2023 Edexcel A-Level Physics (9PH0) 模擬試題連答案詳解

The energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\). The potential difference \(V\) at any time \(t\) during discharge is given by \(V = V_0 e^{-t/RC}\). Substituting \(t = 2\tau = 2RC\) into this equation gives \(V = V_0 e^{-2}\). The energy remaining at this time is therefore \(E = \frac{1}{2} C (V_0 e^{-2})^2 = \frac{1}{2} C V_0^2 e^{-4} = E_0 e^{-4}\). Thus, the fraction of the initial energy remaining is \(e^{-4}\).

[1 mark] Correctly uses the exponential decay formula for potential difference and relates it to the energy stored formula to find the fraction \(e^{-4}\).

By Newton's second law, force is defined as the rate of change of momentum: \(F = \frac{\Delta p}{\Delta t}\). Taking the rebound direction as positive, the initial momentum is \(-mu\) and the final momentum is \(mv\). The change in momentum is \(\Delta p = mv - (-mu) = m(u + v)\). Therefore, the magnitude of the average force is \(F = \frac{m(u + v)}{\Delta t}\).

[1 mark] Recognises that change in momentum is \(m(u+v)\) due to the direction change, and applies Newton's second law to find the force.

An increase in light intensity causes the resistance of the LDR to decrease. Because the total resistance of the series circuit decreases, the current in the circuit increases. According to Ohm's law (\(V = I R\)), since the resistance \(R\) of the fixed resistor remains constant and the current \(I\) increases, the potential difference across the fixed resistor must increase.

[1 mark] Identifies that LDR resistance decreases with light intensity and uses potential divider principles to deduce that the potential difference across the fixed resistor increases.

The neutron (\(n\)) and proton (\(p\)) are baryons and have a lepton number of 0. The electron (\(e^{-}\)) has a lepton number of \(+1\), and the electron antineutrino (\(\overline{\nu}_e\)) has a lepton number of \(-1\). The total lepton number before the decay is 0, and the total lepton number after the decay is \(0 + 1 + (-1) = 0\). Therefore, lepton number is conserved.

[1 mark] Identifies that the total lepton number is 0 before and after the decay, establishing conservation of lepton number.

At the top of the vertical circle, the forces acting downwards are the weight of the water \(mg\) and the normal contact force \(N\) from the bucket. The centripetal force is provided by these downward forces: \(mg + N = \frac{m v^2}{r}\). For the water to stay in the bucket, the normal force must be greater than or equal to zero (\(N \geq 0\)). The minimum speed occurs when \(N = 0\), which yields \(mg = \frac{m v^2}{r}\). Solving for \(v\) gives \(v = \sqrt{gr\)}.

[1 mark] Applies centripetal force equation at the top of the circle and sets the normal force to zero to solve for minimum speed.

According to Faraday's law of electromagnetic induction, the magnitude of the average induced e.m.f. is equal to the rate of change of magnetic flux linkage: \(E = \frac{\Delta \Phi}{\Delta t}\). The initial magnetic flux linkage is \(\Phi_1 = N B A\) (since the coil is perpendicular to the field). The final flux linkage is \(\Phi_2 = 0\) (since the coil is parallel to the field). The change in flux linkage is \(\Delta \Phi = N B A - 0 = N B A\). Therefore, the magnitude of the average induced e.m.f. is \(E = \frac{N B A}{\Delta t}\).

[1 mark] Correctly applies Faraday's law using the change in magnetic flux linkage \(N B A\) over time \(\Delta t\).

The useful work done in lifting the mass is \(W = m g h = 120\text{ kg} \times 9.81\text{ m s}^{-2} \times 15\text{ m} = 17658\text{ J}\). The useful power output of the motor is \(P_{\text{out}} = \frac{W}{t} = \frac{17658\text{ J}}{10\text{ s}} = 1765.8\text{ W}\). Since \(\text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\), the electrical power input is \(P_{\text{in}} = \frac{P_{\text{out}}}{0.75} = \frac{1765.8\text{ W}}{0.75} \approx 2354.4\text{ W} \approx 2.4\text{ kW}\).

[1 mark] Calculates useful power output and divides by efficiency to determine input power of \(2.4\text{ kW}\).

For the oil drop to remain stationary, the upward electric force must balance the downward gravitational force: \(F_{\text{electric}} = F_{\text{gravity}} \Rightarrow q E = m g\). The uniform electric field strength \(E\) between the plates is given by \(E = \frac{V}{d}\). Substituting this expression into the force equation gives \(q \left(\frac{V}{d}\right) = m g\). Rearranging this formula results in \(q V = m g d\).

