of nuclei per unit volume, which increases the frequency of collisions. [1 mark]
- **M4:** Explain that a high collision frequency is required to sustain the reaction rate so that energy output exceeds thermal losses. [1 mark]

1. For a wind turbine to operate continuously, the air must flow through the swept area of the blades.
2. If 100% of the kinetic energy of the incoming wind were extracted, the velocity of the air leaving the turbine blades would be zero (\(v_{\text{exit}} = 0\)).
3. This stationary air would pile up behind the turbine, preventing any further incoming air from passing through the blades.
4. Therefore, to sustain a continuous flow of wind through the turbine, the exiting air must retain some kinetic energy and velocity to carry it away, setting a fundamental theoretical limit on the maximum power extraction efficiency.

- **M1:** State that air must continue to flow past the turbine blades to maintain operation (preventing a blockage or pile-up of air). [1 mark]
- **M2:** State that if 100% of the kinetic energy were extracted, the velocity of the exiting air would be reduced to zero. [1 mark]
- **M3:** Explain that zero-velocity air behind the turbine would stop any subsequent incoming air from passing through. [1 mark]
- **M4:** Conclude that the exiting air must retain some velocity and kinetic energy, establishing a fundamental physical limit on turbine efficiency. [1 mark]

1. For any individual radioactive nucleus, there is a constant probability of decay per unit time, which is represented by the decay constant \(\lambda\).
2. When the sample contains a very large number of nuclei \(N\), the rate of decay (or activity \(A\)) is directly proportional to \(N\), written as \(\frac{dN}{dt} = -\lambda N\).
3. Integrating this rate equation leads directly to the exponential decay law: \(N = N_0 e^{-\lambda t}\) or \(A = A_0 e^{-\lambda t}\).
4. For a sample containing only a few nuclei, random fluctuations are highly significant. We cannot predict exactly when any of the few nuclei will decay, meaning the continuous exponential approximation is no longer valid or reliable.

- **M1:** Define the decay constant \(\lambda\) as the constant probability of decay per unit time for a single nucleus. [1 mark]
- **M2:** State that for large \(N\), the rate of decay (activity) is directly proportional to the number of remaining unstable nuclei (\(A \propto N\) or \(\frac{dN}{dt} = -\lambda N\)). [1 mark]
- **M3:** Explain that integration of this relationship results in the exponential decay law (e.g., \(N = N_0 e^{-\lambda t}\)). [1 mark]
- **M4:** Explain that for a small number of nuclei, statistical fluctuations dominate, so the continuous exponential model cannot reliably predict the decay behavior. [1 mark]

1. In a vacuum, the horizontal velocity component remains constant because there are no horizontal forces. With air resistance, a horizontal drag force acts opposite to the direction of motion, continuously reducing the horizontal velocity component.
2. In a vacuum, the vertical velocity component increases linearly due to constant gravitational acceleration \(g\). With air resistance, the upward vertical drag force reduces the downward acceleration, meaning vertical velocity increases at a slower rate and may approach terminal velocity.
3. The horizontal range of the projectile is significantly reduced because of the loss of horizontal speed.
4. The trajectory is no longer a symmetrical parabola; instead, the path becomes asymmetrical and steeper during the final stage of flight as the horizontal speed drops while falling.

- **M1:** State that air resistance provides a drag force that continuously decreases the horizontal component of velocity (which is constant in a vacuum). [1 mark]
- **M2:** State that drag reduces the downward acceleration, so vertical velocity increases at a slower rate / approaches a terminal velocity (instead of increasing linearly). [1 mark]
- **M3:** State that the horizontal range is reduced compared to motion in a vacuum. [1 mark]
- **M4:** Explain that the trajectory becomes asymmetrical and steeper towards the end of the motion (no longer a symmetrical parabola). [1 mark]

Using the decay equation \( A = A_0 e^{-\lambda t} \), we have \( 40 = 320 e^{-\lambda t} \). This simplifies to \( \frac{40}{320} = e^{-\lambda t} \), so \( \frac{1}{8} = e^{-\lambda t} \). Taking the natural logarithm of both sides: \( \ln(8) = \lambda t \). Given that \( t = 15.0 \text{ hours} = 15.0 \times 3600 \text{ s} = 54000 \text{ s} \), we find \( \lambda = \frac{\ln(8)}{54000} \approx 3.85 \times 10^{-5} \text{ s}^{-1} \).