[1 mark] Equates gravitational and electric forces on the charged drop, substitutes the electric field strength formula, and simplifies to find \(q V = m g d\).

Useful work done on the vehicle by the constant power \( P \) in time \( t \) is given by \( W = P \times t \). Since resistive forces are negligible and the vehicle starts from rest, all this work is converted into the kinetic energy of the vehicle: \( E_k = \frac{1}{2}mv^2 \). Setting work equal to kinetic energy: \( Pt = \frac{1}{2}mv^2 \). Rearranging for velocity \( v \): \( v^2 = \frac{2Pt}{m} \), which gives \( v = \sqrt{\frac{2Pt}{m}} \). This matches option A.

1 mark for selecting the correct option A. Reject all other options as they do not correctly relate the constant power input to the gain in kinetic energy.

According to Faraday's law, the average induced e.m.f. is given by \( \mathcal{E} = \frac{\Delta \Phi_N}{\Delta t} \), where \( \Phi_N = N\Phi \) is the magnetic flux linkage. The initial flux linkage is \( \Phi_{\text{initial}} = N B A = N B L^2 \). The final flux linkage is \( 0 \), so the magnitude of the change in flux linkage is \( \Delta \Phi_N = N B L^2 \). The distance moved by the coil to be completely pulled out is \( L \). At a constant speed \( v \, the time taken is \) \Delta t = \frac{L}{v} \). Therefore, the average induced e.m.f. is \( \mathcal{E} = \frac{N B L^2}{L / v} = N B L v \). This matches option A.

1 mark for selecting the correct option A. Reject all other options because they do not correctly apply Faraday's law of electromagnetic induction to the motion of the square coil.

When the copper disk rotates through the magnetic field of the electromagnet, the magnetic flux linkage through the disk changes. According to Faraday's law of electromagnetic induction, this changing flux linkage induces an electromotive force (e.m.f.) within the disk. Since copper is a good electrical conductor, this induced e.m.f. causes circular currents, known as eddy currents, to flow through the body of the disk. According to Lenz's law, the direction of these induced currents must oppose the change in magnetic flux that created them. These currents interact with the external magnetic field to produce a magnetic force that acts in the opposite direction to the rotation of the disk, resulting in a decelerating torque. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage, which increases as the rotational speed of the disk increases, leading to larger eddy currents and a stronger braking force at higher speeds. If a material with lower electrical conductivity is used, the resistance of the path is greater, which reduces the magnitude of the induced eddy currents for the same induced e.m.f., thereby reducing the braking force.

Mark 1: Mention of changing magnetic flux linkage through the rotating copper disk. Mark 2: Application of Faraday's law to explain the induction of an e.m.f. in the disk. Mark 3: Explanation that eddy currents flow in the disk because copper is a conductor. Mark 4: Application of Lenz's law to show that the induced currents create a force/torque opposing the rotation. Mark 5: Explanation that higher rotational speed increases the rate of change of flux linkage, resulting in larger induced e.m.f., larger currents, and a stronger braking force. Mark 6: Explanation that a lower conductivity material increases resistance, resulting in smaller currents and a weaker braking force.

In the ideal case without air resistance, the horizontal component of velocity remains constant because there are no horizontal forces acting on the ball, and the vertical acceleration is constant at \(g\) downwards. When air resistance is present, a drag force acts in the opposite direction to the ball's velocity vector. Horizontally, this drag force acts against the motion, causing a continuous deceleration, so the horizontal velocity decreases throughout the flight. Vertically, on the way up, both gravity and the vertical component of the drag force act downwards, meaning the vertical deceleration is greater than \(g\). At the peak of the flight, the vertical velocity is momentarily zero, so the vertical drag force is zero, and the acceleration is exactly \(g\). On the descent, the vertical component of the drag force acts upwards, opposing gravity, so the downward acceleration is less than \(g\). As a result of these forces, the ball does not follow a symmetric parabola: the peak of the trajectory is lower and occurs closer horizontally to the launch point, the descent is steeper than the ascent, and the overall horizontal range is significantly reduced.

Mark 1: State that drag acts in the opposite direction to motion and causes continuous horizontal deceleration (horizontal velocity decreases). Mark 2: Explain that on the way up, both gravity and drag act downwards, making the vertical deceleration greater than \(g\). Mark 3: State that at the peak of the flight, the vertical velocity is zero, so the vertical component of drag is zero (acceleration is \(g\)). Mark 4: Explain that on the way down, drag acts upwards, making the downward acceleration less than \(g\). Mark 5: Describe the trajectory as asymmetric, stating that the peak is lower and shifted horizontally towards the launch point. Mark 6: Describe the descent as steeper than the ascent and state that the overall range is reduced.

In an NTC thermistor, which is a semiconductor, increasing the temperature provides thermal energy to the material. This thermal energy liberates a significant number of charge carriers (electrons) by moving them from the valence band to the conduction band, increasing the number density \(n\) of charge carriers. Although the increased temperature also increases lattice vibrations, which leads to more frequent collisions between charge carriers and the lattice, the dramatic increase in the number of charge carriers is the dominant effect. Consequently, the resistance of the thermistor decreases as its temperature increases. In this series circuit, the total resistance of the potential divider decreases, which causes the current \(I\) from the constant DC supply to increase. According to \(V = IR\), since the resistance of the fixed resistor remains constant, the increased current results in a larger potential difference across it. Therefore, the output voltage \(V_{\text{out}}\) across the fixed resistor increases.

Mark 1: State that increasing temperature provides thermal energy which liberates charge carriers in the semiconductor. Mark 2: Explain that this leads to a significant increase in the number density \(n\) of charge carriers. Mark 3: Mention that lattice vibrations increase but the increase in charge carriers is the dominant factor. Mark 4: Conclude that the resistance of the NTC thermistor decreases. Mark 5: Explain that the total circuit resistance decreases, so the circuit current increases. Mark 6: Use \(V = IR\) to explain that because the fixed resistance is constant and the current has increased, the output voltage \(V_{\text{out}}\) increases.

Before the interaction, the incoming gamma-ray photon has zero electric charge. After the interaction, the electron has a charge of \(-1e\) and the positron has a charge of \(+1e\), giving a total charge of zero, which demonstrates the conservation of charge. In the uniform magnetic field, the magnetic force on each particle is given by \(F = Bqv\). Because the particles have opposite charges, they experience forces in opposite directions, causing them to curve in opposite directions, confirming their opposite signs. Regarding conservation of momentum, the incoming photon has momentum in the forward direction. The produced electron and positron must also have a net forward component of momentum to match this, which is why their curved tracks continue in a generally forward direction. As the charged particles travel through the liquid hydrogen, they collide with and ionize the hydrogen atoms, which expends their kinetic energy. As a result, their velocity \(v\) decreases. Since the radius of the circular path in a magnetic field is given by \(r = \frac{mv}{Bq}\), a decrease in velocity \(v\) leads to a continuously decreasing radius of curvature \(r\), causing the tracks to spiral inwards.

Mark 1: State that the initial charge of the photon is zero and the sum of charges of the electron and positron is zero, showing conservation of charge. Mark 2: Explain that the opposite curvature of the paths is due to the magnetic force \(F = Bqv\) acting in opposite directions on opposite charges. Mark 3: State that the forward direction of both tracks demonstrates the conservation of momentum of the incoming photon. Mark 4: Explain that the particles lose kinetic energy/velocity as they travel through the liquid hydrogen due to ionization collisions. Mark 5: State the formula for the radius of a charged particle in a magnetic field, \(r = \frac{mv}{Bq}\) (or equivalent proportional reasoning). Mark 6: Conclude that as velocity \(v\) decreases, the radius of curvature \(r\) decreases, causing the paths to spiral inwards.

(a) For undeflected travel, the magnetic force and electric force must be equal in magnitude and opposite in direction: \(F_E = F_B \implies qE = qvB\). Dividing both sides by \(q\) gives \(E = vB\), which rearranges to \(v = \frac{E}{B}\). (b) The electric field strength is \(E = \frac{V}{d} = \frac{6000\text{ V}}{0.015\text{ m}} = 4.0 \times 10^5\text{ V m}^{-1}\). Using the relation from part (a), the velocity is \(v = \frac{E}{B} = \frac{4.0 \times 10^5\text{ V m}^{-1}}{0.45\text{ T}} = 8.89 \times 10^5\text{ m s}^{-1}\). (c) In the magnetic field, the magnetic force acts as the centripetal force: \(qvB = \frac{mv^2}{r} \implies r = \frac{mv}{Bq}\). Substituting the values: \(r = \frac{6.64 \times 10^{-27}\text{ kg} \times 8.89 \times 10^5\text{ m s}^{-1}}{0.45\text{ T} \times (2 \times 1.60 \times 10^{-19}\text{ C})} = 4.10 \times 10^{-2}\text{ m}\).