1 mark for the correct calculation leading to choice B.

First, calculate the total binding energy of the reactants: \( \text{BE}_{\text{reactants}} = (2 \times 1.11) + (3 \times 2.83) = 2.22 + 8.49 = 10.71 \text{ MeV} \). Next, calculate the total binding energy of the products: \( \text{BE}_{\text{products}} = 4 \times 7.07 = 28.28 \text{ MeV} \) (the free neutron has zero binding energy). The energy released is the difference between the total binding energy of the products and reactants: \( Q = 28.28 - 10.71 = 17.57 \text{ MeV} \).

1 mark for the correct energy difference calculation leading to option C.

The power from a wind turbine is proportional to the swept area and the cube of the wind speed: \( P \propto A v^3 \). Since the swept area is circular, \( A \propto L^2 \), leading to \( P \propto L^2 v^3 \). The new power is proportional to \( (1.2L)^2 \times (1.5v)^3 = 1.44 L^2 \times 3.375 v^3 = 4.86 L^2 v^3 \). Thus, the new power is \( 4.86 P \).

1 mark for using the proportional relation and calculating the factor of 4.86.

The maximum height of a projectile is given by \( H = \frac{u^2 \sin^2\theta}{2g} \) and the horizontal range is given by \( R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g} \). We are given that \( H = \frac{1}{4} R \), so: \( \frac{u^2 \sin^2\theta}{2g} = \frac{1}{4} \left( \frac{2u^2 \sin\theta \cos\theta}{g} \right) \). Simplifying this gives: \( \frac{\sin^2\theta}{2} = \frac{\sin\theta \cos\theta}{2} \). Dividing both sides by \( \frac{1}{2} \sin\theta \) yields \( \sin\theta = \cos\theta \), which means \( \tan\theta = 1 \), so \( \theta = 45^\circ \).

1 mark for setting up the equation for H and R, leading to tan(theta) = 1 and 45 degrees.

At the highest point of the vertical circle, the forces acting on the water are the weight \( mg \) downward and the normal force \( N \) from the bottom of the bucket downward. The net downward force provides the centripetal acceleration: \( mg + N = \frac{m v^2}{r} \). For the water to remain in the bucket, the normal force must be greater than or equal to zero (\( N \ge 0 \)). The minimum speed occurs when \( N = 0 \), which gives: \( mg = \frac{m v_{\text{min}}^2}{r} \implies v_{\text{min}} = \sqrt{g r} \).

1 mark for applying the boundary condition of centripetal force equal to gravity.

The total mechanical energy of a simple harmonic oscillator is given by \( E = \frac{1}{2} k A^2 \). Since the angular frequency is \( \omega = 2\pi f \) and \( k = m \omega^2 \), we can substitute for \( k \) to get: \( k = m (2\pi f)^2 = 4\pi^2 f^2 m \). Therefore, the total energy is: \( E = \frac{1}{2} (4\pi^2 f^2 m) A^2 = 2\pi^2 m f^2 A^2 \).

1 mark for using the total energy formula and substitution of k.

By conservation of momentum, the magnitude of momentum of the alpha particle, \( p_{\alpha} \), is equal to that of the daughter Radon nucleus, \( p_{\text{Rn}} \). Since kinetic energy can be expressed as \( E_k = \frac{p^2}{2m} \), the total kinetic energy released is \( Q = E_{k,\alpha} + E_{k,\text{Rn}} = \frac{p^2}{2 m_{\alpha}} + \frac{p^2}{2 m_{\text{Rn}}} \). This can be factored as \( Q = \frac{p^2}{2 m_{\alpha}} \left( 1 + \frac{m_{\alpha}}{m_{\text{Rn}}} \right) = E_{k,\alpha} \left( 1 + \frac{4}{222} \right) = E_{k,\alpha} \left( \frac{226}{222} \right) \). Solving for \( E_{k,\alpha} \) gives \( E_{k,\alpha} = \frac{222}{226} Q \).

1 mark for applying conservation of momentum and energy share to get 222/226.

The mass of air passing through area \( A \) per second (mass flow rate) is \( \frac{dm}{dt} = \rho A v \). The kinetic energy of mass \( m \) is \( E_k = \frac{1}{2} m v^2 \). Therefore, the rate of kinetic energy (power available in the wind) is \( P = \frac{dE_k}{dt} = \frac{1}{2} \left( \frac{dm}{dt} \right) v^2 = \frac{1}{2} (\rho A v) v^2 = \frac{1}{2} \rho A v^3 \).