(a) Equates electric force \(qE\) and magnetic force \(qvB\) [1 Mark]. Rearranges algebraically to show \(v = E/B\) [1 Mark]. (b) Uses \(E = V/d\) to find \(E = 4.0 \times 10^5\text{ V m}^{-1}\) [1 Mark]. Substitutes \(E\) and \(B\) into \(v = E/B\) [1 Mark]. Obtains \(v = 8.89 \times 10^5\text{ m s}^{-1}\) [1 Mark]. (c) Recalls or derives \(r = mv/Bq\) [1 Mark]. Identifies \(q = 2 \times 1.60 \times 10^{-19}\text{ C} = 3.20 \times 10^{-19}\text{ C}\) [1 Mark]. Correctly calculates \(r = 4.1 \times 10^{-2}\text{ m}\) [1 Mark].

(a) From conservation of momentum: \(m \vec{v}_{A1} = m \vec{v}_{A2} + m \vec{v}_{B2} \implies \vec{v}_{A1} = \vec{v}_{A2} + \vec{v}_{B2}\). Squaring both sides: \(v_{A1}^2 = v_{A2}^2 + v_{B2}^2 + 2\vec{v}_{A2} \cdot \vec{v}_{B2}\). Since the collision is elastic, kinetic energy is conserved: \(\frac{1}{2}mv_{A1}^2 = \frac{1}{2}mv_{A2}^2 + \frac{1}{2}mv_{B2}^2 \implies v_{A1}^2 = v_{A2}^2 + v_{B2}^2\). Equating these yields \(2\vec{v}_{A2} \cdot \vec{v}_{B2} = 0\), which means the velocity vectors are perpendicular (\(90^\circ\) apart). Since stone A is at \(30.0^\circ\), stone B must move at \(\theta = 90.0^\circ - 30.0^\circ = 60.0^\circ\). (b) Using the vector triangle: \(v_{A2} = v_{A1} \cos(30.0^\circ) = 2.40 \times 0.8660 = 2.08\text{ m s}^{-1}\). \(v_{B2} = v_{A1} \sin(30.0^\circ) = 2.40 \times 0.5000 = 1.20\text{ m s}^{-1}\). (c) If kinetic energy is lost, the final kinetic energy is less than the initial: \(v_{A2}^2 + v_{B2}^2 < v_{A1}^2\). From the law of cosines: \(v_{A1}^2 = v_{A2}^2 + v_{B2}^2 + 2 v_{A2} v_{B2} \cos\phi\), where \(\phi\) is the angle between their final directions. Because \(v_{A1}^2 > v_{A2}^2 + v_{B2}^2\), we must have \(\cos\phi > 0\), which means \(\phi < 90^\circ\). Thus, the angle between their paths is less than \(90^\circ\).

(a) Expresses vector momentum conservation and squares it or draws a vector triangle [1 Mark]. Shows \(v_{A1}^2 = v_{A2}^2 + v_{B2}^2\) from KE conservation and concludes dot product is zero (angle is \(90^\circ\)) [1 Mark]. Correctly deduces \(\theta = 60.0^\circ\) [1 Mark]. (b) Relates final speeds to initial speed using trigonometry, e.g., \(v_{A2} = v_{A1}\cos(30^\circ)\) [1 Mark]. Calculates \(v_{A2} = 2.08\text{ m s}^{-1}\) [1 Mark]. Calculates \(v_{B2} = 1.20\text{ m s}^{-1}\) [1 Mark]. (c) Argues that \(v_{A1}^2 > v_{A2}^2 + v_{B2}^2\) when KE is lost [1 Mark]. Explains that the dot product (or \(\cos\phi\)) must be positive, making the angle less than \(90^\circ\) [1 Mark].

(a) The total resistance of the circuit is \(R_{\text{total}} = R + r\). The current is \(I = \frac{\varepsilon}{R + r}\). The power dissipated in the load resistor \(R\) is given by \(P = I^2 R = \left(\frac{\varepsilon}{R + r}\right)^2 R = \frac{\varepsilon^2 R}{(R + r)^2}\). (b) According to the maximum power transfer theorem, maximum power is delivered to the load when the load resistance \(R\) equals the internal resistance \(r\). Therefore, \(r = R = 4.5\ \Omega\). The maximum power is given by: \(P_{\max} = \frac{\varepsilon^2 r}{(r + r)^2} = \frac{\varepsilon^2}{4r}\). Given \(P_{\max} = 1.8\text{ W}\) and \(r = 4.5\ \Omega\): \(1.8 = \frac{\varepsilon^2}{18} \implies \varepsilon^2 = 32.4 \implies \varepsilon = \sqrt{32.4} \approx 5.69\text{ V}\). (c) Efficiency \(\eta\) is the ratio of useful power output to the total power input: \(\eta = \frac{I^2 R}{I^2(R + r)} = \frac{R}{R + r}\). When maximum power is delivered, \(R = r\), so \(\eta = \frac{r}{2r} = 0.50\) or \(50\%\). The remaining \(50\%\) of the energy is wasted as thermal energy in the internal resistance of the cell.