1 mark for the correct derivation of kinetic energy rate.

Let the initial activity of Y be \( A_{0,Y} = A_0 \). This means the initial activity of X is \( A_{0,X} = 2A_0 \). Using the decay equation in terms of half-life, the activity \( A(t) \) of each nuclide at time \( t \) is given by: \( A_X(t) = 2A_0 \times 2^{-t / 3.0} \) and \( A_Y(t) = A_0 \times 2^{-t / 6.0} \). Setting the two activities equal to find when they intersect: \( 2A_0 \times 2^{-t / 3.0} = A_0 \times 2^{-t / 6.0} \). Dividing both sides by \( A_0 \) gives: \( 2 \times 2^{-t / 3.0} = 2^{-t / 6.0} \). Rearranging the exponents: \( 2^1 = 2^{-t / 6.0} \times 2^{t / 3.0} \implies 2^1 = 2^{t / 6.0} \). Equating the exponents: \( 1 = \frac{t}{6.0} \implies t = 6.0 \text{ hours} \).

1 mark for the correct answer. Reject all other options.

To calculate the energy released, we find the difference between the total binding energy of the products and that of the reactants. Total initial binding energy of reactants = \( (2 \times 1.11) + (3 \times 2.83) = 2.22 + 8.49 = 10.71\text{ MeV} \). Total final binding energy of products = \( 4 \times 7.07 = 28.28\text{ MeV} \) (the free neutron has no binding energy). Energy released \( Q = 28.28 - 10.71 = 17.57\text{ MeV} \).

1 mark for the correct answer. Reject all other options.

The available power from wind is given by \( P_w = \frac{1}{2} \rho A v^3 \), where the swept area is \( A = \pi R^2 \). Therefore, electrical power output \( P \propto R^2 v^3 \). The new electrical power output is: \( P' \propto (1.5R)^2 (0.80v)^3 = 2.25 R^2 \times 0.512 v^3 = 1.152 R^2 v^3 \). Thus, \( P' = 1.152 P \approx 1.15 P \).

1 mark for the correct answer. Reject all other options.

The maximum height reached by a projectile is given by \( H = \frac{u^2 \sin^2\theta}{2g} \). The horizontal range is given by \( R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g} \). We are given that \( H = \frac{1}{4} R \). Substituting the formulas: \( \frac{u^2 \sin^2\theta}{2g} = \frac{1}{4} \left( \frac{2u^2 \sin\theta \cos\theta}{g} \right) \). Simplifying this expression yields: \( \frac{\sin^2\theta}{2} = \frac{\sin\theta \cos\theta}{2} \). Dividing both sides by \( \frac{\sin\theta}{2} \) (for \( \theta \neq 0 \)) gives \( \sin\theta = \cos\theta \), which means \( \tan\theta = 1 \). Therefore, \( \theta = 45^\circ \).

1 mark for the correct answer. Reject all other options.

At the highest point of the vertical circle, the forces acting downwards on the water are its weight \( mg \) and the normal contact force \( N \) from the bottom of the bucket. The centripetal force is provided by the sum of these forces: \( \frac{mv^2}{r} = mg + N \). To prevent the water from falling out, the normal reaction force must satisfy \( N \ge 0 \). At the threshold where the water is just about to lose contact, \( N = 0 \). Therefore, \( \frac{mv_{\text{min}}^2}{r} = mg \implies v_{\text{min}} = \sqrt{gr} \).

1 mark for the correct answer. Reject all other options.

The total mechanical energy in simple harmonic motion is constant and is given by \( E_T = \frac{1}{2} k A^2 \), where \( k \) is the stiffness constant. The potential energy at displacement \( x \) is \( E_p = \frac{1}{2} k x^2 \), and the kinetic energy is \( E_k = E_T - E_p = \frac{1}{2} k (A^2 - x^2) \). We are given that \( E_k = 3E_p \). Substituting the expressions: \( \frac{1}{2} k (A^2 - x^2) = 3 \left( \frac{1}{2} k x^2 \right) \). This simplifies to: \( A^2 - x^2 = 3x^2 \implies A^2 = 4x^2 \implies x^2 = \frac{A^2}{4} \). Taking the square root gives \( x = \pm \frac{A}{2} \).

1 mark for the correct answer. Reject all other options.