(a) Expresses current as \(I = \varepsilon / (R+r)\) [1 Mark]. Substitutes \(I\) into \(P = I^2 R\) to obtain the target equation [1 Mark]. (b) Identifies that maximum power occurs when \(R = r\), giving \(r = 4.5\ \Omega\) [1 Mark]. Recalls or derives \(P_{\max} = \varepsilon^2 / 4r\) [1 Mark]. Substitutes values to get \(\varepsilon^2 = 32.4\) [1 Mark]. Calculates \(\varepsilon = 5.7\text{ V}\) (or \(5.69\text{ V\text{)}}) [1 Mark]. (c) Defines efficiency as \)\frac{\text{Power in load}}{\text{Total power}}\) or \(\frac{R}{R+r}\) [1 Mark]. Explains that since \(R = r\), the power is shared equally between the load and the internal resistance, leading to exactly \(50\%\) efficiency [1 Mark].

(a) Find the mass deficit \(\Delta m\): \(\text{Mass of reactants} = 18.998403 + 1.007825 = 20.006228\text{ u}\). \(\text{Mass of products} = 19.001880 + 1.008665 = 20.010545\text{ u}\). \(\Delta m = 20.010545 - 20.006228 = 0.004317\text{ u}\). \(\Delta E = 0.004317 \times 931.5\text{ MeV} = 4.021\text{ MeV}\). (b) At threshold, the products move together with a single final velocity \(v_f\). By conservation of momentum: \(m_{\text{p}} v_i = (m_{\text{F}} + m_{\text{p}}) v_f \implies v_f = \frac{m_{\text{p}}}{m_{\text{F}} + m_{\text{p}}} v_i\). The minimum kinetic energy of the products is: \(E_{kf} = \frac{1}{2}(m_{\text{F}} + m_{\text{p}}) v_f^2 = E_{\text{th}} \left(\frac{m_{\text{p}}}{m_{\text{F}} + m_{\text{p}}}\right)\). By conservation of energy: \(E_{\text{th}} = E_{kf} + \Delta E \implies E_{\text{th}} - E_{\text{th}}\left(\frac{m_{\text{p}}}{m_{\text{F}} + m_{\text{p}}}\right) = \Delta E\). \(E_{\text{th}} \left(\frac{m_{\text{F}}}{m_{\text{F}} + m_{\text{p}}}\right) = \Delta E \implies E_{\text{th}} = \Delta E \left(1 + \frac{m_{\text{p}}}{m_{\text{F}}}\right)\). (c) Substitute the values: \(E_{\text{th}} = 4.021\text{ MeV} \times \left(1 + \frac{1.007825}{18.998403}\right) = 4.021 \times 1.05305 = 4.234\text{ MeV}\).

(a) Calculates mass of reactants and mass of products [1 Mark]. Finds mass deficit \(\Delta m = 0.004317\text{ u}\) [1 Mark]. Calculates \(\Delta E = 4.02\text{ MeV}\) [1 Mark]. (b) Applies conservation of momentum to relate \(v_f\) to \(v_i\) [1 Mark]. Sets up conservation of energy equation: \(E_{\text{th}} = E_{kf} + \Delta E\) [1 Mark]. Performs algebraic substitution and simplification to reach the given formula [1 Mark]. (c) Substitutes masses and \(\Delta E\) into the threshold formula [1 Mark]. Calculates \(E_{\text{th}} = 4.23\text{ MeV}\) [1 Mark].

(a) The discharge formula is \(V = V_0 e^{-t/RC}\). Taking the natural logarithm: \(\ln(V) = \ln(V_0 e^{-t/RC}) = -\frac{1}{RC} t + \ln(V_0)\). This matches the straight-line equation \(y = mx + c\), where the gradient \(m = -\frac{1}{RC}\), and the y-intercept \(c = \ln(V_0)\). (b) The gradient is given as \(-0.0455\text{ s}^{-1}\). Thus: \(-\frac{1}{RC} = -0.0455 \implies RC = 21.98\text{ s}\). Given \(C = 220 \times 10^{-6}\text{ F}\): \(R = \frac{21.98}{220 \times 10^{-6}\text{ F}} = 99900\ \Omega \approx 1.00 \times 10^5\ \Omega\). (c) The initial energy stored is \(E_0 = \frac{1}{2} C V_0^2 = 0.5 \times (220 \times 10^{-6}) \times 12.0^2 = 0.01584\text{ J}\). After two time constants (\(t = 2\tau\)), the potential difference is \(V = V_0 e^{-2}\). The energy remaining is \(E = \frac{1}{2} C V^2 = E_0 e^{-4} = 0.01584 \times e^{-4} = 2.90 \times 10^{-4}\text{ J}\).

(a) Recalls the capacitor discharging formula \(V = V_0 e^{-t/RC}\) [1 Mark]. Correctly applies logarithmic rules to obtain \(\ln(V) = -\frac{1}{RC} t + \ln(V_0)\) and links to \(y = mx + c\) [1 Mark]. (b) Equates gradient to \(-1/RC\) [1 Mark]. Substitutes \(C = 220 \times 10^{-6}\text{ F}\) [1 Mark]. Obtains \(R = 1.0 \times 10^5\ \Omega\) (accept range \(9.9 \times 10^4\) to \(1.0 \times 10^5\ \Omega\)) [1 Mark]. (c) Calculates initial energy \(E_0 = 0.0158\text{ J}\) or recognizes energy scales as \(e^{-4}\) [1 Mark]. Applies \(V = V_0 e^{-2}\) to find \(V = 1.62\text{ V}\) [1 Mark]. Calculates final energy \(E = 2.9 \times 10^{-4}\text{ J}\) [1 Mark].

(a) Horizontal displacement: \(x = (v_0 \cos\alpha) t\). Vertical displacement: \(y = (v_0 \sin\alpha) t - \frac{1}{2} g t^2\). (b) Substituting \(x\) and \(y\) into \(y = -x \tan\beta\): \((v_0 \sin\alpha) t - \frac{1}{2} g t^2 = -(v_0 \cos\alpha) t \tan\beta\). Since \(t > 0\) at landing, we can divide by \(t\): \(v_0 \sin\alpha - \frac{1}{2} g t = -v_0 \cos\alpha \tan\beta\). Rearranging to solve for \(t\): \(\frac{1}{2} g t = v_0 (\sin\alpha + \cos\alpha \tan\beta) \implies t = \frac{2 v_0 (\sin\alpha + \cos\alpha \tan\beta)}{g}\). (c) Calculate \(t\): \(t = \frac{2 \times 15.0 \times (\sin(20.0^\circ) + \cos(20.0^\circ)\tan(30.0^\circ))}{9.81} = \frac{30.0 \times 0.88455}{9.81} = 2.705\text{ s}\). Calculate horizontal distance \(x = 15.0 \times \cos(20.0^\circ) \times 2.705 = 38.13\text{ m}\). The distance \(d\) along the slope is related to \(x\) by \(x = d \cos\beta \implies d = \frac{38.13}{\cos(30.0^\circ)} = 44.03\text{ m}\).

(a) Writes correct horizontal kinematic equation [1 Mark]. Writes correct vertical kinematic equation with \(-g\) [1 Mark]. (b) Substitutes \(x\) and \(y\) into \(y = -x \tan\beta\) [1 Mark]. Correctly divides by \(t\) and rearranges [1 Mark]. Shows the target equation clearly [1 Mark]. (c) Calculates time of flight \(t = 2.71\text{ s}\) [1 Mark]. Calculates horizontal distance \(x = 38.1\text{ m}\) or vertical distance \(y = -22.0\text{ m}\) [1 Mark]. Uses trigonometry or Pythagoras' theorem to find slope distance \(d = 44.0\text{ m}\) (accept range \(43.8\text{ m}\) to \(44.2\text{ m}\)) [1 Mark].

\(\textbf{Part (a)}\)
The magnetic force provides the necessary centripetal force for circular motion:
\( F_B = Bqv \)
\( F_c = \frac{mv^2}{r} \)
Equating these forces:
\( Bqv = \frac{mv^2}{r} \)
Rearranging for \( r \):
\( r = \frac{mv}{Bq} \) (as required).

\(\textbf{Part (b)}\)
An electric field exists only in the gap between the dees. As the ion crosses the gap, the electric field does work on it, increasing its kinetic energy and speed. Inside the dees, there is no electric field (as they act as Faraday cages/shields), so the magnetic field causes the ion to follow a semi-circular path at constant speed. To ensure the ion is accelerated every time it crosses the gap, the electric field must reverse direction in the time it takes the ion to complete a semi-circle.
The time \( t \) to complete one semi-circle is:
\( t = \frac{\pi r}{v} \)
Substituting \( r = \frac{mv}{Bq} \):
\( t = \frac{\pi m}{Bq} \)
The period \( T \) of the AC supply must equal the time for a full orbit:
\( T = 2t = \frac{2\pi m}{Bq} \)
Since frequency \( f = 1/T \):
\( f = \frac{Bq}{2\pi m} \) (as required).

\(\textbf{Part (c)(i)}\)
\( f = \frac{1.50 \times 1.60 \times 10^{-19}}{2 \pi \times 3.34 \times 10^{-27}} \)
\( f = 1.1436 \times 10^7\text{ Hz} \approx 11.4\text{ MHz} \) (as required).

\(\textbf{Part (c)(ii)}\)
In one complete orbit, the ion crosses the gap twice.
For 50 orbits, there are \( 50 \times 2 = 100 \) crossings.
Energy gained per crossing \( = qV_0 = 1.60 \times 10^{-19}\text{ C} \times 80.0 \times 10^3\text{ V} = 1.28 \times 10^{-14}\text{ J} \).
Total energy gained \( E_k = 100 \times 1.28 \times 10^{-14}\text{ J} = 1.28 \times 10^{-12}\text{ J} \approx 1.3 \times 10^{-12}\text{ J} \) (as required).

\(\textbf{Part (c)(iii)}\)
Using the unrounded kinetic energy \( E_k = 1.28 \times 10^{-12}\text{ J} \):
\( E_k = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2E_k}{m}} \)
\( v = \sqrt{\frac{2 \times 1.28 \times 10^{-12}}{3.34 \times 10^{-27}}} = 2.77 \times 10^7\text{ m s}^{-1} \)
Radius \( r = \frac{mv}{Bq} = \frac{3.34 \times 10^{-27} \times 2.77 \times 10^7}{1.50 \times 1.60 \times 10^{-19}} = 0.385\text{ m} \)
(If using \( 1.3 \times 10^{-12}\text{ J} \), \( v = 2.79 \times 10^7\text{ m s}^{-1} \implies r = 0.388\text{ m} \approx 0.39\text{ m} \)).

\(\textbf{Part (a) [3 Marks]}\)
• [1M] Equates magnetic force to centripetal force: \( Bqv = \frac{mv^2}{r} \).
• [1M] Correctly identifies that \( F_B = Bqv \) and \( F_c = \frac{mv^2}{r} \).
• [1M] Clear algebraic manipulation leading to \( r = \frac{mv}{Bq} \).

\(\textbf{Part (b) [4 Marks]}\)
• [1M] Explains that the electric field in the gap does work / increases kinetic energy and speed.
• [1M] Explains that there is no electric field inside the dees, where magnetic field causes circular motion at constant speed.
• [1M] Relates the time for half an orbit \( t = \frac{\pi r}{v} \) to the period \( T = 2t \).
• [1M] Substitutes \( r \) to show \( T = \frac{2\pi m}{Bq} \) and hence \( f = \frac{Bq}{2\pi m} \).

\(\textbf{Part (c)(i) [2 Marks]}\)
• [1M] Correct substitution of values into frequency formula: \( f = \frac{1.50 \times 1.60 \times 10^{-19}}{2 \pi \times 3.34 \times 10^{-27}} \).
• [1M] Evaluates to \( 1.14 \times 10^7\text{ Hz} \) or \( 11.4\text{ MHz} \) (must show at least 3 s.f. before rounding to 11.4 MHz).

\(\textbf{Part (c)(ii) [2 Marks]}\)
• [1M] Multiplies orbits by 2 to find 100 crossings, and states energy per crossing is \( qV_0 \).
• [1M] Shows calculation resulting in \( 1.28 \times 10^{-12}\text{ J} \) (which rounds to \( 1.3 \times 10^{-12}\text{ J} \)).

\(\textbf{Part (c)(iii) [3 Marks]}\)
• [1M] Relates \( E_k \) to velocity, calculating \( v \approx 2.77 \times 10^7\text{ m s}^{-1} \) (or \( 2.79 \times 10^7\text{ m s}^{-1} \)).
• [1M] Correct substitution of \( v \), \( m \), \( B \), and \( q \) into \( r = \frac{mv}{Bq} \).
• [1M] Correct final value: \( 0.385\text{ m} \) to \( 0.39\text{ m} \) with appropriate units.

\(\textbf{Part (a)}\)
For circular orbit, the gravitational force provides the centripetal force:
\( \frac{G M_E m}{r^2} = \frac{m v^2}{r} \implies v^2 = \frac{G M_E}{r} \)
The speed \( v \) is also related to the orbital period \( T \) by:
\( v = \frac{2 \pi r}{T} \)
Squaring both sides:
\( v^2 = \frac{4 \pi^2 r^2}{T^2} \)
Equating the expressions for \( v^2 \):
\( \frac{G M_E}{r} = \frac{4 \pi^2 r^2}{T^2} \)
Rearranging gives:
\( T^2 = \frac{4\pi^2 r^3}{G M_E} \) (as required).

\(\textbf{Part (b)(i)}\)
Gravitational potential energy:
\( E_p = -\frac{G M_E m}{r_1} \)
\( E_p = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1200}{6.80 \times 10^6} = -7.027 \times 10^{10}\text{ J} \approx -7.0 \times 10^{10}\text{ J} \) (as required).

\(\textbf{Part (b)(ii)}\)
Total energy \( E_{\text{total}} = E_k + E_p \)
Since \( v^2 = \frac{G M_E}{r} \), the kinetic energy is:
\( E_k = \frac{1}{2} m v^2 = \frac{G M_E m}{2r} \)
Thus:
\( E_{\text{total}} = \frac{G M_E m}{2r} - \frac{G M_E m}{r} = -\frac{G M_E m}{2r} \) (as required).
For the parking orbit:
\( E_{\text{total}} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1200}{2 \times 6.80 \times 10^6} = -3.514 \times 10^{10}\text{ J} \).

\(\textbf{Part (b)(iii)}\)
Total energy in the geostationary orbit \( r_2 \):
\( E_{\text{total}, 2} = -\frac{G M_E m}{2r_2} \)
\( E_{\text{total}, 2} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1200}{2 \times 4.22 \times 10^7} = -5.659 \times 10^9\text{ J} \)
The additional energy required is:
\( \Delta E = E_{\text{total}, 2} - E_{\text{total}, 1} = -5.659 \times 10^9\text{ J} - (-3.514 \times 10^{10}\text{ J}) = 2.948 \times 10^{10}\text{ J} \approx 2.9 \times 10^{10}\text{ J} \) (as required).

\(\textbf{Part (b)(iv)}\)
To move the probe from a lower orbit \( r_1 \) to a higher orbit \( r_2 \), it must move further away from the Earth's centre.
This requires work to be done against the attractive gravitational force of the Earth, which increases the gravitational potential energy \( E_p \).
Although the kinetic energy \( E_k \) decreases in a higher orbit, the gain in \( E_p \) is twice the loss in \( E_k \), resulting in an overall increase in total mechanical energy. This energy must be supplied by the engines (positive work done on the probe).

\(\textbf{Part (a) [3 Marks]}\)
• [1M] Sets gravitational force equal to centripetal force: \( \frac{G M_E m}{r^2} = \frac{m v^2}{r} \).
• [1M] Recalls \( v = \frac{2\pi r}{T} \) (or substitutes \( \omega = \frac{2\pi}{T} \) into \( a = \omega^2 r \)).
• [1M] Correct algebraic steps to obtain \( T^2 = \frac{4\pi^2 r^3}{G M_E} \).

\(\textbf{Part (b)(i) [2 Marks]}\)
• [1M] Correct substitution into the formula \( E_p = -\frac{G M_E m}{r} \).
• [1M] Correctly calculates \( -7.03 \times 10^{10}\text{ J} \) and rounds to show \( -7.0 \times 10^{10}\text{ J} \).

\(\textbf{Part (b)(ii) [3 Marks]}\)
• [1M] Uses circular motion condition to show \( E_k = \frac{G M_E m}{2r} \).
• [1M] Adds \( E_k \) and \( E_p \) algebraically to obtain \( E_{\text{total}} = -\frac{G M_E m}{2r} \).
• [1M] Calculates \( E_{\text{total}, 1} = -3.51 \times 10^{10}\text{ J} \) (allow \( -3.5 \times 10^{10}\text{ J} \)).

\(\textbf{Part (b)(iii) [3 Marks]}\)
• [1M] Calculates total energy at geostationary orbit \( E_{\text{total}, 2} \approx -5.66 \times 10^9\text{ J} \).
• [1M] Computes difference \( \Delta E = E_{\text{total}, 2} - E_{\text{total}, 1} \).
• [1M] Obtains \( 2.95 \times 10^{10}\text{ J} \) and states this is approximately \( 2.9 \times 10^{10}\text{ J} \).

\(\textbf{Part (b)(iv) [3 Marks]}\)
• [1M] Identifies that moving to a larger radius requires work done against the Earth's gravitational pull.
• [1M] Explains that the gravitational potential energy increases (becomes less negative).
• [1M] Explains that although kinetic energy decreases, the increase in potential energy is larger in magnitude, so total mechanical energy must increase.